{"id":1199,"date":"2017-04-10T15:10:12","date_gmt":"2017-04-10T19:10:12","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/?post_type=chapter&#038;p=1199"},"modified":"2017-04-10T15:10:13","modified_gmt":"2017-04-10T19:10:13","slug":"23-12-rlc-series-ac-circuits","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/chapter\/23-12-rlc-series-ac-circuits\/","title":{"raw":"23.12 RLC Series AC Circuits","rendered":"23.12 RLC Series AC Circuits"},"content":{"raw":"<div>RLC Series AC Circuits<\/div>\n<div>\n<ul><li>Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and\/or current in a RLC series circuit.<\/li>\n \t<li>Draw the circuit diagram for an RLC series circuit.<\/li>\n \t<li>Explain the significance of the resonant frequency.<\/li>\n<\/ul><\/div>\n<section id=\"fs-id1169736628865\"><h1>Impedance<\/h1>\n<p id=\"fs-id1169737742205\">When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other\u2019s effect. <a href=\"#import-auto-id1169736621511\" class=\"autogenerated-content\">[link]<\/a> shows an <em>RLC <\/em>series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an <em>RLC<\/em> circuit is the frequency dependence of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {}, and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important \u201cresonance\u201d features that are the basis of many applications, such as radio tuners.<\/p>\n\n<figure id=\"import-auto-id1169736621511\"><figcaption>An <em>RLC<\/em> series circuit with an AC voltage source.<\/figcaption><span id=\"import-auto-id1169737820340\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_01a.jpg\" alt=\"The figure describes an R LC series circuit. It shows a resistor R connected in series with an inductor L, connected to a capacitor C in series to an A C source V. The voltage of the A C source is given by V equals V zero sine two pi f t. The voltage across R is V R, across L is V L and across C is V C.\" width=\"275\"\/><\/span>\n\n<\/figure><p id=\"import-auto-id1169738181435\">The combined effect of resistance RR size 12{R} {}, inductive reactance XLXL size 12{X rSub { size 8{L} } } {}, and capacitive reactance XCXC size 12{X rSub { size 8{C} } } {} is defined to be <span>impedance<\/span>, an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an <em>RLC<\/em> circuit are related by an AC version of Ohm\u2019s law:<\/p>\n\n<div class=\"equation\">I0=V0Z or Irms=VrmsZ.I0=V0Z or Irms=VrmsZ. size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } \" or \"I rSub { size 8{ ital \"rms\"} } = { {V rSub { size 8{ ital \"rms\"} } } over {Z} } \".\" } {}<\/div>\n<p id=\"import-auto-id1169737764601\">Here I0I0 size 12{I rSub { size 8{0} } } {} is the peak current, V0V0 size 12{V rSub { size 8{0} } } {} the peak source voltage, and ZZ is the impedance of the circuit. The units of impedance are ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for ZZ size 12{Z} {} in terms of RR, XLXL size 12{X rSub { size 8{L} } } {}, and XCXC size 12{X rSub { size 8{C} } } {}, we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled VRVR size 12{V rSub { size 8{R} } } {}, VLVL size 12{V rSub { size 8{L} } } {}, and VCVC size 12{V rSub { size 8{C} } } {} in <a href=\"#import-auto-id1169736621511\" class=\"autogenerated-content\">[link]<\/a>.<\/p>\nConservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {} are equal and in phase. But we know from the preceding section that the voltage across the inductor VLVL size 12{V rSub { size 8{L} } } {} leads the current by one-fourth of a cycle, the voltage across the capacitor VCVC size 12{V rSub { size 8{C} } } {} follows the current by one-fourth of a cycle, and the voltage across the resistor VRVR size 12{V rSub { size 8{R} } } {} is exactly in phase with the current. <a href=\"#import-auto-id1169738164070\" class=\"autogenerated-content\">[link]<\/a> shows these relationships in one graph, as well as showing the total voltage around the circuit V=VR+VL+VCV=VR+VL+VC size 12{V=V rSub { size 8{R} } +V rSub { size 8{L} } +V rSub { size 8{C} } } {}, where all four voltages are the instantaneous values. According to Kirchhoff\u2019s loop rule, the total voltage around the circuit VV is also the voltage of the source.\n<p id=\"import-auto-id1169738092269\">You can see from <a href=\"#import-auto-id1169738164070\" class=\"autogenerated-content\">[link]<\/a> that while VRVR size 12{V rSub { size 8{R} } } {} is in phase with the current, VLVL size 12{V rSub { size 8{L} } } {} leads by 90\u00ba90\u00ba, and VCVC size 12{V rSub { size 8{C} } } {} follows by 90\u00ba90\u00ba. Thus VLVL size 12{V rSub { size 8{L} } } {} and VCVC size 12{V rSub { size 8{C} } } {} are 180\u00ba180\u00ba out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage V0V0 size 12{V rSub { size 8{0} } } {} of the source does <em>not<\/em> equal the sum of the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}. The actual relationship is<\/p>\n\n<div class=\"equation\">V0=V0R2+(V0L\u2212V0C)2,V0=V0R2+(V0L\u2212V0C)2, size 12{V rSub { size 8{0} } = sqrt {V rSub { size 8{0R} } \"\" lSup { size 8{2} } + ( V rSub { size 8{0L} } - V rSub { size 8{0C} } ) rSup { size 8{2} } } ,} {}<\/div>\n<p id=\"import-auto-id1169738246882\">where V0RV0R size 12{V rSub { size 8{0R} } } {}, V0LV0L size 12{V rSub { size 8{0L} } } {}, and V0CV0C size 12{V rSub { size 8{0C} } } {} are the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}, respectively. Now, using Ohm\u2019s law and definitions from <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4\">Reactance, Inductive and Capacitive<\/a>, we substitute V0=I0ZV0=I0Z size 12{V rSub { size 8{0} } =I rSub { size 8{0} } Z} {} into the above, as well as V0R=I0RV0R=I0R size 12{V rSub { size 8{0R} } =I rSub { size 8{0} } R} {}, V0L=I0XLV0L=I0XL size 12{V rSub { size 8{0L} } =I rSub { size 8{0} } X rSub { size 8{L} } } {}, and V0C=I0XCV0C=I0XC size 12{V rSub { size 8{0C} } =I rSub { size 8{0} } X rSub { size 8{C} } } {}, yielding<\/p>\n\n<div class=\"equation\">I0Z=I02R2+(I0XL\u2212I0XC)2=I0R2+(XL\u2212XC)2.I0Z=I02R2+(I0XL\u2212I0XC)2=I0R2+(XL\u2212XC)2. size 12{I rSub { size 8{0} } Z= sqrt {I rSub { size 8{0} rSup { size 8{2} } } R rSup { size 8{2} } + ( I rSub { size 8{0} } X rSub { size 8{L} } - I rSub { size 8{0} } X rSub { size 8{C} } ) rSup { size 8{2} } } =I rSub { size 8{0} } sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {}<\/div>\n<p id=\"import-auto-id1169736584681\">I0I0 size 12{I rSub { size 8{0} } } {} cancels to yield an expression for ZZ:<\/p>\n\n<div class=\"equation\" id=\"eip-415\">Z=R2+(XL\u2212XC)2,Z=R2+(XL\u2212XC)2, size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {}<\/div>\n<p id=\"import-auto-id1169736710957\">which is the impedance of an <em>RLC<\/em> series AC circuit. For circuits without a resistor, take R=0R=0; for those without an inductor, take XL=0XL=0 size 12{X rSub { size 8{L} } =0} {}; and for those without a capacitor, take XC=0XC=0 size 12{X rSub { size 8{C} } =0} {}.<\/p>\n\n<figure id=\"import-auto-id1169738164070\"><figcaption>This graph shows the relationships of the voltages in an <em>RLC<\/em> circuit to the current. The voltages across the circuit elements add to equal the voltage of the source, which is seen to be out of phase with the current.<\/figcaption><span id=\"import-auto-id1169737733064\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_02a.jpg\" alt=\"The figure shows graphs showing the relationships of the voltages in an RLC circuit to the current. It has five graphs on the left and two graphs on the right. The first graph on the right is for current I versus time t. Current is plotted along Y axis and time is along X axis. The curve is a smooth progressive sine wave. The second graph is on the right is for voltage V R versus time t. Voltage V R is plotted along Y axis and time is along X axis. The curve is a smooth progressive sine wave. The third graph is on the right is for voltage V L versus time t. Voltage V L is plotted along Y axis and time is along X axis. The curve is a smooth progressive cosine wave. The fourth graph is on the right is for voltage V C versus time t. Voltage V C is plotted along Y axis and time t is along X axis. The curve is a smooth progressive cosine wave starting from negative Y axis. The fifth graph shows the voltage V verses time t for the R L C circuit. Voltage V is plotted along Y axis and time t is along X axis. The curve is a smooth progressive sine wave starting from a point near to origin on negative X axis. The first and the fifth graphs are again shown on the right and their amplitudes and phases compared. The current graph is shown to have a lesser amplitude.\" width=\"300\"\/><\/span>\n\n<\/figure><div class=\"example\" id=\"fs-id1169737723572\">\n<div class=\"title\">Calculating Impedance and Current<\/div>\n<p id=\"import-auto-id1169737041418\">An <em>RLC <\/em>series circuit has a 40.0 \u03a940.0 \u03a9 resistor, a 3.00 mH inductor, and a\n5.00 \u03bcF5.00 \u03bcF capacitor. (a) Find the circuit\u2019s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for\nLL and\nCC\nare the same as in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736972664\" class=\"autogenerated-content\">[link]<\/a> and <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736597928\" class=\"autogenerated-content\">[link]<\/a>. (b) If the voltage source has Vrms=120VVrms=120V size 12{V rSub { size 8{\"rms\"} } =\"120\"`V} {}, what is IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at each frequency?<\/p>\n<p id=\"import-auto-id1169735477467\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169738239044\">For each frequency, we use Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {} to find the impedance and then Ohm\u2019s law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again.<\/p>\n<p id=\"import-auto-id1169738147909\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738028359\">At 60.0 Hz, the values of the reactances were found in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736972664\" class=\"autogenerated-content\">[link]<\/a> to be XL=1.13\u03a9XL=1.13\u03a9 size 12{X rSub { size 8{L} } =1 \".\" \"13\" %OMEGA } {} and in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736597928\" class=\"autogenerated-content\">[link]<\/a> to be XC=531 \u03a9XC=531 \u03a9 size 12{X rSub { size 8{C} } =\"531 \" %OMEGA } {}. Entering these and the given 40.0 \u03a940.0 \u03a9 for resistance into Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {} yields<\/p>\n\n<div class=\"equation\" id=\"eip-331\">Z=R2+(XL\u2212XC)2=(40.0\u03a9)2+(1.13\u03a9\u2212531\u03a9)2=531\u03a9\u00a0at\u00a060.0 Hz.Z=R2+(XL\u2212XC)2=(40.0\u03a9)2+(1.13\u03a9\u2212531\u03a9)2=531\u03a9\u00a0at\u00a060.0 Hz.alignl { stack {\nsize 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {} #\n\" \"= sqrt { ( \"40\" \".\" 0` %OMEGA ) rSup { size 8{2} } + ( 1 \".\" \"13\" %OMEGA - \"531\" %OMEGA ) rSup { size 8{2} } } {} #\n\" \"=\"531\" %OMEGA \" at 60\" \".\" \"0 Hz\" {}\n} } {}<\/div>\n<p id=\"import-auto-id1169738036576\">Similarly, at 10.0 kHz, XL=188\u03a9XL=188\u03a9 size 12{X rSub { size 8{L} } =\"188\" %OMEGA } {} and XC=3.18\u03a9XC=3.18\u03a9 size 12{X rSub { size 8{C} } =3 \".\" \"18\" %OMEGA } {}, so that<\/p>\n\n<div class=\"equation\">Z=(40.0\u03a9)2+(188\u03a9\u22123.18\u03a9)2=190\u03a9\u00a0at\u00a010.0 kHz.Z=(40.0\u03a9)2+(188\u03a9\u22123.18\u03a9)2=190\u03a9\u00a0at\u00a010.0 kHz.alignl { stack {\nsize 12{Z= sqrt { ( \"40\" \".\" 0` %OMEGA ) rSup { size 8{2} } + ( \"188\" %OMEGA - 3 \".\" \"18\" %OMEGA ) rSup { size 8{2} } } } {} #\n\" \"=\"190\" %OMEGA \" at 10\" \".\" \"0 kHz\" {}\n} } {}<\/div>\n<p id=\"import-auto-id1169737738196\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169736768845\">In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that XLXL size 12{X rSub { size 8{L} } } {} dominates at high frequency and XCXC size 12{X rSub { size 8{C} } } {} dominates at low frequency.<\/p>\n<p id=\"import-auto-id1169737904584\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737710224\">The current IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} can be found using the AC version of Ohm\u2019s law in Equation Irms=Vrms\/ZIrms=Vrms\/Z size 12{I rSub { size 8{\"rms\"} } =V rSub { size 8{\"rms\"} } \/Z} {}:<\/p>\nIrms=VrmsZ=120 V531 \u03a9=0.226 AIrms=VrmsZ=120 V531 \u03a9=0.226 A size 12{I rSub { size 8{\"rms\"} } = { {V rSub { size 8{\"rms\"} } } over {Z} } = { {\"120\"\" V\"} over {\"531 \" %OMEGA } } =0 \".\" \"226\"\" A\"} {} at 60.0 Hz\n<p id=\"import-auto-id1169737758253\">Finally, at 10.0 kHz, we find<\/p>\n<p id=\"import-auto-id1169738110287\">Irms=VrmsZ=120 V190 \u03a9=0.633 AIrms=VrmsZ=120 V190 \u03a9=0.633 A size 12{I rSub { size 8{\"rms\"} } = { {V rSub { size 8{\"rms\"} } } over {Z} } = { {\"120\"\" V\"} over {\"190 \" %OMEGA } } =0 \".\" \"633\"\" A\"} {} at 10.0 kHz<\/p>\n<p id=\"import-auto-id1169737861583\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738198520\">The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736597928\" class=\"autogenerated-content\">[link]<\/a>. The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736972664\" class=\"autogenerated-content\">[link]<\/a>. The inductor dominates at high frequency.<\/p>\n\n<\/div>\n<\/section><section id=\"fs-id1169738011926\"><h1>Resonance in <em>RLC<\/em> Series AC Circuits<\/h1>\n<p id=\"import-auto-id1169736708835\">How does an <em>RLC<\/em> circuit behave as a function of the frequency of the driving voltage source? Combining Ohm\u2019s law, Irms=Vrms\/ZIrms=Vrms\/Z size 12{I rSub { size 8{\"rms\"} } =V rSub { size 8{\"rms\"} } \/Z} {}, and the expression for impedance ZZ from Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {} gives<\/p>\n\n<div class=\"equation\">Irms=VrmsR2+(XL\u2212XC)2.Irms=VrmsR2+(XL\u2212XC)2. size 12{I rSub { size 8{\"rms\"} } = { {V rSub { size 8{\"rms\"} } } over { sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } } } {}<\/div>\n<p id=\"import-auto-id1169737965627\">The reactances vary with frequency, with XLXL size 12{X rSub { size 8{L} } } {} large at high frequencies and XCXC size 12{X rSub { size 8{C} } } {} large at low frequencies, as we have seen in three previous examples. At some intermediate frequency f0f0 size 12{f rSub { size 8{0} } } {}, the reactances will be equal and cancel, giving <em>Z=RZ=R size 12{Z=R} {}<\/em> \u2014this is a minimum value for impedance, and a maximum value for IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} results. We can get an expression for f0f0 size 12{f rSub { size 8{0} } } {} by taking<\/p>\n\n<div class=\"equation\">XL=XC.XL=XC. size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}<\/div>\n<p id=\"import-auto-id1169738064647\">Substituting the definitions of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {},<\/p>\n\n<div class=\"equation\">2\u03c0f0L=12\u03c0f0C.2\u03c0f0L=12\u03c0f0C. size 12{2\u03c0f rSub { size 8{0} } L= { {1} over {2\u03c0f rSub { size 8{0} } C} } } {}<\/div>\n<p id=\"import-auto-id1169737796056\">Solving this expression for f0f0 size 12{f rSub { size 8{0} } } {} yields<\/p>\n\n<div class=\"equation\">f0=12\u03c0LC,f0=12\u03c0LC, size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital \"LC\"} } } } {}<\/div>\n<p id=\"import-auto-id1169737949912\">where f0f0 size 12{f rSub { size 8{0} } } {} is the <span id=\"import-auto-id1169737994451\">resonant frequency<\/span> of an <em>RLC<\/em> series circuit. This is also the <em>natural frequency<\/em> at which the circuit would oscillate if not driven by the voltage source. At f0f0 size 12{f rSub { size 8{0} } } {}, the effects of the inductor and capacitor cancel, so that <em>Z=RZ=R size 12{Z=R} {}<\/em>, and IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} is a maximum.<\/p>\n<p id=\"import-auto-id1169738136783\">Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation\u2014in this case, forced by the voltage source\u2014at the natural frequency of the system. The receiver in a radio is an <em>RLC<\/em> circuit that oscillates best at its f0f0 size 12{f rSub { size 8{0} } } {}. A variable capacitor is often used to adjust f0f0 size 12{f rSub { size 8{0} } } {} to receive a desired frequency and to reject others. <a href=\"#import-auto-id1169738205664\" class=\"autogenerated-content\">[link]<\/a> is a graph of current as a function of frequency, illustrating a resonant peak in IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at f0f0 size 12{f rSub { size 8{0} } } {}. The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radio receiver, for example.<\/p>\n\n<figure id=\"import-auto-id1169738205664\"><figcaption>A graph of current versus frequency for two <em>RLC<\/em> series circuits differing only in the amount of resistance. Both have a resonance at f0f0 size 12{f rSub { size 8{0} } } {}, but that for the higher resistance is lower and broader. The driving AC voltage source has a fixed amplitude V0V0 size 12{V rSub { size 8{0} } } {}.<\/figcaption><span id=\"import-auto-id1169737919303\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_03a.jpg\" alt=\"The figure describes a graph of current I versus frequency f. Current I r m s is plotted along Y axis and frequency f is plotted along X axis. Two curves are shown. The upper curve is for small resistance and lower curve is for large resistance. Both the curves have a smooth rise and a fall. The peaks are marked for frequency f zero. The curve for smaller resistance has a higher value of peak than the curve for large resistance.\" width=\"225\"\/><\/span>\n\n<\/figure><div class=\"example\" id=\"fs-id1169738045330\">\n<div class=\"title\">Calculating Resonant Frequency and Current<\/div>\n<p id=\"import-auto-id1169738007674\">For the same <em>RLC<\/em> series circuit having a 40.0 \u03a940.0 \u03a9 resistor, a 3.00 mH inductor, and a 5.00 \u03bcF5.00 \u03bcF capacitor: (a) Find the resonant frequency. (b) Calculate IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at resonance if VrmsVrms size 12{V rSub { size 8{\"rms\"} } } {} is 120 V.<\/p>\n<p id=\"import-auto-id1169738107664\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169738239161\">The resonant frequency is found by using the expression in f0=12\u03c0LCf0=12\u03c0LC size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital \"LC\"} } } } {}. The current at that frequency is the same as if the resistor alone were in the circuit.<\/p>\n<p id=\"import-auto-id1169737853725\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738043142\">Entering the given values for LL and CC into the expression given for f0f0 size 12{f rSub { size 8{0} } } {} in f0=12\u03c0LCf0=12\u03c0LC size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital \"LC\"} } } } {} yields<\/p>\n\n<div class=\"equation\">f0=12\u03c0LC=12\u03c0(3.00\u00d710\u22123 H)(5.00\u00d710\u22126 F)=1.30 kHz.f0=12\u03c0LC=12\u03c0(3.00\u00d710\u22123 H)(5.00\u00d710\u22126 F)=1.30 kHz.alignl { stack {\nsize 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital \"LC\"} } } } {} #\n\" \"= { {1} over {2\u03c0 sqrt { ( 3 \".\" \"00\" times \"10\" rSup { size 8{ - 3} } \" H\" ) ( 5 \".\" \"00\" times \"10\" rSup { size 8{ - 6} } \" F\" ) } } } =1 \".\" \"30\"\" kHz\" {}\n} } {}<\/div>\n<p id=\"import-auto-id1169738008646\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738087141\">We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.<\/p>\n<p id=\"import-auto-id1169736708769\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737846339\">The current is given by Ohm\u2019s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus,<\/p>\n\n<div class=\"equation\">Irms=VrmsZ=120 V40.0 \u03a9=3.00 A.Irms=VrmsZ=120 V40.0 \u03a9=3.00 A. size 12{I rSub { size 8{\"rms\"} } = { {V rSub { size 8{\"rms\"} } } over {Z} } = { {\"120\"\" V\"} over {\"40\" \".\" \"0 \" %OMEGA } } =3 \".\" \"00\"\" A\"} {}<\/div>\n<p id=\"import-auto-id1169737909774\"><strong>Discussion for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169736843029\">At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.<\/p>\n\n<\/div>\n<\/section><section id=\"fs-id1169737725407\"><h1>Power in <em>RLC<\/em> Series AC Circuits<\/h1>\n<p id=\"import-auto-id1169737969148\">If current varies with frequency in an <em>RLC<\/em> circuit, then the power delivered to it also varies with frequency. But the average power is not simply current times voltage, as it is in purely resistive circuits. As was seen in <a href=\"#import-auto-id1169738164070\" class=\"autogenerated-content\">[link]<\/a>, voltage and current are out of phase in an <em>RLC<\/em> circuit. There is a <span id=\"import-auto-id1169738085250\">phase angle<\/span> \u03d5\u03d5 size 12{\u03d5} {} between the source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from<\/p>\n\n<div class=\"equation\" id=\"eip-543\">cos\u03d5=RZ.cos\u03d5=RZ. size 12{\"cos\"\u03d5= { {R} over {Z} } } {}<\/div>\n<p id=\"import-auto-id1169738181411\">For example, at the resonant frequency or in a purely resistive circuit <em>Z=RZ=R size 12{Z=R} {}<\/em>, so that cos\u03d5=1cos\u03d5=1 size 12{\"cos\"\u03d5=1} {}. This implies that \u03d5=0\u00ba\u03d5=0\u00ba size 12{\u03d5=0 rSup { size 8{ circ } } } {} and that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage and current are out of phase and because IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} is lower. The fact that source voltage and current are out of phase affects the power delivered to the circuit. It can be shown that the <em><em>average power<\/em><\/em> is<\/p>\n\n<div class=\"equation\">Pave=IrmsVrmscos\u03d5,Pave=IrmsVrmscos\u03d5, size 12{P rSub { size 8{\"ave\"} } =I rSub { size 8{\"rms\"} } V rSub { size 8{\"rms\"} } \"cos\"\u03d5} {}<\/div>\n<p id=\"import-auto-id1169738060586\">Thus cos\u03d5cos\u03d5 size 12{\"cos\"\u03d5} {} is called the <span id=\"import-auto-id1169736719487\">power factor<\/span>, which can range from 0 to 1. Power factors near 1 are desirable when designing an efficient motor, for example. At the resonant frequency, cos\u03d5=1cos\u03d5=1 size 12{\"cos\"\u03d5=1} {}.<\/p>\n\n<div class=\"example\" id=\"fs-id1169737987472\">\n<div class=\"title\">Calculating the Power Factor and Power<\/div>\n<p id=\"import-auto-id1169737973055\">For the same <em>RLC<\/em> series circuit having a 40.0 \u03a940.0 \u03a9 resistor, a 3.00 mH inductor, a\n5.00 \u03bcF5.00 \u03bcF capacitor, and a voltage source with a VrmsVrms of 120 V: (a) Calculate the power factor and phase angle for f=60.0Hzf=60.0Hz size 12{f=\"60\" \".\" 0`\"Hz\"} {}. (b) What is the average power at 50.0 Hz? (c) Find the average power at the circuit\u2019s resonant frequency.<\/p>\n<p id=\"import-auto-id1169738250320\"><strong>Strategy and Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738181417\">The power factor at 60.0 Hz is found from<\/p>\n\n<div class=\"equation\" id=\"eip-248\">cos\u03d5=RZ.cos\u03d5=RZ. size 12{\"cos\"\u03d5= { {R} over {Z} } } {}<\/div>\n<p id=\"import-auto-id1169738075830\">We know Z= 531 \u03a9Z= 531 \u03a9 from <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a>, so that<\/p>\n\n<div class=\"equation\">cos\u03d5=40.0\u03a9531 \u03a9=0.0753 at 60.0 Hz.cos\u03d5=40.0\u03a9531 \u03a9=0.0753 at 60.0 Hz. size 12{\"cos\"\u00d8= { {\"40\" \".\" 0 %OMEGA } over {5\"31 \" %OMEGA } } =0 \".\" \"0753\"} {}<\/div>\n<p id=\"import-auto-id1169736844129\">This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is<\/p>\n\n<div class=\"equation\">\u03d5=cos\u221210.0753=85.7\u00ba at 60.0 Hz.\u03d5=cos\u221210.0753=85.7\u00ba at 60.0 Hz. size 12{\u03d5=\"cos\" rSup { size 8{ - 1} } 0 \".\" \"0753\"=\"85\" \".\" 7 rSup { size 8{ circ } } } {}<\/div>\n<p id=\"import-auto-id1169738148048\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169737789995\">The phase angle is close to\n90\u00ba90\u00ba, consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure <em>RC<\/em> circuit has its voltage and current 90\u00ba90\u00ba out of phase).<\/p>\n<p id=\"import-auto-id1169737851510\"><strong>Strategy and Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737851513\">The average power at 60.0 Hz is<\/p>\n\n<div class=\"equation\">Pave=IrmsVrmscos\u03d5.Pave=IrmsVrmscos\u03d5. size 12{P rSub { size 8{\"ave\"} } =I rSub { size 8{\"rms\"} } V rSub { size 8{\"rms\"} } \"cos\"\u03d5} {}<\/div>\n<p id=\"import-auto-id1169736627643\">IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} was found to be 0.226 A in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a>. Entering the known values gives<\/p>\n\n<div class=\"equation\">Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz.Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz. size 12{P rSub { size 8{\"ave\"} } = ( 0 \".\" \"226\"\" A\" ) ( \"120\"\" V\" ) ( 0 \".\" \"0753\" ) =2 \".\" \"04\"\" W\"} {}<\/div>\n<p id=\"import-auto-id1169737923237\"><strong>Strategy and Solution for (c)<\/strong><\/p>\n<p id=\"import-auto-id1169737923241\">At the resonant frequency, we know cos\u03d5=1cos\u03d5=1 size 12{\"cos\"\u03d5=1} {}, and IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} was found to be 6.00 A in <a href=\"#fs-id1169738045330\" class=\"autogenerated-content\">[link]<\/a>. Thus,<\/p>\n<p id=\"import-auto-id1169735477311\">Pave=(3.00 A)(120 V)(1)=360 WPave=(3.00 A)(120 V)(1)=360 W size 12{P rSub { size 8{\"ave\"} } = ( 3 \".\" \"00\"\" A\" ) ( \"120\"\" V\" ) ( 1 ) =\"350\"\" W\"} {} at resonance (1.30 kHz)<\/p>\n<p id=\"import-auto-id1169738080554\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id1169738080557\">Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.<\/p>\n\n<\/div>\n<p id=\"import-auto-id1169738164030\">Power delivered to an <em>RLC<\/em> series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy input and output but do not dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistor dissipating exactly what the voltage source puts into the circuit. This assumes no significant electromagnetic radiation from the inductor and capacitor, such as radio waves. Such radiation can happen and may even be desired, as we will see in the next chapter on electromagnetic radiation, but it can also be suppressed as is the case in this chapter. The circuit is analogous to the wheel of a car driven over a corrugated road as shown in <a href=\"#import-auto-id1169736885804\" class=\"autogenerated-content\">[link]<\/a>. The regularly spaced bumps in the road are analogous to the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance damping and limiting the amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current, and energy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in the electric field of a capacitor). The amplitude of the wheels\u2019 motion is a maximum if the bumps in the road are hit at the resonant frequency.<\/p>\n\n<figure id=\"import-auto-id1169736885804\"><figcaption>The forced but damped motion of the wheel on the car spring is analogous to an <em>RLC<\/em> series AC circuit. The shock absorber damps the motion and dissipates energy, analogous to the resistance in an <em>RLC<\/em> circuit. The mass and spring determine the resonant frequency.<\/figcaption><span id=\"import-auto-id1169736885805\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_04a.jpg\" alt=\"The figure describes the path of motion of a wheel of a car. The front wheel of a car is shown. A shock absorber attached to the wheel is also shown. The path of motion is shown as vertically up and down.\" width=\"225\"\/><\/span>\n\n<\/figure><p id=\"import-auto-id1169738211177\">A pure <em>LC<\/em> circuit with negligible resistance oscillates at f0f0 size 12{f rSub { size 8{0} } } {}, the same resonant frequency as an <em>RLC<\/em> circuit. It can serve as a frequency standard or clock circuit\u2014for example, in a digital wristwatch. With a very small resistance, only a very small energy input is necessary to maintain the oscillations. The circuit is analogous to a car with no shock absorbers. Once it starts oscillating, it continues at its natural frequency for some time. <a href=\"#import-auto-id1169738257733\" class=\"autogenerated-content\">[link]<\/a> shows the analogy between an <em>LC<\/em> circuit and a mass on a spring.<\/p>\n\n<figure id=\"import-auto-id1169738257733\"><figcaption>An <em>LC<\/em> circuit is analogous to a mass oscillating on a spring with no friction and no driving force. Energy moves back and forth between the inductor and capacitor, just as it moves from kinetic to potential in the mass-spring system.<\/figcaption><span id=\"import-auto-id1169737987631\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_05a.jpg\" alt=\"The figure describes four stages of an L C oscillation circuit compared to a mass oscillating on a spring. Part a of the figure shows a mass attached to a horizontal spring. The spring is attached to a fixed support on the left. The mass is at rest as shown by velocity v equals zero. The energy of the spring is shown as potential energy. This is compared with a circuit containing a capacitor C and inductor L connected together. The energy is shown as stored in the electric field E of the capacitor between the plates. One plate is shown to have a negative polarity and other plate is shown to have a positive polarity. Part b of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The mass is shown to move horizontal toward the fixed support with velocity v. The energy here is stored as the kinetic energy of the spring. This is compared with a circuit containing a capacitor C and inductor L connected together. A current is shown in the circuit and energy is stored as magnetic field B in the inductor. Part c of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The spring is shown as not stretched and the energy is shown as potential energy of the spring. The mass is show to have displaced toward left. This is compared with a circuit containing a capacitor C and inductor L connected together. The energy is shown as stored in the electric field E of the capacitor between the plates. One plate is shown to have a negative polarity and other plate is shown to have a positive polarity. But the polarities are reverse of the first case in part a. Part d of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The mass is shown to move toward right with velocity v. the energy of the spring is kinetic energy. This is compared with a circuit containing a capacitor C and inductor L connected together. A current is shown in the circuit opposite to that in part b and energy is stored as magnetic field B in the inductor.\" width=\"350\"\/><\/span>\n\n<\/figure><\/section><div class=\"note\">\n<div class=\"title\">PhET Explorations: Circuit Construction Kit (AC+DC), Virtual Lab<\/div>\nBuild circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instruments such as voltmeters and ammeters.\n\n<figure id=\"eip-id1732613\"><figcaption><a href=\"\/resources\/a810be47772627ab26601b90a2c46bf50d610b25\/circuit-construction-kit-ac-virtual-lab_en.jar\">Circuit Construction Kit (AC+DC), Virtual Lab<\/a><\/figcaption>\u00a0\n\n<a href=\"\/resources\/a810be47772627ab26601b90a2c46bf50d610b25\/circuit-construction-kit-ac-virtual-lab_en.jar\"><img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/PhET_Icon-56.png\" alt=\"\" width=\"450\"\/><\/a>\n\n\u00a0\n\n\u00a0\n\n\u00a0\n\n<\/figure><\/div>\n<section id=\"fs-id1169736711049\" class=\"section-summary\"><h1>Section Summary<\/h1>\n<ul id=\"fs-id1169738178781\"><li id=\"import-auto-id1169737979451\">The AC analogy to resistance is impedance ZZ, the combined effect of resistors, inductors, and capacitors, defined by the AC version of Ohm\u2019s law:\n<div class=\"equation\" id=\"eip-111\">I0=V0Z or Irms=VrmsZ,I0=V0Z or Irms=VrmsZ, size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } \" or \"I rSub { size 8{ ital \"rms\"} } = { {V rSub { size 8{ ital \"rms\"} } } over {Z} } ,} {}<\/div>\nwhere I0I0 size 12{I rSub { size 8{0} } } {} is the peak current and V0V0 size 12{V rSub { size 8{0} } } {} is the peak source voltage.<\/li>\n \t<li id=\"import-auto-id1169737742046\">Impedance has units of ohms and is given by Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {}.<\/li>\n \t<li id=\"import-auto-id1169737961915\">The resonant frequency f0f0 size 12{f rSub { size 8{0} } } {}, at which XL=XCXL=XC size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}, is\n<div class=\"equation\" id=\"eip-766\">f0=12\u03c0LC.f0=12\u03c0LC. size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital \"LC\"} } } } {}<\/div><\/li>\n \t<li id=\"import-auto-id1169736843344\">In an AC circuit, there is a phase angle <em>\u03d5\u03d5 size 12{\u03d5} {}<\/em> between source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from\n<div class=\"equation\">cos\u03d5=RZ,cos\u03d5=RZ, size 12{\"cos\"\u03d5= { {R} over {Z} } } {}<\/div><\/li>\n \t<li id=\"import-auto-id1169736615857\">\u03d5=0\u00ba\u03d5=0\u00ba size 12{\u03d5=0 rSup { size 8{ circ } } } {} for a purely resistive circuit or an <em>RLC<\/em> circuit at resonance.<\/li>\n \t<li id=\"import-auto-id1169736821483\">The average power delivered to an <em>RLC<\/em> circuit is affected by the phase angle and is given by\n<div class=\"equation\">Pave=IrmsVrmscos\u03d5,Pave=IrmsVrmscos\u03d5, size 12{P rSub { size 8{\"ave\"} } =I rSub { size 8{\"rms\"} } V rSub { size 8{\"rms\"} } \"cos\"\u03d5} {}<\/div>\ncos\u03d5cos\u03d5 size 12{\"cos\"\u03d5} {} is called the power factor, which ranges from 0 to 1.<\/li>\n<\/ul><\/section><section id=\"fs-id1169738150522\" class=\"conceptual-questions\"><h1>Conceptual Questions<\/h1>\n<div class=\"exercise\" id=\"fs-id1169736614731\">\n<div class=\"problem\" id=\"fs-id1169736614734\">\n<p id=\"import-auto-id1169737909824\">Does the resonant frequency of an AC circuit depend on the peak voltage of the AC source? Explain why or why not.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738073810\">\n<div class=\"problem\" id=\"fs-id1169738073813\">\n<p id=\"import-auto-id1169736702937\">Suppose you have a motor with a power factor significantly less than 1. Explain why it would be better to improve the power factor as a method of improving the motor\u2019s output, rather than to increase the voltage input.<\/p>\n\n<\/div>\n<\/div>\n<\/section><section class=\"problems-exercises\"><h1>Problems &amp; Exercises<\/h1>\n<div class=\"exercise\" id=\"fs-id1169736657142\">\n<div class=\"problem\" id=\"fs-id1169738198661\">\n<p id=\"import-auto-id1169737936131\">An <em>RL<\/em> circuit consists of a\n40.0 \u03a940.0 \u03a9 resistor and a\n3.00 mH inductor. (a) Find its impedance ZZ at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ with those found in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a> in which there was also a capacitor.<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736895358\">\n<p id=\"import-auto-id1169738134318\">(a) 40.02 \u03a940.02 \u03a9 at 60.0 Hz, 193 \u03a9193 \u03a9 at 10.0 kHz<\/p>\n<p id=\"eip-id3570199\">(b) At 60 Hz, with a capacitor, Z=531 \u03a9Z=531 \u03a9, over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z=190 \u03a9Z=190 \u03a9, about the same as without the capacitor. The capacitor has a smaller effect at high frequencies.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737930730\">\n<div class=\"problem\" id=\"fs-id1169737118482\">\n<p id=\"import-auto-id1169738086377\">An <em>RC<\/em> circuit consists of a\n40.0 \u03a940.0 \u03a9 resistor and a\n5.00 \u03bcF5.00 \u03bcF capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ with those found in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a>, in which there was also an inductor.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738012738\">\n<div class=\"problem\" id=\"fs-id1169737787125\">\n<p id=\"import-auto-id1169738143727\">An <em>LC<\/em> circuit consists of a 3.00mH3.00mH size 12{3 \".\" \"00\" \u03bcH} {} inductor and a 5.00\u03bcF5.00\u03bcF size 12{5 \".\" \"00\" \u03bcF} {} capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ size 12{Z} {} with those found in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a> in which there was also a resistor.<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736844109\">\n<p id=\"import-auto-id1169738246107\">(a) 529 \u03a9529 \u03a9 at 60.0 Hz, 185 \u03a9185 \u03a9 at 10.0 kHz<\/p>\n<p id=\"eip-id2226884\">(b) These values are close to those obtained in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a> because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737794135\">\n<div class=\"problem\" id=\"fs-id1169737794139\">\n<p id=\"import-auto-id1169737793235\">What is the resonant frequency of a 0.500 mH inductor connected to a\n40.0 \u03bcF40.0 \u03bcF capacitor?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738072090\">\n<div class=\"problem\" id=\"fs-id1169738071811\">\n<p id=\"import-auto-id1169737728540\">To receive AM radio, you want an <em>RLC<\/em> circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00 \u03bcH1.00 \u03bcH inductor connected to a variable capacitor. What range of capacitance is needed?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738181296\">\n<p id=\"import-auto-id1169737756327\">9.30 nF to 101 nF<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737830882\">\n<div class=\"problem\" id=\"fs-id1169738086090\">\n<p id=\"import-auto-id1169738055925\">Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and capacitors ranging from 1.00 pF to 0.100 F. What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738244135\">\n<div class=\"problem\" id=\"fs-id1169738087540\">\n<p id=\"import-auto-id1169738043322\">What capacitance do you need to produce a resonant frequency of 1.00 GHz, when using an 8.00 nH inductor?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737909755\">\n<p id=\"import-auto-id1169738043327\">3.17 pF<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737763566\">\n<div class=\"problem\" id=\"fs-id1169738007185\">\n<p id=\"import-auto-id1169737000533\">What inductance do you need to produce a resonant frequency of 60.0 Hz, when using a 2.00 \u03bcF2.00 \u03bcF capacitor?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738136983\">\n<div class=\"problem\" id=\"fs-id1169737787019\">\n<p id=\"import-auto-id1169737939475\">The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737821329\">\n<p id=\"import-auto-id1169737939482\">(a) 1.31 \u03bcH1.31 \u03bcH<\/p>\n<p id=\"import-auto-id1169738145328\">(b) 1.66 pF<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738187525\">\n<div class=\"problem\" id=\"fs-id1169737917711\">\n<p id=\"import-auto-id1169738038225\">An <em>RLC<\/em> series circuit has a\n2.50 \u03a92.50 \u03a9 resistor, a\n100 \u03bcH100 \u03bcH inductor, and an\n80.0 \u03bcF80.0 \u03bcF capacitor.(a) Find the circuit\u2019s impedance at 120 Hz. (b) Find the circuit\u2019s impedance at 5.00 kHz. (c) If the voltage source has Vrms=5.60VVrms=5.60V size 12{V rSub { size 8{\"rms\"} } =5 \".\" \"60\"`V} {}, what is IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at resonance?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738137511\">\n<div class=\"problem\" id=\"fs-id1169737924005\">\n<p id=\"import-auto-id1169737763460\">An <em>RLC<\/em> series circuit has a 1.00 k\u03a91.00 k\u03a9 resistor, a 150 \u03bcH150 \u03bcH inductor, and a 25.0 nF capacitor. (a) Find the circuit\u2019s impedance at 500 Hz. (b) Find the circuit\u2019s impedance at 7.50 kHz. (c) If the voltage source has Vrms=408VVrms=408V size 12{V rSub { size 8{\"rms\"} } =\"408\"`V} {}, what is IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{\"rms\"} } } {} at resonance?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737844573\">\n<p id=\"import-auto-id1169737827160\">(a) 12.8 k\u03a912.8 k\u03a9<\/p>\n<p id=\"import-auto-id1169737827162\">(b) 1.31 k\u03a91.31 k\u03a9<\/p>\n<p id=\"import-auto-id1169737827164\">(c) 31.9 mA at 500 Hz, 312 mA at 7.50 kHz<\/p>\n<p id=\"import-auto-id1169737827167\">(d) 82.2 kHz<\/p>\n<p id=\"import-auto-id1169737788984\">(e) 0.408 A<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738220311\">\n<div class=\"problem\" id=\"fs-id1169738220316\">\n<p id=\"import-auto-id1169737788992\">An <em>RLC<\/em> series circuit has a\n2.50 \u03a92.50 \u03a9 resistor, a\n100 \u03bcH100 \u03bcH inductor, and an\n80.0 \u03bcF80.0 \u03bcF capacitor. (a) Find the power factor at f=120 Hzf=120 Hz. (b) What is the phase angle at 120 Hz? (c) What is the average power at 120 Hz? (d) Find the average power at the circuit\u2019s resonant frequency.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737939490\">\n<div class=\"problem\" id=\"fs-id1169737939495\">\n<p id=\"import-auto-id1169737762694\">An <em>RLC<\/em> series circuit has a\n1.00 k\u03a91.00 k\u03a9 resistor, a\n150 \u03bcH150 \u03bcH inductor, and a 25.0 nF capacitor. (a) Find the power factor at f=7.50 Hzf=7.50 Hz. (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit\u2019s resonant frequency.<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736621311\">\n<p id=\"import-auto-id1169738052466\">(a) 0.159<\/p>\n<p id=\"import-auto-id1169738052468\">(b) 80.9\u00ba80.9\u00ba<\/p>\n<p id=\"import-auto-id1169738052470\">(c) 26.4 W<\/p>\n<p id=\"import-auto-id1169738052473\">(d) 166 W<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737047401\">\n<div class=\"problem\" id=\"fs-id1169737047405\">\n<p id=\"import-auto-id1169736617346\">An <em>RLC<\/em> series circuit has a\n200 \u03a9200 \u03a9\nresistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0\u00ba45.0\u00ba. (a) What is the impedance? (b) Find the circuit\u2019s capacitance. (c) If Vrms=408VVrms=408V size 12{V rSub { size 8{\"rms\"} } =\"408\"`V} {} is applied, what is the average power supplied?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738224784\">\n<div class=\"problem\" id=\"fs-id1169737050938\">\n<p id=\"import-auto-id1169738038100\">Referring to <a href=\"#fs-id1169737987472\" class=\"autogenerated-content\">[link]<\/a>, find the average power at 10.0 kHz.<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736932443\">\n<p id=\"import-auto-id1169738038104\">16.0 W<\/p>\n\n<\/div>\n<\/div>\n<\/section><div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1169737780982\" class=\"definition\"><dt>impedance<\/dt>\n \t<dd id=\"fs-id1169737764264\">the AC analogue to resistance in a DC circuit; it is the combined effect of resistance, inductive reactance, and capacitive reactance in the form Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } - X rSub { size 8{C} } ) rSup { size 8{2} } } } {}<\/dd>\n<\/dl><dl id=\"import-auto-id1169737780990\" class=\"definition\"><dt>resonant frequency<\/dt>\n \t<dd id=\"fs-id1169738011017\">the frequency at which the impedance in a circuit is at a minimum, and also the frequency at which the circuit would oscillate if not driven by a voltage source; calculated by f0=12\u03c0LCf0=12\u03c0LC size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital \"LC\"} } } } {}<\/dd>\n<\/dl><dl id=\"import-auto-id1169738055916\" class=\"definition\"><dt>phase angle<\/dt>\n \t<dd id=\"fs-id1169736622298\">denoted by <em>\u03d5\u03d5 size 12{\u03d5} {}<\/em>, the amount by which the voltage and current are out of phase with each other in a circuit<\/dd>\n<\/dl><dl id=\"import-auto-id1169738202247\" class=\"definition\"><dt>power factor<\/dt>\n \t<dd id=\"fs-id1169736971054\">the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase; calculated by cos\u03d5cos\u03d5 size 12{\"cos\"\u03d5} {}<\/dd>\n<\/dl><\/div>","rendered":"<div>RLC Series AC Circuits<\/div>\n<div>\n<ul>\n<li>Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and\/or current in a RLC series circuit.<\/li>\n<li>Draw the circuit diagram for an RLC series circuit.<\/li>\n<li>Explain the significance of the resonant frequency.<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1169736628865\">\n<h1>Impedance<\/h1>\n<p id=\"fs-id1169737742205\">When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other\u2019s effect. <a href=\"#import-auto-id1169736621511\" class=\"autogenerated-content\">[link]<\/a> shows an <em>RLC <\/em>series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an <em>RLC<\/em> circuit is the frequency dependence of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {}, and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important \u201cresonance\u201d features that are the basis of many applications, such as radio tuners.<\/p>\n<figure id=\"import-auto-id1169736621511\"><figcaption>An <em>RLC<\/em> series circuit with an AC voltage source.<\/figcaption><span id=\"import-auto-id1169737820340\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_01a.jpg\" alt=\"The figure describes an R LC series circuit. It shows a resistor R connected in series with an inductor L, connected to a capacitor C in series to an A C source V. The voltage of the A C source is given by V equals V zero sine two pi f t. The voltage across R is V R, across L is V L and across C is V C.\" width=\"275\" \/><\/span><\/p>\n<\/figure>\n<p id=\"import-auto-id1169738181435\">The combined effect of resistance RR size 12{R} {}, inductive reactance XLXL size 12{X rSub { size 8{L} } } {}, and capacitive reactance XCXC size 12{X rSub { size 8{C} } } {} is defined to be <span>impedance<\/span>, an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an <em>RLC<\/em> circuit are related by an AC version of Ohm\u2019s law:<\/p>\n<div class=\"equation\">I0=V0Z or Irms=VrmsZ.I0=V0Z or Irms=VrmsZ. size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } &#8221; or &#8220;I rSub { size 8{ ital &#8220;rms&#8221;} } = { {V rSub { size 8{ ital &#8220;rms&#8221;} } } over {Z} } &#8220;.&#8221; } {}<\/div>\n<p id=\"import-auto-id1169737764601\">Here I0I0 size 12{I rSub { size 8{0} } } {} is the peak current, V0V0 size 12{V rSub { size 8{0} } } {} the peak source voltage, and ZZ is the impedance of the circuit. The units of impedance are ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for ZZ size 12{Z} {} in terms of RR, XLXL size 12{X rSub { size 8{L} } } {}, and XCXC size 12{X rSub { size 8{C} } } {}, we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled VRVR size 12{V rSub { size 8{R} } } {}, VLVL size 12{V rSub { size 8{L} } } {}, and VCVC size 12{V rSub { size 8{C} } } {} in <a href=\"#import-auto-id1169736621511\" class=\"autogenerated-content\">[link]<\/a>.<\/p>\n<p>Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {} are equal and in phase. But we know from the preceding section that the voltage across the inductor VLVL size 12{V rSub { size 8{L} } } {} leads the current by one-fourth of a cycle, the voltage across the capacitor VCVC size 12{V rSub { size 8{C} } } {} follows the current by one-fourth of a cycle, and the voltage across the resistor VRVR size 12{V rSub { size 8{R} } } {} is exactly in phase with the current. <a href=\"#import-auto-id1169738164070\" class=\"autogenerated-content\">[link]<\/a> shows these relationships in one graph, as well as showing the total voltage around the circuit V=VR+VL+VCV=VR+VL+VC size 12{V=V rSub { size 8{R} } +V rSub { size 8{L} } +V rSub { size 8{C} } } {}, where all four voltages are the instantaneous values. According to Kirchhoff\u2019s loop rule, the total voltage around the circuit VV is also the voltage of the source.<\/p>\n<p id=\"import-auto-id1169738092269\">You can see from <a href=\"#import-auto-id1169738164070\" class=\"autogenerated-content\">[link]<\/a> that while VRVR size 12{V rSub { size 8{R} } } {} is in phase with the current, VLVL size 12{V rSub { size 8{L} } } {} leads by 90\u00ba90\u00ba, and VCVC size 12{V rSub { size 8{C} } } {} follows by 90\u00ba90\u00ba. Thus VLVL size 12{V rSub { size 8{L} } } {} and VCVC size 12{V rSub { size 8{C} } } {} are 180\u00ba180\u00ba out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage V0V0 size 12{V rSub { size 8{0} } } {} of the source does <em>not<\/em> equal the sum of the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}. The actual relationship is<\/p>\n<div class=\"equation\">V0=V0R2+(V0L\u2212V0C)2,V0=V0R2+(V0L\u2212V0C)2, size 12{V rSub { size 8{0} } = sqrt {V rSub { size 8{0R} } &#8220;&#8221; lSup { size 8{2} } + ( V rSub { size 8{0L} } &#8211; V rSub { size 8{0C} } ) rSup { size 8{2} } } ,} {}<\/div>\n<p id=\"import-auto-id1169738246882\">where V0RV0R size 12{V rSub { size 8{0R} } } {}, V0LV0L size 12{V rSub { size 8{0L} } } {}, and V0CV0C size 12{V rSub { size 8{0C} } } {} are the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}, respectively. Now, using Ohm\u2019s law and definitions from <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4\">Reactance, Inductive and Capacitive<\/a>, we substitute V0=I0ZV0=I0Z size 12{V rSub { size 8{0} } =I rSub { size 8{0} } Z} {} into the above, as well as V0R=I0RV0R=I0R size 12{V rSub { size 8{0R} } =I rSub { size 8{0} } R} {}, V0L=I0XLV0L=I0XL size 12{V rSub { size 8{0L} } =I rSub { size 8{0} } X rSub { size 8{L} } } {}, and V0C=I0XCV0C=I0XC size 12{V rSub { size 8{0C} } =I rSub { size 8{0} } X rSub { size 8{C} } } {}, yielding<\/p>\n<div class=\"equation\">I0Z=I02R2+(I0XL\u2212I0XC)2=I0R2+(XL\u2212XC)2.I0Z=I02R2+(I0XL\u2212I0XC)2=I0R2+(XL\u2212XC)2. size 12{I rSub { size 8{0} } Z= sqrt {I rSub { size 8{0} rSup { size 8{2} } } R rSup { size 8{2} } + ( I rSub { size 8{0} } X rSub { size 8{L} } &#8211; I rSub { size 8{0} } X rSub { size 8{C} } ) rSup { size 8{2} } } =I rSub { size 8{0} } sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {}<\/div>\n<p id=\"import-auto-id1169736584681\">I0I0 size 12{I rSub { size 8{0} } } {} cancels to yield an expression for ZZ:<\/p>\n<div class=\"equation\" id=\"eip-415\">Z=R2+(XL\u2212XC)2,Z=R2+(XL\u2212XC)2, size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {}<\/div>\n<p id=\"import-auto-id1169736710957\">which is the impedance of an <em>RLC<\/em> series AC circuit. For circuits without a resistor, take R=0R=0; for those without an inductor, take XL=0XL=0 size 12{X rSub { size 8{L} } =0} {}; and for those without a capacitor, take XC=0XC=0 size 12{X rSub { size 8{C} } =0} {}.<\/p>\n<figure id=\"import-auto-id1169738164070\"><figcaption>This graph shows the relationships of the voltages in an <em>RLC<\/em> circuit to the current. The voltages across the circuit elements add to equal the voltage of the source, which is seen to be out of phase with the current.<\/figcaption><span id=\"import-auto-id1169737733064\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_02a.jpg\" alt=\"The figure shows graphs showing the relationships of the voltages in an RLC circuit to the current. It has five graphs on the left and two graphs on the right. The first graph on the right is for current I versus time t. Current is plotted along Y axis and time is along X axis. The curve is a smooth progressive sine wave. The second graph is on the right is for voltage V R versus time t. Voltage V R is plotted along Y axis and time is along X axis. The curve is a smooth progressive sine wave. The third graph is on the right is for voltage V L versus time t. Voltage V L is plotted along Y axis and time is along X axis. The curve is a smooth progressive cosine wave. The fourth graph is on the right is for voltage V C versus time t. Voltage V C is plotted along Y axis and time t is along X axis. The curve is a smooth progressive cosine wave starting from negative Y axis. The fifth graph shows the voltage V verses time t for the R L C circuit. Voltage V is plotted along Y axis and time t is along X axis. The curve is a smooth progressive sine wave starting from a point near to origin on negative X axis. The first and the fifth graphs are again shown on the right and their amplitudes and phases compared. The current graph is shown to have a lesser amplitude.\" width=\"300\" \/><\/span><\/p>\n<\/figure>\n<div class=\"example\" id=\"fs-id1169737723572\">\n<div class=\"title\">Calculating Impedance and Current<\/div>\n<p id=\"import-auto-id1169737041418\">An <em>RLC <\/em>series circuit has a 40.0 \u03a940.0 \u03a9 resistor, a 3.00 mH inductor, and a<br \/>\n5.00 \u03bcF5.00 \u03bcF capacitor. (a) Find the circuit\u2019s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for<br \/>\nLL and<br \/>\nCC<br \/>\nare the same as in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736972664\" class=\"autogenerated-content\">[link]<\/a> and <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736597928\" class=\"autogenerated-content\">[link]<\/a>. (b) If the voltage source has Vrms=120VVrms=120V size 12{V rSub { size 8{&#8220;rms&#8221;} } =&#8221;120&#8243;`V} {}, what is IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at each frequency?<\/p>\n<p id=\"import-auto-id1169735477467\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169738239044\">For each frequency, we use Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {} to find the impedance and then Ohm\u2019s law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again.<\/p>\n<p id=\"import-auto-id1169738147909\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738028359\">At 60.0 Hz, the values of the reactances were found in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736972664\" class=\"autogenerated-content\">[link]<\/a> to be XL=1.13\u03a9XL=1.13\u03a9 size 12{X rSub { size 8{L} } =1 &#8220;.&#8221; &#8220;13&#8221; %OMEGA } {} and in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736597928\" class=\"autogenerated-content\">[link]<\/a> to be XC=531 \u03a9XC=531 \u03a9 size 12{X rSub { size 8{C} } =&#8221;531 &#8221; %OMEGA } {}. Entering these and the given 40.0 \u03a940.0 \u03a9 for resistance into Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {} yields<\/p>\n<div class=\"equation\" id=\"eip-331\">Z=R2+(XL\u2212XC)2=(40.0\u03a9)2+(1.13\u03a9\u2212531\u03a9)2=531\u03a9\u00a0at\u00a060.0 Hz.Z=R2+(XL\u2212XC)2=(40.0\u03a9)2+(1.13\u03a9\u2212531\u03a9)2=531\u03a9\u00a0at\u00a060.0 Hz.alignl { stack {<br \/>\nsize 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {} #<br \/>\n&#8221; &#8220;= sqrt { ( &#8220;40&#8221; &#8220;.&#8221; 0` %OMEGA ) rSup { size 8{2} } + ( 1 &#8220;.&#8221; &#8220;13&#8221; %OMEGA &#8211; &#8220;531&#8221; %OMEGA ) rSup { size 8{2} } } {} #<br \/>\n&#8221; &#8220;=&#8221;531&#8243; %OMEGA &#8221; at 60&#8243; &#8220;.&#8221; &#8220;0 Hz&#8221; {}<br \/>\n} } {}<\/div>\n<p id=\"import-auto-id1169738036576\">Similarly, at 10.0 kHz, XL=188\u03a9XL=188\u03a9 size 12{X rSub { size 8{L} } =&#8221;188&#8243; %OMEGA } {} and XC=3.18\u03a9XC=3.18\u03a9 size 12{X rSub { size 8{C} } =3 &#8220;.&#8221; &#8220;18&#8221; %OMEGA } {}, so that<\/p>\n<div class=\"equation\">Z=(40.0\u03a9)2+(188\u03a9\u22123.18\u03a9)2=190\u03a9\u00a0at\u00a010.0 kHz.Z=(40.0\u03a9)2+(188\u03a9\u22123.18\u03a9)2=190\u03a9\u00a0at\u00a010.0 kHz.alignl { stack {<br \/>\nsize 12{Z= sqrt { ( &#8220;40&#8221; &#8220;.&#8221; 0` %OMEGA ) rSup { size 8{2} } + ( &#8220;188&#8221; %OMEGA &#8211; 3 &#8220;.&#8221; &#8220;18&#8221; %OMEGA ) rSup { size 8{2} } } } {} #<br \/>\n&#8221; &#8220;=&#8221;190&#8243; %OMEGA &#8221; at 10&#8243; &#8220;.&#8221; &#8220;0 kHz&#8221; {}<br \/>\n} } {}<\/div>\n<p id=\"import-auto-id1169737738196\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169736768845\">In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that XLXL size 12{X rSub { size 8{L} } } {} dominates at high frequency and XCXC size 12{X rSub { size 8{C} } } {} dominates at low frequency.<\/p>\n<p id=\"import-auto-id1169737904584\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737710224\">The current IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} can be found using the AC version of Ohm\u2019s law in Equation Irms=Vrms\/ZIrms=Vrms\/Z size 12{I rSub { size 8{&#8220;rms&#8221;} } =V rSub { size 8{&#8220;rms&#8221;} } \/Z} {}:<\/p>\n<p>Irms=VrmsZ=120 V531 \u03a9=0.226 AIrms=VrmsZ=120 V531 \u03a9=0.226 A size 12{I rSub { size 8{&#8220;rms&#8221;} } = { {V rSub { size 8{&#8220;rms&#8221;} } } over {Z} } = { {&#8220;120&#8243;&#8221; V&#8221;} over {&#8220;531 &#8221; %OMEGA } } =0 &#8220;.&#8221; &#8220;226&#8221;&#8221; A&#8221;} {} at 60.0 Hz<\/p>\n<p id=\"import-auto-id1169737758253\">Finally, at 10.0 kHz, we find<\/p>\n<p id=\"import-auto-id1169738110287\">Irms=VrmsZ=120 V190 \u03a9=0.633 AIrms=VrmsZ=120 V190 \u03a9=0.633 A size 12{I rSub { size 8{&#8220;rms&#8221;} } = { {V rSub { size 8{&#8220;rms&#8221;} } } over {Z} } = { {&#8220;120&#8243;&#8221; V&#8221;} over {&#8220;190 &#8221; %OMEGA } } =0 &#8220;.&#8221; &#8220;633&#8221;&#8221; A&#8221;} {} at 10.0 kHz<\/p>\n<p id=\"import-auto-id1169737861583\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738198520\">The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736597928\" class=\"autogenerated-content\">[link]<\/a>. The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4#fs-id1169736972664\" class=\"autogenerated-content\">[link]<\/a>. The inductor dominates at high frequency.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1169738011926\">\n<h1>Resonance in <em>RLC<\/em> Series AC Circuits<\/h1>\n<p id=\"import-auto-id1169736708835\">How does an <em>RLC<\/em> circuit behave as a function of the frequency of the driving voltage source? Combining Ohm\u2019s law, Irms=Vrms\/ZIrms=Vrms\/Z size 12{I rSub { size 8{&#8220;rms&#8221;} } =V rSub { size 8{&#8220;rms&#8221;} } \/Z} {}, and the expression for impedance ZZ from Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {} gives<\/p>\n<div class=\"equation\">Irms=VrmsR2+(XL\u2212XC)2.Irms=VrmsR2+(XL\u2212XC)2. size 12{I rSub { size 8{&#8220;rms&#8221;} } = { {V rSub { size 8{&#8220;rms&#8221;} } } over { sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } } } {}<\/div>\n<p id=\"import-auto-id1169737965627\">The reactances vary with frequency, with XLXL size 12{X rSub { size 8{L} } } {} large at high frequencies and XCXC size 12{X rSub { size 8{C} } } {} large at low frequencies, as we have seen in three previous examples. At some intermediate frequency f0f0 size 12{f rSub { size 8{0} } } {}, the reactances will be equal and cancel, giving <em>Z=RZ=R size 12{Z=R} {}<\/em> \u2014this is a minimum value for impedance, and a maximum value for IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} results. We can get an expression for f0f0 size 12{f rSub { size 8{0} } } {} by taking<\/p>\n<div class=\"equation\">XL=XC.XL=XC. size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}<\/div>\n<p id=\"import-auto-id1169738064647\">Substituting the definitions of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {},<\/p>\n<div class=\"equation\">2\u03c0f0L=12\u03c0f0C.2\u03c0f0L=12\u03c0f0C. size 12{2\u03c0f rSub { size 8{0} } L= { {1} over {2\u03c0f rSub { size 8{0} } C} } } {}<\/div>\n<p id=\"import-auto-id1169737796056\">Solving this expression for f0f0 size 12{f rSub { size 8{0} } } {} yields<\/p>\n<div class=\"equation\">f0=12\u03c0LC,f0=12\u03c0LC, size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital &#8220;LC&#8221;} } } } {}<\/div>\n<p id=\"import-auto-id1169737949912\">where f0f0 size 12{f rSub { size 8{0} } } {} is the <span id=\"import-auto-id1169737994451\">resonant frequency<\/span> of an <em>RLC<\/em> series circuit. This is also the <em>natural frequency<\/em> at which the circuit would oscillate if not driven by the voltage source. At f0f0 size 12{f rSub { size 8{0} } } {}, the effects of the inductor and capacitor cancel, so that <em>Z=RZ=R size 12{Z=R} {}<\/em>, and IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} is a maximum.<\/p>\n<p id=\"import-auto-id1169738136783\">Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation\u2014in this case, forced by the voltage source\u2014at the natural frequency of the system. The receiver in a radio is an <em>RLC<\/em> circuit that oscillates best at its f0f0 size 12{f rSub { size 8{0} } } {}. A variable capacitor is often used to adjust f0f0 size 12{f rSub { size 8{0} } } {} to receive a desired frequency and to reject others. <a href=\"#import-auto-id1169738205664\" class=\"autogenerated-content\">[link]<\/a> is a graph of current as a function of frequency, illustrating a resonant peak in IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at f0f0 size 12{f rSub { size 8{0} } } {}. The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radio receiver, for example.<\/p>\n<figure id=\"import-auto-id1169738205664\"><figcaption>A graph of current versus frequency for two <em>RLC<\/em> series circuits differing only in the amount of resistance. Both have a resonance at f0f0 size 12{f rSub { size 8{0} } } {}, but that for the higher resistance is lower and broader. The driving AC voltage source has a fixed amplitude V0V0 size 12{V rSub { size 8{0} } } {}.<\/figcaption><span id=\"import-auto-id1169737919303\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_03a.jpg\" alt=\"The figure describes a graph of current I versus frequency f. Current I r m s is plotted along Y axis and frequency f is plotted along X axis. Two curves are shown. The upper curve is for small resistance and lower curve is for large resistance. Both the curves have a smooth rise and a fall. The peaks are marked for frequency f zero. The curve for smaller resistance has a higher value of peak than the curve for large resistance.\" width=\"225\" \/><\/span><\/p>\n<\/figure>\n<div class=\"example\" id=\"fs-id1169738045330\">\n<div class=\"title\">Calculating Resonant Frequency and Current<\/div>\n<p id=\"import-auto-id1169738007674\">For the same <em>RLC<\/em> series circuit having a 40.0 \u03a940.0 \u03a9 resistor, a 3.00 mH inductor, and a 5.00 \u03bcF5.00 \u03bcF capacitor: (a) Find the resonant frequency. (b) Calculate IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at resonance if VrmsVrms size 12{V rSub { size 8{&#8220;rms&#8221;} } } {} is 120 V.<\/p>\n<p id=\"import-auto-id1169738107664\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169738239161\">The resonant frequency is found by using the expression in f0=12\u03c0LCf0=12\u03c0LC size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital &#8220;LC&#8221;} } } } {}. The current at that frequency is the same as if the resistor alone were in the circuit.<\/p>\n<p id=\"import-auto-id1169737853725\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738043142\">Entering the given values for LL and CC into the expression given for f0f0 size 12{f rSub { size 8{0} } } {} in f0=12\u03c0LCf0=12\u03c0LC size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital &#8220;LC&#8221;} } } } {} yields<\/p>\n<div class=\"equation\">f0=12\u03c0LC=12\u03c0(3.00\u00d710\u22123 H)(5.00\u00d710\u22126 F)=1.30 kHz.f0=12\u03c0LC=12\u03c0(3.00\u00d710\u22123 H)(5.00\u00d710\u22126 F)=1.30 kHz.alignl { stack {<br \/>\nsize 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital &#8220;LC&#8221;} } } } {} #<br \/>\n&#8221; &#8220;= { {1} over {2\u03c0 sqrt { ( 3 &#8220;.&#8221; &#8220;00&#8221; times &#8220;10&#8221; rSup { size 8{ &#8211; 3} } &#8221; H&#8221; ) ( 5 &#8220;.&#8221; &#8220;00&#8221; times &#8220;10&#8221; rSup { size 8{ &#8211; 6} } &#8221; F&#8221; ) } } } =1 &#8220;.&#8221; &#8220;30&#8221;&#8221; kHz&#8221; {}<br \/>\n} } {}<\/div>\n<p id=\"import-auto-id1169738008646\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738087141\">We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.<\/p>\n<p id=\"import-auto-id1169736708769\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737846339\">The current is given by Ohm\u2019s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus,<\/p>\n<div class=\"equation\">Irms=VrmsZ=120 V40.0 \u03a9=3.00 A.Irms=VrmsZ=120 V40.0 \u03a9=3.00 A. size 12{I rSub { size 8{&#8220;rms&#8221;} } = { {V rSub { size 8{&#8220;rms&#8221;} } } over {Z} } = { {&#8220;120&#8243;&#8221; V&#8221;} over {&#8220;40&#8221; &#8220;.&#8221; &#8220;0 &#8221; %OMEGA } } =3 &#8220;.&#8221; &#8220;00&#8221;&#8221; A&#8221;} {}<\/div>\n<p id=\"import-auto-id1169737909774\"><strong>Discussion for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169736843029\">At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1169737725407\">\n<h1>Power in <em>RLC<\/em> Series AC Circuits<\/h1>\n<p id=\"import-auto-id1169737969148\">If current varies with frequency in an <em>RLC<\/em> circuit, then the power delivered to it also varies with frequency. But the average power is not simply current times voltage, as it is in purely resistive circuits. As was seen in <a href=\"#import-auto-id1169738164070\" class=\"autogenerated-content\">[link]<\/a>, voltage and current are out of phase in an <em>RLC<\/em> circuit. There is a <span id=\"import-auto-id1169738085250\">phase angle<\/span> \u03d5\u03d5 size 12{\u03d5} {} between the source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from<\/p>\n<div class=\"equation\" id=\"eip-543\">cos\u03d5=RZ.cos\u03d5=RZ. size 12{&#8220;cos&#8221;\u03d5= { {R} over {Z} } } {}<\/div>\n<p id=\"import-auto-id1169738181411\">For example, at the resonant frequency or in a purely resistive circuit <em>Z=RZ=R size 12{Z=R} {}<\/em>, so that cos\u03d5=1cos\u03d5=1 size 12{&#8220;cos&#8221;\u03d5=1} {}. This implies that \u03d5=0\u00ba\u03d5=0\u00ba size 12{\u03d5=0 rSup { size 8{ circ } } } {} and that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage and current are out of phase and because IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} is lower. The fact that source voltage and current are out of phase affects the power delivered to the circuit. It can be shown that the <em><em>average power<\/em><\/em> is<\/p>\n<div class=\"equation\">Pave=IrmsVrmscos\u03d5,Pave=IrmsVrmscos\u03d5, size 12{P rSub { size 8{&#8220;ave&#8221;} } =I rSub { size 8{&#8220;rms&#8221;} } V rSub { size 8{&#8220;rms&#8221;} } &#8220;cos&#8221;\u03d5} {}<\/div>\n<p id=\"import-auto-id1169738060586\">Thus cos\u03d5cos\u03d5 size 12{&#8220;cos&#8221;\u03d5} {} is called the <span id=\"import-auto-id1169736719487\">power factor<\/span>, which can range from 0 to 1. Power factors near 1 are desirable when designing an efficient motor, for example. At the resonant frequency, cos\u03d5=1cos\u03d5=1 size 12{&#8220;cos&#8221;\u03d5=1} {}.<\/p>\n<div class=\"example\" id=\"fs-id1169737987472\">\n<div class=\"title\">Calculating the Power Factor and Power<\/div>\n<p id=\"import-auto-id1169737973055\">For the same <em>RLC<\/em> series circuit having a 40.0 \u03a940.0 \u03a9 resistor, a 3.00 mH inductor, a<br \/>\n5.00 \u03bcF5.00 \u03bcF capacitor, and a voltage source with a VrmsVrms of 120 V: (a) Calculate the power factor and phase angle for f=60.0Hzf=60.0Hz size 12{f=&#8221;60&#8243; &#8220;.&#8221; 0`&#8221;Hz&#8221;} {}. (b) What is the average power at 50.0 Hz? (c) Find the average power at the circuit\u2019s resonant frequency.<\/p>\n<p id=\"import-auto-id1169738250320\"><strong>Strategy and Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738181417\">The power factor at 60.0 Hz is found from<\/p>\n<div class=\"equation\" id=\"eip-248\">cos\u03d5=RZ.cos\u03d5=RZ. size 12{&#8220;cos&#8221;\u03d5= { {R} over {Z} } } {}<\/div>\n<p id=\"import-auto-id1169738075830\">We know Z= 531 \u03a9Z= 531 \u03a9 from <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a>, so that<\/p>\n<div class=\"equation\">cos\u03d5=40.0\u03a9531 \u03a9=0.0753 at 60.0 Hz.cos\u03d5=40.0\u03a9531 \u03a9=0.0753 at 60.0 Hz. size 12{&#8220;cos&#8221;\u00d8= { {&#8220;40&#8221; &#8220;.&#8221; 0 %OMEGA } over {5&#8243;31 &#8221; %OMEGA } } =0 &#8220;.&#8221; &#8220;0753&#8221;} {}<\/div>\n<p id=\"import-auto-id1169736844129\">This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is<\/p>\n<div class=\"equation\">\u03d5=cos\u221210.0753=85.7\u00ba at 60.0 Hz.\u03d5=cos\u221210.0753=85.7\u00ba at 60.0 Hz. size 12{\u03d5=&#8221;cos&#8221; rSup { size 8{ &#8211; 1} } 0 &#8220;.&#8221; &#8220;0753&#8221;=&#8221;85&#8243; &#8220;.&#8221; 7 rSup { size 8{ circ } } } {}<\/div>\n<p id=\"import-auto-id1169738148048\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169737789995\">The phase angle is close to<br \/>\n90\u00ba90\u00ba, consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure <em>RC<\/em> circuit has its voltage and current 90\u00ba90\u00ba out of phase).<\/p>\n<p id=\"import-auto-id1169737851510\"><strong>Strategy and Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737851513\">The average power at 60.0 Hz is<\/p>\n<div class=\"equation\">Pave=IrmsVrmscos\u03d5.Pave=IrmsVrmscos\u03d5. size 12{P rSub { size 8{&#8220;ave&#8221;} } =I rSub { size 8{&#8220;rms&#8221;} } V rSub { size 8{&#8220;rms&#8221;} } &#8220;cos&#8221;\u03d5} {}<\/div>\n<p id=\"import-auto-id1169736627643\">IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} was found to be 0.226 A in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a>. Entering the known values gives<\/p>\n<div class=\"equation\">Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz.Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz. size 12{P rSub { size 8{&#8220;ave&#8221;} } = ( 0 &#8220;.&#8221; &#8220;226&#8221;&#8221; A&#8221; ) ( &#8220;120&#8221;&#8221; V&#8221; ) ( 0 &#8220;.&#8221; &#8220;0753&#8221; ) =2 &#8220;.&#8221; &#8220;04&#8221;&#8221; W&#8221;} {}<\/div>\n<p id=\"import-auto-id1169737923237\"><strong>Strategy and Solution for (c)<\/strong><\/p>\n<p id=\"import-auto-id1169737923241\">At the resonant frequency, we know cos\u03d5=1cos\u03d5=1 size 12{&#8220;cos&#8221;\u03d5=1} {}, and IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} was found to be 6.00 A in <a href=\"#fs-id1169738045330\" class=\"autogenerated-content\">[link]<\/a>. Thus,<\/p>\n<p id=\"import-auto-id1169735477311\">Pave=(3.00 A)(120 V)(1)=360 WPave=(3.00 A)(120 V)(1)=360 W size 12{P rSub { size 8{&#8220;ave&#8221;} } = ( 3 &#8220;.&#8221; &#8220;00&#8221;&#8221; A&#8221; ) ( &#8220;120&#8221;&#8221; V&#8221; ) ( 1 ) =&#8221;350&#8243;&#8221; W&#8221;} {} at resonance (1.30 kHz)<\/p>\n<p id=\"import-auto-id1169738080554\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id1169738080557\">Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.<\/p>\n<\/div>\n<p id=\"import-auto-id1169738164030\">Power delivered to an <em>RLC<\/em> series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy input and output but do not dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistor dissipating exactly what the voltage source puts into the circuit. This assumes no significant electromagnetic radiation from the inductor and capacitor, such as radio waves. Such radiation can happen and may even be desired, as we will see in the next chapter on electromagnetic radiation, but it can also be suppressed as is the case in this chapter. The circuit is analogous to the wheel of a car driven over a corrugated road as shown in <a href=\"#import-auto-id1169736885804\" class=\"autogenerated-content\">[link]<\/a>. The regularly spaced bumps in the road are analogous to the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance damping and limiting the amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current, and energy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in the electric field of a capacitor). The amplitude of the wheels\u2019 motion is a maximum if the bumps in the road are hit at the resonant frequency.<\/p>\n<figure id=\"import-auto-id1169736885804\"><figcaption>The forced but damped motion of the wheel on the car spring is analogous to an <em>RLC<\/em> series AC circuit. The shock absorber damps the motion and dissipates energy, analogous to the resistance in an <em>RLC<\/em> circuit. The mass and spring determine the resonant frequency.<\/figcaption><span id=\"import-auto-id1169736885805\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_04a.jpg\" alt=\"The figure describes the path of motion of a wheel of a car. The front wheel of a car is shown. A shock absorber attached to the wheel is also shown. The path of motion is shown as vertically up and down.\" width=\"225\" \/><\/span><\/p>\n<\/figure>\n<p id=\"import-auto-id1169738211177\">A pure <em>LC<\/em> circuit with negligible resistance oscillates at f0f0 size 12{f rSub { size 8{0} } } {}, the same resonant frequency as an <em>RLC<\/em> circuit. It can serve as a frequency standard or clock circuit\u2014for example, in a digital wristwatch. With a very small resistance, only a very small energy input is necessary to maintain the oscillations. The circuit is analogous to a car with no shock absorbers. Once it starts oscillating, it continues at its natural frequency for some time. <a href=\"#import-auto-id1169738257733\" class=\"autogenerated-content\">[link]<\/a> shows the analogy between an <em>LC<\/em> circuit and a mass on a spring.<\/p>\n<figure id=\"import-auto-id1169738257733\"><figcaption>An <em>LC<\/em> circuit is analogous to a mass oscillating on a spring with no friction and no driving force. Energy moves back and forth between the inductor and capacitor, just as it moves from kinetic to potential in the mass-spring system.<\/figcaption><span id=\"import-auto-id1169737987631\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_12_05a.jpg\" alt=\"The figure describes four stages of an L C oscillation circuit compared to a mass oscillating on a spring. Part a of the figure shows a mass attached to a horizontal spring. The spring is attached to a fixed support on the left. The mass is at rest as shown by velocity v equals zero. The energy of the spring is shown as potential energy. This is compared with a circuit containing a capacitor C and inductor L connected together. The energy is shown as stored in the electric field E of the capacitor between the plates. One plate is shown to have a negative polarity and other plate is shown to have a positive polarity. Part b of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The mass is shown to move horizontal toward the fixed support with velocity v. The energy here is stored as the kinetic energy of the spring. This is compared with a circuit containing a capacitor C and inductor L connected together. A current is shown in the circuit and energy is stored as magnetic field B in the inductor. Part c of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The spring is shown as not stretched and the energy is shown as potential energy of the spring. The mass is show to have displaced toward left. This is compared with a circuit containing a capacitor C and inductor L connected together. The energy is shown as stored in the electric field E of the capacitor between the plates. One plate is shown to have a negative polarity and other plate is shown to have a positive polarity. But the polarities are reverse of the first case in part a. Part d of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The mass is shown to move toward right with velocity v. the energy of the spring is kinetic energy. This is compared with a circuit containing a capacitor C and inductor L connected together. A current is shown in the circuit opposite to that in part b and energy is stored as magnetic field B in the inductor.\" width=\"350\" \/><\/span><\/p>\n<\/figure>\n<\/section>\n<div class=\"note\">\n<div class=\"title\">PhET Explorations: Circuit Construction Kit (AC+DC), Virtual Lab<\/div>\n<p>Build circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instruments such as voltmeters and ammeters.<\/p>\n<figure id=\"eip-id1732613\"><figcaption><a href=\"\/resources\/a810be47772627ab26601b90a2c46bf50d610b25\/circuit-construction-kit-ac-virtual-lab_en.jar\">Circuit Construction Kit (AC+DC), Virtual Lab<\/a><\/figcaption>\u00a0<\/p>\n<p><a href=\"\/resources\/a810be47772627ab26601b90a2c46bf50d610b25\/circuit-construction-kit-ac-virtual-lab_en.jar\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/PhET_Icon-56.png\" alt=\"\" width=\"450\" \/><\/a><\/p>\n<p>\u00a0<\/p>\n<p>\u00a0<\/p>\n<p>\u00a0<\/p>\n<\/figure>\n<\/div>\n<section id=\"fs-id1169736711049\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<ul id=\"fs-id1169738178781\">\n<li id=\"import-auto-id1169737979451\">The AC analogy to resistance is impedance ZZ, the combined effect of resistors, inductors, and capacitors, defined by the AC version of Ohm\u2019s law:\n<div class=\"equation\" id=\"eip-111\">I0=V0Z or Irms=VrmsZ,I0=V0Z or Irms=VrmsZ, size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } &#8221; or &#8220;I rSub { size 8{ ital &#8220;rms&#8221;} } = { {V rSub { size 8{ ital &#8220;rms&#8221;} } } over {Z} } ,} {}<\/div>\n<p>where I0I0 size 12{I rSub { size 8{0} } } {} is the peak current and V0V0 size 12{V rSub { size 8{0} } } {} is the peak source voltage.<\/li>\n<li id=\"import-auto-id1169737742046\">Impedance has units of ohms and is given by Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {}.<\/li>\n<li id=\"import-auto-id1169737961915\">The resonant frequency f0f0 size 12{f rSub { size 8{0} } } {}, at which XL=XCXL=XC size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}, is\n<div class=\"equation\" id=\"eip-766\">f0=12\u03c0LC.f0=12\u03c0LC. size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital &#8220;LC&#8221;} } } } {}<\/div>\n<\/li>\n<li id=\"import-auto-id1169736843344\">In an AC circuit, there is a phase angle <em>\u03d5\u03d5 size 12{\u03d5} {}<\/em> between source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from\n<div class=\"equation\">cos\u03d5=RZ,cos\u03d5=RZ, size 12{&#8220;cos&#8221;\u03d5= { {R} over {Z} } } {}<\/div>\n<\/li>\n<li id=\"import-auto-id1169736615857\">\u03d5=0\u00ba\u03d5=0\u00ba size 12{\u03d5=0 rSup { size 8{ circ } } } {} for a purely resistive circuit or an <em>RLC<\/em> circuit at resonance.<\/li>\n<li id=\"import-auto-id1169736821483\">The average power delivered to an <em>RLC<\/em> circuit is affected by the phase angle and is given by\n<div class=\"equation\">Pave=IrmsVrmscos\u03d5,Pave=IrmsVrmscos\u03d5, size 12{P rSub { size 8{&#8220;ave&#8221;} } =I rSub { size 8{&#8220;rms&#8221;} } V rSub { size 8{&#8220;rms&#8221;} } &#8220;cos&#8221;\u03d5} {}<\/div>\n<p>cos\u03d5cos\u03d5 size 12{&#8220;cos&#8221;\u03d5} {} is called the power factor, which ranges from 0 to 1.<\/li>\n<\/ul>\n<\/section>\n<section id=\"fs-id1169738150522\" class=\"conceptual-questions\">\n<h1>Conceptual Questions<\/h1>\n<div class=\"exercise\" id=\"fs-id1169736614731\">\n<div class=\"problem\" id=\"fs-id1169736614734\">\n<p id=\"import-auto-id1169737909824\">Does the resonant frequency of an AC circuit depend on the peak voltage of the AC source? Explain why or why not.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738073810\">\n<div class=\"problem\" id=\"fs-id1169738073813\">\n<p id=\"import-auto-id1169736702937\">Suppose you have a motor with a power factor significantly less than 1. Explain why it would be better to improve the power factor as a method of improving the motor\u2019s output, rather than to increase the voltage input.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"problems-exercises\">\n<h1>Problems &amp; Exercises<\/h1>\n<div class=\"exercise\" id=\"fs-id1169736657142\">\n<div class=\"problem\" id=\"fs-id1169738198661\">\n<p id=\"import-auto-id1169737936131\">An <em>RL<\/em> circuit consists of a<br \/>\n40.0 \u03a940.0 \u03a9 resistor and a<br \/>\n3.00 mH inductor. (a) Find its impedance ZZ at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ with those found in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a> in which there was also a capacitor.<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736895358\">\n<p id=\"import-auto-id1169738134318\">(a) 40.02 \u03a940.02 \u03a9 at 60.0 Hz, 193 \u03a9193 \u03a9 at 10.0 kHz<\/p>\n<p id=\"eip-id3570199\">(b) At 60 Hz, with a capacitor, Z=531 \u03a9Z=531 \u03a9, over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z=190 \u03a9Z=190 \u03a9, about the same as without the capacitor. The capacitor has a smaller effect at high frequencies.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737930730\">\n<div class=\"problem\" id=\"fs-id1169737118482\">\n<p id=\"import-auto-id1169738086377\">An <em>RC<\/em> circuit consists of a<br \/>\n40.0 \u03a940.0 \u03a9 resistor and a<br \/>\n5.00 \u03bcF5.00 \u03bcF capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ with those found in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a>, in which there was also an inductor.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738012738\">\n<div class=\"problem\" id=\"fs-id1169737787125\">\n<p id=\"import-auto-id1169738143727\">An <em>LC<\/em> circuit consists of a 3.00mH3.00mH size 12{3 &#8220;.&#8221; &#8220;00&#8221; \u03bcH} {} inductor and a 5.00\u03bcF5.00\u03bcF size 12{5 &#8220;.&#8221; &#8220;00&#8221; \u03bcF} {} capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ size 12{Z} {} with those found in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a> in which there was also a resistor.<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736844109\">\n<p id=\"import-auto-id1169738246107\">(a) 529 \u03a9529 \u03a9 at 60.0 Hz, 185 \u03a9185 \u03a9 at 10.0 kHz<\/p>\n<p id=\"eip-id2226884\">(b) These values are close to those obtained in <a href=\"#fs-id1169737723572\" class=\"autogenerated-content\">[link]<\/a> because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737794135\">\n<div class=\"problem\" id=\"fs-id1169737794139\">\n<p id=\"import-auto-id1169737793235\">What is the resonant frequency of a 0.500 mH inductor connected to a<br \/>\n40.0 \u03bcF40.0 \u03bcF capacitor?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738072090\">\n<div class=\"problem\" id=\"fs-id1169738071811\">\n<p id=\"import-auto-id1169737728540\">To receive AM radio, you want an <em>RLC<\/em> circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00 \u03bcH1.00 \u03bcH inductor connected to a variable capacitor. What range of capacitance is needed?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738181296\">\n<p id=\"import-auto-id1169737756327\">9.30 nF to 101 nF<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737830882\">\n<div class=\"problem\" id=\"fs-id1169738086090\">\n<p id=\"import-auto-id1169738055925\">Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and capacitors ranging from 1.00 pF to 0.100 F. What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738244135\">\n<div class=\"problem\" id=\"fs-id1169738087540\">\n<p id=\"import-auto-id1169738043322\">What capacitance do you need to produce a resonant frequency of 1.00 GHz, when using an 8.00 nH inductor?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737909755\">\n<p id=\"import-auto-id1169738043327\">3.17 pF<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737763566\">\n<div class=\"problem\" id=\"fs-id1169738007185\">\n<p id=\"import-auto-id1169737000533\">What inductance do you need to produce a resonant frequency of 60.0 Hz, when using a 2.00 \u03bcF2.00 \u03bcF capacitor?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738136983\">\n<div class=\"problem\" id=\"fs-id1169737787019\">\n<p id=\"import-auto-id1169737939475\">The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737821329\">\n<p id=\"import-auto-id1169737939482\">(a) 1.31 \u03bcH1.31 \u03bcH<\/p>\n<p id=\"import-auto-id1169738145328\">(b) 1.66 pF<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738187525\">\n<div class=\"problem\" id=\"fs-id1169737917711\">\n<p id=\"import-auto-id1169738038225\">An <em>RLC<\/em> series circuit has a<br \/>\n2.50 \u03a92.50 \u03a9 resistor, a<br \/>\n100 \u03bcH100 \u03bcH inductor, and an<br \/>\n80.0 \u03bcF80.0 \u03bcF capacitor.(a) Find the circuit\u2019s impedance at 120 Hz. (b) Find the circuit\u2019s impedance at 5.00 kHz. (c) If the voltage source has Vrms=5.60VVrms=5.60V size 12{V rSub { size 8{&#8220;rms&#8221;} } =5 &#8220;.&#8221; &#8220;60&#8221;`V} {}, what is IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at resonance?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738137511\">\n<div class=\"problem\" id=\"fs-id1169737924005\">\n<p id=\"import-auto-id1169737763460\">An <em>RLC<\/em> series circuit has a 1.00 k\u03a91.00 k\u03a9 resistor, a 150 \u03bcH150 \u03bcH inductor, and a 25.0 nF capacitor. (a) Find the circuit\u2019s impedance at 500 Hz. (b) Find the circuit\u2019s impedance at 7.50 kHz. (c) If the voltage source has Vrms=408VVrms=408V size 12{V rSub { size 8{&#8220;rms&#8221;} } =&#8221;408&#8243;`V} {}, what is IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{&#8220;rms&#8221;} } } {} at resonance?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737844573\">\n<p id=\"import-auto-id1169737827160\">(a) 12.8 k\u03a912.8 k\u03a9<\/p>\n<p id=\"import-auto-id1169737827162\">(b) 1.31 k\u03a91.31 k\u03a9<\/p>\n<p id=\"import-auto-id1169737827164\">(c) 31.9 mA at 500 Hz, 312 mA at 7.50 kHz<\/p>\n<p id=\"import-auto-id1169737827167\">(d) 82.2 kHz<\/p>\n<p id=\"import-auto-id1169737788984\">(e) 0.408 A<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738220311\">\n<div class=\"problem\" id=\"fs-id1169738220316\">\n<p id=\"import-auto-id1169737788992\">An <em>RLC<\/em> series circuit has a<br \/>\n2.50 \u03a92.50 \u03a9 resistor, a<br \/>\n100 \u03bcH100 \u03bcH inductor, and an<br \/>\n80.0 \u03bcF80.0 \u03bcF capacitor. (a) Find the power factor at f=120 Hzf=120 Hz. (b) What is the phase angle at 120 Hz? (c) What is the average power at 120 Hz? (d) Find the average power at the circuit\u2019s resonant frequency.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737939490\">\n<div class=\"problem\" id=\"fs-id1169737939495\">\n<p id=\"import-auto-id1169737762694\">An <em>RLC<\/em> series circuit has a<br \/>\n1.00 k\u03a91.00 k\u03a9 resistor, a<br \/>\n150 \u03bcH150 \u03bcH inductor, and a 25.0 nF capacitor. (a) Find the power factor at f=7.50 Hzf=7.50 Hz. (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit\u2019s resonant frequency.<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736621311\">\n<p id=\"import-auto-id1169738052466\">(a) 0.159<\/p>\n<p id=\"import-auto-id1169738052468\">(b) 80.9\u00ba80.9\u00ba<\/p>\n<p id=\"import-auto-id1169738052470\">(c) 26.4 W<\/p>\n<p id=\"import-auto-id1169738052473\">(d) 166 W<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737047401\">\n<div class=\"problem\" id=\"fs-id1169737047405\">\n<p id=\"import-auto-id1169736617346\">An <em>RLC<\/em> series circuit has a<br \/>\n200 \u03a9200 \u03a9<br \/>\nresistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0\u00ba45.0\u00ba. (a) What is the impedance? (b) Find the circuit\u2019s capacitance. (c) If Vrms=408VVrms=408V size 12{V rSub { size 8{&#8220;rms&#8221;} } =&#8221;408&#8243;`V} {} is applied, what is the average power supplied?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738224784\">\n<div class=\"problem\" id=\"fs-id1169737050938\">\n<p id=\"import-auto-id1169738038100\">Referring to <a href=\"#fs-id1169737987472\" class=\"autogenerated-content\">[link]<\/a>, find the average power at 10.0 kHz.<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736932443\">\n<p id=\"import-auto-id1169738038104\">16.0 W<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1169737780982\" class=\"definition\">\n<dt>impedance<\/dt>\n<dd id=\"fs-id1169737764264\">the AC analogue to resistance in a DC circuit; it is the combined effect of resistance, inductive reactance, and capacitive reactance in the form Z=R2+(XL\u2212XC)2Z=R2+(XL\u2212XC)2 size 12{Z= sqrt {R rSup { size 8{2} } + ( X rSub { size 8{L} } &#8211; X rSub { size 8{C} } ) rSup { size 8{2} } } } {}<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1169737780990\" class=\"definition\">\n<dt>resonant frequency<\/dt>\n<dd id=\"fs-id1169738011017\">the frequency at which the impedance in a circuit is at a minimum, and also the frequency at which the circuit would oscillate if not driven by a voltage source; calculated by f0=12\u03c0LCf0=12\u03c0LC size 12{f rSub { size 8{0} } = { {1} over {2\u03c0 sqrt { ital &#8220;LC&#8221;} } } } {}<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1169738055916\" class=\"definition\">\n<dt>phase angle<\/dt>\n<dd id=\"fs-id1169736622298\">denoted by <em>\u03d5\u03d5 size 12{\u03d5} {}<\/em>, the amount by which the voltage and current are out of phase with each other in a circuit<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1169738202247\" class=\"definition\">\n<dt>power factor<\/dt>\n<dd id=\"fs-id1169736971054\">the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase; calculated by cos\u03d5cos\u03d5 size 12{&#8220;cos&#8221;\u03d5} {}<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":103,"menu_order":13,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1199","chapter","type-chapter","status-publish","hentry"],"part":1129,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1199","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/users\/103"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1199\/revisions"}],"predecessor-version":[{"id":1945,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1199\/revisions\/1945"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/parts\/1129"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1199\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/media?parent=1199"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapter-type?post=1199"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/contributor?post=1199"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/license?post=1199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}