{"id":1192,"date":"2017-04-10T15:10:11","date_gmt":"2017-04-10T19:10:11","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/?post_type=chapter&#038;p=1192"},"modified":"2017-04-10T15:10:11","modified_gmt":"2017-04-10T19:10:11","slug":"23-11-reactance-inductive-and-capacitive","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/chapter\/23-11-reactance-inductive-and-capacitive\/","title":{"raw":"23.11 Reactance, Inductive and Capacitive","rendered":"23.11 Reactance, Inductive and Capacitive"},"content":{"raw":"<div>Reactance, Inductive and Capacitive<\/div>\n<div>\n<ul><li>Sketch voltage and current versus time in simple inductive, capacitive, and resistive circuits.<\/li>\n \t<li>Calculate inductive and capacitive reactance.<\/li>\n \t<li>Calculate current and\/or voltage in simple inductive, capacitive, and resistive circuits.<\/li>\n<\/ul><\/div>\n<p id=\"import-auto-id1169737788365\">Many circuits also contain capacitors and inductors, in addition to resistors and an AC voltage source. We have seen how capacitors and inductors respond to DC voltage when it is switched on and off. We will now explore how inductors and capacitors react to sinusoidal AC voltage.<\/p>\n\n<section id=\"fs-id1169738105248\"><h1>Inductors and Inductive Reactance<\/h1>\nSuppose an inductor is connected directly to an AC voltage source, as shown in <a href=\"#import-auto-id1169736696969\" class=\"autogenerated-content\">[link]<\/a>. It is reasonable to assume negligible resistance, since in practice we can make the resistance of an inductor so small that it has a negligible effect on the circuit. Also shown is a graph of voltage and current as functions of time.\n\n<figure id=\"import-auto-id1169736696969\"><figcaption>(a) An AC voltage source in series with an inductor having negligible resistance. (b) Graph of current and voltage across the inductor as functions of time.<\/figcaption><span id=\"import-auto-id1169738014869\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_01a.jpg\" alt=\"Part a of the figure describes an A C voltage source V connected across an inductor L. The voltage across the inductance is shown as V L. Part b of the figure describes a graph showing the variation of current and voltage across the inductance as a function of time. The voltage V L and current I L is plotted along the Y axis and the time t is along the X axis. The graph for current is a progressive sine wave from the origin. The graph for voltage V is a cosine wave and an amplitude slightly less than the current wave.\" width=\"400\"\/><\/span>\n\n<\/figure><p id=\"import-auto-id1169737804173\">The graph in <a href=\"#import-auto-id1169736696969\" class=\"autogenerated-content\">[link]<\/a>(b) starts with voltage at a maximum. Note that the current starts at zero and rises to its peak <em>after<\/em> the voltage that drives it, just as was the case when DC voltage was switched on in the preceding section. When the voltage becomes negative at point a, the current begins to decrease; it becomes zero at point b, where voltage is its most negative. The current then becomes negative, again following the voltage. The voltage becomes positive at point c and begins to make the current less negative. At point d, the current goes through zero just as the voltage reaches its positive peak to start another cycle. This behavior is summarized as follows:<\/p>\n\n<div class=\"note\" id=\"fs-id1169735571584\">\n<div class=\"title\">AC Voltage in an Inductor<\/div>\n<p id=\"import-auto-id1169738084033\">When a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a\n90\u00ba90\u00ba phase angle.<\/p>\n\n<\/div>\n<p id=\"import-auto-id1169738086892\">Current lags behind voltage, since inductors oppose change in current. Changing current induces a back emf V=\u2212L(\u0394I\/\u0394t)V=\u2212L(\u0394I\/\u0394t) size 12{V= - L ( \u0394I\/\u0394t ) } {}. This is considered to be an effective resistance of the inductor to AC. The rms current II size 12{I} {} through an inductor LL size 12{L} {} is given by a version of Ohm\u2019s law:<\/p>\n\n<div class=\"equation\">I=VXL,I=VXL, size 12{I= { {V} over {X rSub { size 8{L} } } } } {}<\/div>\n<p id=\"import-auto-id1169738061327\">where\nVV is the rms voltage across the inductor and XLXL size 12{X rSub { size 8{L} } } {} is defined to be<\/p>\n\n<div class=\"equation\">XL=2\u03c0fL,XL=2\u03c0fL, size 12{X rSub { size 8{L} } =2\u03c0 ital \"fL\"} {}<\/div>\n<p id=\"import-auto-id1169737988958\">with ff size 12{f} {} the frequency of the AC voltage source in hertz (An analysis of the circuit using Kirchhoff\u2019s loop rule and calculus actually produces this expression). XLXL size 12{X rSub { size 8{L} } } {} is called the <span id=\"import-auto-id1169737804525\">inductive reactance<\/span>, because the inductor reacts to impede the current. XLXL size 12{X rSub { size 8{L} } } {} has units of ohms (1 H=1 \u03a9\u22c5s1 H=1 \u03a9\u22c5s, so that frequency times inductance has units of (cycles\/s)(\u03a9\u22c5s)=\u03a9(cycles\/s)(\u03a9\u22c5s)=\u03a9 size 12{ ( \"cycles\/s\" ) ( ` %OMEGA cdot s ) = %OMEGA } {}), consistent with its role as an effective resistance. It makes sense that XLXL size 12{X rSub { size 8{L} } } {} is proportional to LL size 12{L} {}, since the greater the induction the greater its resistance to change. It is also reasonable that XLXL size 12{X rSub { size 8{L} } } {} is proportional to frequency ff size 12{f} {}, since greater frequency means greater change in current. That is, \u0394I\/\u0394t\u0394I\/\u0394t size 12{\u0394I} {} is large for large frequencies (large ff size 12{f} {}<em>,<\/em> small \u0394t\u0394t size 12{\u0394t} {}). The greater the change, the greater the opposition of an inductor.<\/p>\n\n<div class=\"example\" id=\"fs-id1169736972664\">\n<div class=\"title\">Calculating Inductive Reactance and then Current<\/div>\n<p id=\"import-auto-id1169738234775\">(a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current at each frequency if the applied rms voltage is 120 V?<\/p>\n<p id=\"import-auto-id1169737771210\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169738010624\">The inductive reactance is found directly from the expression XL=2\u03c0fLXL=2\u03c0fL size 12{X rSub { size 8{L} } =2\u03c0 ital \"fL\"} {}. Once XLXL size 12{X rSub { size 8{L} } } {} has been found at each frequency, Ohm\u2019s law as stated in the Equation I=V\/XLI=V\/XL size 12{I=V\/X rSub { size 8{L} } } {} can be used to find the current at each frequency.<\/p>\n<p id=\"import-auto-id1169738082082\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169736659051\">Entering the frequency and inductance into Equation XL=2\u03c0fLXL=2\u03c0fL size 12{X rSub { size 8{L} } =2\u03c0 ital \"fL\"} {} gives<\/p>\n\n<div class=\"equation\">XL=2\u03c0fL=6.28(60.0\/ s)(3.00 mH)=1.13 \u03a9\u00a0at\u00a060 Hz.XL=2\u03c0fL=6.28(60.0\/ s)(3.00 mH)=1.13 \u03a9\u00a0at\u00a060 Hz.<\/div>\n<p id=\"import-auto-id1169737806696\">Similarly, at 10 kHz,<\/p>\n\n<div class=\"equation\" id=\"eip-126\">XL=2\u03c0fL=6.28(1.00\u00d7104\/s)(3.00 mH)=188 \u03a9 at 10 kHz.XL=2\u03c0fL=6.28(1.00\u00d7104\/s)(3.00 mH)=188 \u03a9 at 10 kHz. size 12{X rSub { size 8{L} } =2\u03c0 ital \"fL\"=6 \".\" \"28\" ( 3 \".\" \"00\"\" mH\" ) =\"188\" %OMEGA } {}<\/div>\n<p id=\"import-auto-id1169736864032\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169736584158\">The rms current is now found using the version of Ohm\u2019s law in Equation I=V\/XLI=V\/XL size 12{I=V\/X rSub { size 8{L} } } {}, given the applied rms voltage is 120 V. For the first frequency, this yields<\/p>\n\n<div class=\"equation\">I=VXL=120 V1.13 \u03a9=106 A at 60 Hz.I=VXL=120 V1.13 \u03a9=106 A at 60 Hz.<\/div>\n<p id=\"import-auto-id1169738134009\">Similarly, at 10 kHz,<\/p>\n\n<div class=\"equation\">I=VXL=120 V188 \u03a9=0.637 A at 10 kHz.I=VXL=120 V188 \u03a9=0.637 A at 10 kHz. size 12{I= { {V} over {X rSub { size 8{L} } } } = { {\"120\"\" V\"} over {\"188 \" %OMEGA } } =0 \".\" \"637\"\" A\"} {}<\/div>\n<p id=\"import-auto-id1169737711680\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id1169738145157\">The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and the current is small, consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most. Inductors can be used to filter out high frequencies; for example, a large inductor can be put in series with a sound reproduction system or in series with your home computer to reduce high-frequency sound output from your speakers or high-frequency power spikes into your computer.<\/p>\n\n<\/div>\n<p id=\"import-auto-id1169737053571\">Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes its flow. With AC, there is no time for the current to become extremely large.<\/p>\n\n<\/section><section id=\"fs-id1169738031883\"><h1>Capacitors and Capacitive Reactance<\/h1>\n<p id=\"import-auto-id1169736684871\">Consider the capacitor connected directly to an AC voltage source as shown in <a href=\"#import-auto-id1169738082254\" class=\"autogenerated-content\">[link]<\/a>. The resistance of a circuit like this can be made so small that it has a negligible effect compared with the capacitor, and so we can assume negligible resistance. Voltage across the capacitor and current are graphed as functions of time in the figure.<\/p>\n\n<figure id=\"import-auto-id1169738082254\"><figcaption>(a) An AC voltage source in series with a capacitor <em>C<\/em> having negligible resistance. (b) Graph of current and voltage across the capacitor as functions of time.<\/figcaption><span id=\"import-auto-id1169738087944\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_02a.jpg\" alt=\"Part a of the figure shows a capacitor C connected across an A C voltage source V. The voltage across the capacitor is given by V C. Part b of the diagram shows a graph for the variation of current and voltage across the capacitor as functions of time. The voltage V C and current I C is plotted along the Y axis and the time t is along the X axis. The graph for current is a progressive sine wave from the origin starting with a wave along the negative Y axis. The graph for voltage is a cosine wave and amplitude slightly less than the current wave.\" width=\"400\"\/><\/span>\n\n<\/figure><p id=\"import-auto-id1169737765333\">The graph in <a href=\"#import-auto-id1169738082254\" class=\"autogenerated-content\">[link]<\/a> starts with voltage across the capacitor at a maximum. The current is zero at this point, because the capacitor is fully charged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges. At point a, the capacitor has fully discharged (<em>Q=0Q=0 size 12{Q=0} {}<\/em> on it) and the voltage across it is zero. The current remains negative between points a and b, causing the voltage on the capacitor to reverse. This is complete at point b, where the current is zero and the voltage has its most negative value. The current becomes positive after point b, neutralizing the charge on the capacitor and bringing the voltage to zero at point c, which allows the current to reach its maximum. Between points c and d, the current drops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltage follows what the current is doing by one-fourth of a cycle:<\/p>\n\n<div class=\"note\" id=\"fs-id1169737784517\">\n<div class=\"title\">AC Voltage in a Capacitor<\/div>\n<p id=\"import-auto-id1169737796643\">When a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a\n90\u00ba90\u00ba\nphase angle.<\/p>\n\n<\/div>\nThe capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied, there is an rms current, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC, and so the rms current II size 12{I} {} in the circuit containing only a capacitor CC size 12{C} {} is given by another version of Ohm\u2019s law to be\n<div class=\"equation\" id=\"eip-604\">I=VXC,I=VXC, size 12{I= { {V} over {X rSub { size 8{C} } } } } {}<\/div>\n<p id=\"import-auto-id1169737917721\">where VV size 12{V} {} is the rms voltage and XCXC size 12{X rSub { size 8{C} } } {} is defined (As with XLXL size 12{X rSub { size 8{L} } } {}, this expression for XCXC size 12{X rSub { size 8{C} } } {} results from an analysis of the circuit using Kirchhoff\u2019s rules and calculus)<sup> to be<\/sup><\/p>\n\n<div class=\"equation\">XC=12\u03c0fC,XC=12\u03c0fC, size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } } {}<\/div>\n<p id=\"import-auto-id1169736615624\">where XCXC size 12{X rSub { size 8{C} } } {} is called the <span id=\"import-auto-id1169737819555\">capacitive reactance<\/span>, because the capacitor reacts to impede the current. XCXC size 12{X rSub { size 8{C} } } {} has units of ohms (verification left as an exercise for the reader). XCXC size 12{X rSub { size 8{C} } } {} is inversely proportional to the capacitance CC size 12{C} {}; the larger the capacitor, the greater the charge it can store and the greater the current that can flow. It is also inversely proportional to the frequency ff size 12{f} {}; the greater the frequency, the less time there is to fully charge the capacitor, and so it impedes current less.<\/p>\n\n<div class=\"example\" id=\"fs-id1169736597928\">\n<div class=\"title\">Calculating Capacitive Reactance and then Current<\/div>\n<p id=\"import-auto-id1169737895572\">(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current if the applied rms voltage is 120 V?<\/p>\n<p id=\"import-auto-id1169737939050\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169737714880\">The capacitive reactance is found directly from the expression in XC=12\u03c0fCXC=12\u03c0fC size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } } {}. Once\nXCXC<\/p>\nhas been found at each frequency, Ohm\u2019s law stated as I=V\/XCI=V\/XC size 12{I=V\/X rSub { size 8{C} } } {} can be used to find the current at each frequency.\n<p id=\"import-auto-id1169738179315\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738094747\">Entering the frequency and capacitance into XC=12\u03c0fCXC=12\u03c0fC size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } } {} gives<\/p>\n\n<div class=\"equation\">XC=12\u03c0fC=16.28(60.0\/s)(5.00\u00a0\u03bcF)=531 \u03a9\u00a0at\u00a060 Hz.XC=12\u03c0fC=16.28(60.0\/s)(5.00\u00a0\u03bcF)=531 \u03a9\u00a0at\u00a060 Hz.alignl { stack {\nsize 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } } {} #\n\" \"= { {1} over {6 \".\" \"28\" ( \"60\" \".\" 0\/s ) ( 5 \".\" \"00\" \u03bcF ) } } =\"531 \" %OMEGA \" at 60 Hz\" {}\n} } {}<\/div>\n<p id=\"import-auto-id1169738052092\">Similarly, at 10 kHz,<\/p>\n\n<div class=\"equation\" id=\"eip-399\">XC=12\u03c0fC=16.28(1.00\u00d7104\/s)(5.00\u00a0\u03bcF)=3.18 \u03a9\u00a0at\u00a010 kHz.XC=12\u03c0fC=16.28(1.00\u00d7104\/s)(5.00\u00a0\u03bcF)=3.18 \u03a9\u00a0at\u00a010 kHz.alignl { stack {\nsize 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } = { {1} over {6 \".\" \"28\" ( 1 \".\" \"00\" times \"10\" rSup { size 8{4} } \/s ) ( 5 \".\" \"00\" \u03bcF ) } } } {} #\n\" \"=3 \".\" \"18\" %OMEGA \" at 10 kHz\" {}\n} } {}<\/div>\n<p id=\"import-auto-id1169736590721\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169738251850\">The rms current is now found using the version of Ohm\u2019s law in I=V\/XCI=V\/XC size 12{I=V\/X rSub { size 8{C} } } {}, given the applied rms voltage is 120 V. For the first frequency, this yields<\/p>\n\n<div class=\"equation\">I=VXC=120 V531 \u03a9=0.226 A at 60 Hz.I=VXC=120 V531 \u03a9=0.226 A at 60 Hz. size 12{I= { {V} over {X rSub { size 8{C} } } } = { {\"120\"\" V\"} over {\"531 \" %OMEGA } } =0 \".\" \"226\"\" A\"} {}<\/div>\n<p id=\"import-auto-id1169736894272\">Similarly, at 10 kHz,<\/p>\n\n<div class=\"equation\">I=VXC=120 V3.18 \u03a9=37.7 A at 10 kHz.I=VXC=120 V3.18 \u03a9=37.7 A at 10 kHz. size 12{I= { {V} over {X rSub { size 8{C} } } } = { {\"120\"\" V\"} over {3 \".\" \"18 \" %OMEGA } } =\"37\" \".\" 7\" A\"} {}<\/div>\n<p id=\"import-auto-id1169737827258\"><strong>Discussion<\/strong><\/p>\nThe capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum.\n\n<\/div>\n<p id=\"import-auto-id1169738176743\">Although a capacitor is basically an open circuit, there is an rms current in a circuit with an AC voltage applied to a capacitor. This is because the voltage is continually reversing, charging and discharging the capacitor. If the frequency goes to zero (DC), XCXC size 12{X rSub { size 8{C} } } {} tends to infinity, and the current is zero once the capacitor is charged. At very high frequencies, the capacitor\u2019s reactance tends to zero\u2014it has a negligible reactance and does not impede the current (it acts like a simple wire). <em><em>Capacitors have the opposite effect on AC circuits that inductors have<\/em><\/em>.<\/p>\n\n<\/section><section id=\"fs-id1169737961091\"><h1>Resistors in an AC Circuit<\/h1>\n<p id=\"fs-id1169737906325\">Just as a reminder, consider <a href=\"#import-auto-id1169737758127\" class=\"autogenerated-content\">[link]<\/a>, which shows an AC voltage applied to a resistor and a graph of voltage and current versus time. The voltage and current are exactly <em>in <\/em><em><em>phase<\/em><\/em> in a resistor. There is no frequency dependence to the behavior of plain resistance in a circuit:<\/p>\n\n<figure id=\"import-auto-id1169737758127\"><figcaption>(a) An AC voltage source in series with a resistor. (b) Graph of current and voltage across the resistor as functions of time, showing them to be exactly in phase.<\/figcaption><span id=\"import-auto-id1169737763020\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_03a.jpg\" alt=\"Part a of the diagram shows a resistor R connected across an A C voltage source V. The voltage drop across the resistor R is given by V R.Part b of the diagram shows a graph showing the variation of voltage V R and current I R with time t. the V R and current I R are plotted along Y axis and time t is along the X axis. Both I and V are progressive cosine waves. The amplitude of I wave is more than V wave.\" width=\"400\"\/><\/span>\n\n<\/figure><div class=\"note\" id=\"fs-id1169737730746\">\n<div class=\"title\">AC Voltage in a Resistor<\/div>\n<p id=\"import-auto-id1169736702911\">When a sinusoidal voltage is applied to a resistor, the voltage is exactly in phase with the current\u2014they have a 0\u00ba0\u00ba phase angle.<\/p>\n\n<\/div>\n<\/section><section id=\"fs-id1169737784611\" class=\"section-summary\"><h1>Section Summary<\/h1>\n<ul id=\"fs-id1169738048183\"><li id=\"import-auto-id1169737725415\">For inductors in AC circuits, we find that when a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a 90\u00ba90\u00ba phase angle.<\/li>\n \t<li id=\"import-auto-id1169736844107\">The opposition of an inductor to a change in current is expressed as a type of AC resistance.<\/li>\n \t<li id=\"import-auto-id1169736583869\">Ohm\u2019s law for an inductor is\n<div class=\"equation\">I=VXL,I=VXL, size 12{I= { {V} over {X rSub { size 8{L} } } } } {}<\/div>\nwhere VV size 12{V} {} is the rms voltage across the inductor.<\/li>\n \t<li id=\"import-auto-id1169737882424\">XLXL size 12{X rSub { size 8{L} } } {} is defined to be the inductive reactance, given by\n<div class=\"equation\">XL=2\u03c0fL,XL=2\u03c0fL, size 12{X rSub { size 8{L} } =2\u03c0 ital \"fL\"} {}<\/div>\nwith ff size 12{f} {} the frequency of the AC voltage source in hertz.<\/li>\n \t<li id=\"import-auto-id1169738208811\">Inductive reactance XLXL size 12{X rSub { size 8{L} } } {} has units of ohms and is greatest at high frequencies.<\/li>\n \t<li id=\"import-auto-id1169736713656\">For capacitors, we find that when a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a 90\u00ba90\u00ba phase angle.<\/li>\n \t<li id=\"import-auto-id1169738012850\">Since a capacitor can stop current when fully charged, it limits current and offers another form of AC resistance; Ohm\u2019s law for a capacitor is\n<div class=\"equation\">I=VXC,I=VXC, size 12{I= { {V} over {X rSub { size 8{C} } } } } {}<\/div>\nwhere VV size 12{V} {} is the rms voltage across the capacitor.<\/li>\n \t<li id=\"import-auto-id1169737821215\">XCXC size 12{X rSub { size 8{C} } } {} is defined to be the capacitive reactance, given by\n<div class=\"equation\">XC=12\u03c0fC.XC=12\u03c0fC. size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } } {}<\/div><\/li>\n \t<li id=\"import-auto-id1169738016153\">XCXC size 12{X rSub { size 8{C} } } {} has units of ohms and is greatest at low frequencies.<\/li>\n<\/ul><\/section><section id=\"fs-id1169738011145\" class=\"conceptual-questions\"><h1>Conceptual Questions<\/h1>\n<div class=\"exercise\" id=\"fs-id1169738010807\">\n<div class=\"problem\" id=\"fs-id1169738250359\">\n<p id=\"import-auto-id1169737050603\">Presbycusis is a hearing loss due to age that progressively affects higher frequencies. A hearing aid amplifier is designed to amplify all frequencies equally. To adjust its output for presbycusis, would you put a capacitor in series or parallel with the hearing aid\u2019s speaker? Explain.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737854523\">\n<div class=\"problem\" id=\"fs-id1169738105398\">\n<p id=\"import-auto-id1169738054334\">Would you use a large inductance or a large capacitance in series with a system to filter out low frequencies, such as the 100 Hz hum in a sound system? Explain.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737930275\">\n<div class=\"problem\" id=\"fs-id1169737968751\">\n<p id=\"import-auto-id1169738211558\">High-frequency noise in AC power can damage computers. Does the plug-in unit designed to prevent this damage use a large inductance or a large capacitance (in series with the computer) to filter out such high frequencies? Explain.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"problem\" id=\"fs-id1169738116308\">\n<p id=\"import-auto-id1169738012609\">Does inductance depend on current, frequency, or both? What about inductive reactance?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737785016\">\n<div class=\"problem\" id=\"fs-id1169737861279\">\n<p id=\"import-auto-id1169738173969\">Explain why the capacitor in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(a) acts as a low-frequency filter between the two circuits, whereas that in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(b) acts as a high-frequency filter.<\/p>\n\n<\/div>\n<\/div>\n<figure id=\"import-auto-id1169738245387\"><figcaption>Capacitors and inductors. Capacitor with high frequency and low frequency.<\/figcaption><span id=\"import-auto-id1169737793629\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_04a.jpg\" alt=\"The figure describes two circuits with two different connections. The first part of the diagram shows circuit one and circuit two connected in series and a capacitor C is connected between them. Both the circuits are shown as grounded. The second part of the diagram shows two circuits circuit one and circuit two connected to each other. At the point of connection one end of the capacitor is connected and the other end of the capacitor is grounded. Both the circuits are shown as grounded.\" width=\"250\"\/><\/span>\n\n<\/figure><div class=\"exercise\" id=\"fs-id1169736669333\">\n<div class=\"problem\" id=\"fs-id1169738050827\">\n<p id=\"import-auto-id1169737869873\">If the capacitors in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a> are replaced by inductors, which acts as a low-frequency filter and which as a high-frequency filter?<\/p>\n\n<\/div>\n<\/div>\n<\/section><section id=\"fs-id1169737726120\" class=\"problems-exercises\"><h1>Problems &amp; Exercises<\/h1>\n<div class=\"exercise\" id=\"fs-id1169738117280\">\n<div class=\"problem\" id=\"fs-id1169737793629\">\n<p id=\"import-auto-id1169738200256\">At what frequency will a 30.0 mH inductor have a reactance of\n100 \u03a9100 \u03a9?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737917875\">\n<p id=\"import-auto-id1169737728465\">531 Hz<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737786419\">\n<div class=\"problem\" id=\"fs-id1169736656664\">\n<p id=\"import-auto-id1169737812535\">What value of inductance should be used if a\n20.0 k\u03a920.0 k\u03a9\nreactance is needed at a frequency of 500 Hz?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738077769\">\n<div class=\"problem\" id=\"fs-id1169738072176\">\n<p id=\"import-auto-id1169736606594\">What capacitance should be used to produce a\n2.00 M\u03a92.00 M\u03a9 reactance at 60.0 Hz?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737909282\">\n<p id=\"import-auto-id1169738045746\">1.33 nF<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738198876\">\n<div class=\"problem\" id=\"fs-id1169737757642\">\n<p id=\"import-auto-id1169738036322\">At what frequency will an 80.0 mF capacitor have a reactance of 0.250 \u03a90.250 \u03a9?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737729245\">\n<div class=\"problem\" id=\"fs-id1169737754951\">\n<p id=\"import-auto-id1169736825551\">(a) Find the current through a 0.500 H inductor connected to a 60.0 Hz, 480 V AC source. (b) What would the current be at 100 kHz?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738193476\">\n<p id=\"import-auto-id1169737772039\">(a) 2.55 A<\/p>\n<p id=\"import-auto-id1169737826940\">(b) 1.53 mA<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738086496\">\n<div class=\"problem\" id=\"fs-id1169737812738\">\n<p id=\"import-auto-id1169737825035\">(a) What current flows when a 60.0 Hz, 480 V AC source is connected to a\n0.250 \u03bcF0.250 \u03bcF\ncapacitor? (b) What would the current be at 25.0 kHz?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738198521\">\n<div class=\"problem\" id=\"fs-id1169737848534\">\n<p id=\"import-auto-id1169737972545\">A 20.0 kHz, 16.0 V source connected to an inductor produces a 2.00 A current. What is the inductance?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738077151\">\n\n63.7 \u00b5H63.7 \u00b5H\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737994347\">\n<div class=\"problem\" id=\"fs-id1169737764442\">\n<p id=\"import-auto-id1169736658894\">A 20.0 Hz, 16.0 V source produces a 2.00 mA current when connected to a capacitor. What is the capacitance?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738250258\">\n<div class=\"problem\" id=\"fs-id1169737802646\">\n<p id=\"import-auto-id1169737734640\">(a) An inductor designed to filter high-frequency noise from power supplied to a personal computer is placed in series with the computer. What minimum inductance should it have to produce a\n2.00 k\u03a92.00 k\u03a9 reactance for 15.0 kHz noise? (b) What is its reactance at 60.0 Hz?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736983958\">\n<p id=\"import-auto-id1169737814409\">(a) 21.2 mH<\/p>\n<p id=\"import-auto-id1169737854280\">(b) 8.00 \u03a98.00 \u03a9<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738162931\">\n<div class=\"problem\" id=\"fs-id1169736713656\">\n<p id=\"import-auto-id1169736583393\">The capacitor in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(a) is designed to filter low-frequency signals, impeding their transmission between circuits. (a) What capacitance is needed to produce a\n100 k\u03a9100 k\u03a9 reactance at a frequency of 120 Hz? (b) What would its reactance be at 1.00 MHz? (c) Discuss the implications of your answers to (a) and (b).<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736611033\">\n<div class=\"problem\" id=\"fs-id1169736583899\">\n<p id=\"import-auto-id1169738116034\">The capacitor in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(b) will filter high-frequency signals by shorting them to earth\/ground. (a) What capacitance is needed to produce a reactance of\n10.0 m\u03a910.0 m\u03a9 for a 5.00 kHz signal? (b) What would its reactance be at 3.00 Hz? (c) Discuss the implications of your answers to (a) and (b).<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738110286\">\n<p id=\"import-auto-id1169738138113\">(a) 3.18 mF<\/p>\n<p id=\"import-auto-id1169737965361\">(b) 16.7 \u03a916.7 \u03a9<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737802631\">\n<div class=\"problem\" id=\"fs-id1169736756883\">\n<p id=\"import-auto-id1169737819164\"><strong>Unreasonable Results<\/strong><\/p>\nIn a recording of voltages due to brain activity (an EEG), a 10.0 mV signal with a 0.500 Hz frequency is applied to a capacitor, producing a current of 100 mA. Resistance is negligible. (a) What is the capacitance? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736919851\">\n<div class=\"problem\" id=\"fs-id1169736620528\">\n<p id=\"import-auto-id1169738131438\"><strong>Construct Your Own Problem<\/strong><\/p>\nConsider the use of an inductor in series with a computer operating on 60 Hz electricity. Construct a problem in which you calculate the relative reduction in voltage of incoming high frequency noise compared to 60 Hz voltage. Among the things to consider are the acceptable series reactance of the inductor for 60 Hz power and the likely frequencies of noise coming through the power lines.\n\n<\/div>\n<\/div>\n<\/section><div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1169738074090\" class=\"definition\"><dt>inductive reactance<\/dt>\n \t<dd>the opposition of an inductor to a change in current; calculated by XL=2\u03c0fLXL=2\u03c0fL size 12{X rSub { size 8{L} } =2\u03c0 ital \"fL\"} {}<\/dd>\n<\/dl><dl id=\"import-auto-id1169736611323\" class=\"definition\"><dt>capacitive reactance<\/dt>\n \t<dd id=\"fs-id1169737724984\">the opposition of a capacitor to a change in current; calculated by XC=12\u03c0fCXC=12\u03c0fC size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital \"fC\"} } } {}<\/dd>\n<\/dl><\/div>","rendered":"<div>Reactance, Inductive and Capacitive<\/div>\n<div>\n<ul>\n<li>Sketch voltage and current versus time in simple inductive, capacitive, and resistive circuits.<\/li>\n<li>Calculate inductive and capacitive reactance.<\/li>\n<li>Calculate current and\/or voltage in simple inductive, capacitive, and resistive circuits.<\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id1169737788365\">Many circuits also contain capacitors and inductors, in addition to resistors and an AC voltage source. We have seen how capacitors and inductors respond to DC voltage when it is switched on and off. We will now explore how inductors and capacitors react to sinusoidal AC voltage.<\/p>\n<section id=\"fs-id1169738105248\">\n<h1>Inductors and Inductive Reactance<\/h1>\n<p>Suppose an inductor is connected directly to an AC voltage source, as shown in <a href=\"#import-auto-id1169736696969\" class=\"autogenerated-content\">[link]<\/a>. It is reasonable to assume negligible resistance, since in practice we can make the resistance of an inductor so small that it has a negligible effect on the circuit. Also shown is a graph of voltage and current as functions of time.<\/p>\n<figure id=\"import-auto-id1169736696969\"><figcaption>(a) An AC voltage source in series with an inductor having negligible resistance. (b) Graph of current and voltage across the inductor as functions of time.<\/figcaption><span id=\"import-auto-id1169738014869\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_01a.jpg\" alt=\"Part a of the figure describes an A C voltage source V connected across an inductor L. The voltage across the inductance is shown as V L. Part b of the figure describes a graph showing the variation of current and voltage across the inductance as a function of time. The voltage V L and current I L is plotted along the Y axis and the time t is along the X axis. The graph for current is a progressive sine wave from the origin. The graph for voltage V is a cosine wave and an amplitude slightly less than the current wave.\" width=\"400\" \/><\/span><\/p>\n<\/figure>\n<p id=\"import-auto-id1169737804173\">The graph in <a href=\"#import-auto-id1169736696969\" class=\"autogenerated-content\">[link]<\/a>(b) starts with voltage at a maximum. Note that the current starts at zero and rises to its peak <em>after<\/em> the voltage that drives it, just as was the case when DC voltage was switched on in the preceding section. When the voltage becomes negative at point a, the current begins to decrease; it becomes zero at point b, where voltage is its most negative. The current then becomes negative, again following the voltage. The voltage becomes positive at point c and begins to make the current less negative. At point d, the current goes through zero just as the voltage reaches its positive peak to start another cycle. This behavior is summarized as follows:<\/p>\n<div class=\"note\" id=\"fs-id1169735571584\">\n<div class=\"title\">AC Voltage in an Inductor<\/div>\n<p id=\"import-auto-id1169738084033\">When a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a<br \/>\n90\u00ba90\u00ba phase angle.<\/p>\n<\/div>\n<p id=\"import-auto-id1169738086892\">Current lags behind voltage, since inductors oppose change in current. Changing current induces a back emf V=\u2212L(\u0394I\/\u0394t)V=\u2212L(\u0394I\/\u0394t) size 12{V= &#8211; L ( \u0394I\/\u0394t ) } {}. This is considered to be an effective resistance of the inductor to AC. The rms current II size 12{I} {} through an inductor LL size 12{L} {} is given by a version of Ohm\u2019s law:<\/p>\n<div class=\"equation\">I=VXL,I=VXL, size 12{I= { {V} over {X rSub { size 8{L} } } } } {}<\/div>\n<p id=\"import-auto-id1169738061327\">where<br \/>\nVV is the rms voltage across the inductor and XLXL size 12{X rSub { size 8{L} } } {} is defined to be<\/p>\n<div class=\"equation\">XL=2\u03c0fL,XL=2\u03c0fL, size 12{X rSub { size 8{L} } =2\u03c0 ital &#8220;fL&#8221;} {}<\/div>\n<p id=\"import-auto-id1169737988958\">with ff size 12{f} {} the frequency of the AC voltage source in hertz (An analysis of the circuit using Kirchhoff\u2019s loop rule and calculus actually produces this expression). XLXL size 12{X rSub { size 8{L} } } {} is called the <span id=\"import-auto-id1169737804525\">inductive reactance<\/span>, because the inductor reacts to impede the current. XLXL size 12{X rSub { size 8{L} } } {} has units of ohms (1 H=1 \u03a9\u22c5s1 H=1 \u03a9\u22c5s, so that frequency times inductance has units of (cycles\/s)(\u03a9\u22c5s)=\u03a9(cycles\/s)(\u03a9\u22c5s)=\u03a9 size 12{ ( &#8220;cycles\/s&#8221; ) ( ` %OMEGA cdot s ) = %OMEGA } {}), consistent with its role as an effective resistance. It makes sense that XLXL size 12{X rSub { size 8{L} } } {} is proportional to LL size 12{L} {}, since the greater the induction the greater its resistance to change. It is also reasonable that XLXL size 12{X rSub { size 8{L} } } {} is proportional to frequency ff size 12{f} {}, since greater frequency means greater change in current. That is, \u0394I\/\u0394t\u0394I\/\u0394t size 12{\u0394I} {} is large for large frequencies (large ff size 12{f} {}<em>,<\/em> small \u0394t\u0394t size 12{\u0394t} {}). The greater the change, the greater the opposition of an inductor.<\/p>\n<div class=\"example\" id=\"fs-id1169736972664\">\n<div class=\"title\">Calculating Inductive Reactance and then Current<\/div>\n<p id=\"import-auto-id1169738234775\">(a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current at each frequency if the applied rms voltage is 120 V?<\/p>\n<p id=\"import-auto-id1169737771210\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169738010624\">The inductive reactance is found directly from the expression XL=2\u03c0fLXL=2\u03c0fL size 12{X rSub { size 8{L} } =2\u03c0 ital &#8220;fL&#8221;} {}. Once XLXL size 12{X rSub { size 8{L} } } {} has been found at each frequency, Ohm\u2019s law as stated in the Equation I=V\/XLI=V\/XL size 12{I=V\/X rSub { size 8{L} } } {} can be used to find the current at each frequency.<\/p>\n<p id=\"import-auto-id1169738082082\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169736659051\">Entering the frequency and inductance into Equation XL=2\u03c0fLXL=2\u03c0fL size 12{X rSub { size 8{L} } =2\u03c0 ital &#8220;fL&#8221;} {} gives<\/p>\n<div class=\"equation\">XL=2\u03c0fL=6.28(60.0\/ s)(3.00 mH)=1.13 \u03a9\u00a0at\u00a060 Hz.XL=2\u03c0fL=6.28(60.0\/ s)(3.00 mH)=1.13 \u03a9\u00a0at\u00a060 Hz.<\/div>\n<p id=\"import-auto-id1169737806696\">Similarly, at 10 kHz,<\/p>\n<div class=\"equation\" id=\"eip-126\">XL=2\u03c0fL=6.28(1.00\u00d7104\/s)(3.00 mH)=188 \u03a9 at 10 kHz.XL=2\u03c0fL=6.28(1.00\u00d7104\/s)(3.00 mH)=188 \u03a9 at 10 kHz. size 12{X rSub { size 8{L} } =2\u03c0 ital &#8220;fL&#8221;=6 &#8220;.&#8221; &#8220;28&#8221; ( 3 &#8220;.&#8221; &#8220;00&#8221;&#8221; mH&#8221; ) =&#8221;188&#8243; %OMEGA } {}<\/div>\n<p id=\"import-auto-id1169736864032\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169736584158\">The rms current is now found using the version of Ohm\u2019s law in Equation I=V\/XLI=V\/XL size 12{I=V\/X rSub { size 8{L} } } {}, given the applied rms voltage is 120 V. For the first frequency, this yields<\/p>\n<div class=\"equation\">I=VXL=120 V1.13 \u03a9=106 A at 60 Hz.I=VXL=120 V1.13 \u03a9=106 A at 60 Hz.<\/div>\n<p id=\"import-auto-id1169738134009\">Similarly, at 10 kHz,<\/p>\n<div class=\"equation\">I=VXL=120 V188 \u03a9=0.637 A at 10 kHz.I=VXL=120 V188 \u03a9=0.637 A at 10 kHz. size 12{I= { {V} over {X rSub { size 8{L} } } } = { {&#8220;120&#8243;&#8221; V&#8221;} over {&#8220;188 &#8221; %OMEGA } } =0 &#8220;.&#8221; &#8220;637&#8221;&#8221; A&#8221;} {}<\/div>\n<p id=\"import-auto-id1169737711680\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id1169738145157\">The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and the current is small, consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most. Inductors can be used to filter out high frequencies; for example, a large inductor can be put in series with a sound reproduction system or in series with your home computer to reduce high-frequency sound output from your speakers or high-frequency power spikes into your computer.<\/p>\n<\/div>\n<p id=\"import-auto-id1169737053571\">Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes its flow. With AC, there is no time for the current to become extremely large.<\/p>\n<\/section>\n<section id=\"fs-id1169738031883\">\n<h1>Capacitors and Capacitive Reactance<\/h1>\n<p id=\"import-auto-id1169736684871\">Consider the capacitor connected directly to an AC voltage source as shown in <a href=\"#import-auto-id1169738082254\" class=\"autogenerated-content\">[link]<\/a>. The resistance of a circuit like this can be made so small that it has a negligible effect compared with the capacitor, and so we can assume negligible resistance. Voltage across the capacitor and current are graphed as functions of time in the figure.<\/p>\n<figure id=\"import-auto-id1169738082254\"><figcaption>(a) An AC voltage source in series with a capacitor <em>C<\/em> having negligible resistance. (b) Graph of current and voltage across the capacitor as functions of time.<\/figcaption><span id=\"import-auto-id1169738087944\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_02a.jpg\" alt=\"Part a of the figure shows a capacitor C connected across an A C voltage source V. The voltage across the capacitor is given by V C. Part b of the diagram shows a graph for the variation of current and voltage across the capacitor as functions of time. The voltage V C and current I C is plotted along the Y axis and the time t is along the X axis. The graph for current is a progressive sine wave from the origin starting with a wave along the negative Y axis. The graph for voltage is a cosine wave and amplitude slightly less than the current wave.\" width=\"400\" \/><\/span><\/p>\n<\/figure>\n<p id=\"import-auto-id1169737765333\">The graph in <a href=\"#import-auto-id1169738082254\" class=\"autogenerated-content\">[link]<\/a> starts with voltage across the capacitor at a maximum. The current is zero at this point, because the capacitor is fully charged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges. At point a, the capacitor has fully discharged (<em>Q=0Q=0 size 12{Q=0} {}<\/em> on it) and the voltage across it is zero. The current remains negative between points a and b, causing the voltage on the capacitor to reverse. This is complete at point b, where the current is zero and the voltage has its most negative value. The current becomes positive after point b, neutralizing the charge on the capacitor and bringing the voltage to zero at point c, which allows the current to reach its maximum. Between points c and d, the current drops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltage follows what the current is doing by one-fourth of a cycle:<\/p>\n<div class=\"note\" id=\"fs-id1169737784517\">\n<div class=\"title\">AC Voltage in a Capacitor<\/div>\n<p id=\"import-auto-id1169737796643\">When a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a<br \/>\n90\u00ba90\u00ba<br \/>\nphase angle.<\/p>\n<\/div>\n<p>The capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied, there is an rms current, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC, and so the rms current II size 12{I} {} in the circuit containing only a capacitor CC size 12{C} {} is given by another version of Ohm\u2019s law to be<\/p>\n<div class=\"equation\" id=\"eip-604\">I=VXC,I=VXC, size 12{I= { {V} over {X rSub { size 8{C} } } } } {}<\/div>\n<p id=\"import-auto-id1169737917721\">where VV size 12{V} {} is the rms voltage and XCXC size 12{X rSub { size 8{C} } } {} is defined (As with XLXL size 12{X rSub { size 8{L} } } {}, this expression for XCXC size 12{X rSub { size 8{C} } } {} results from an analysis of the circuit using Kirchhoff\u2019s rules and calculus)<sup> to be<\/sup><\/p>\n<div class=\"equation\">XC=12\u03c0fC,XC=12\u03c0fC, size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } } {}<\/div>\n<p id=\"import-auto-id1169736615624\">where XCXC size 12{X rSub { size 8{C} } } {} is called the <span id=\"import-auto-id1169737819555\">capacitive reactance<\/span>, because the capacitor reacts to impede the current. XCXC size 12{X rSub { size 8{C} } } {} has units of ohms (verification left as an exercise for the reader). XCXC size 12{X rSub { size 8{C} } } {} is inversely proportional to the capacitance CC size 12{C} {}; the larger the capacitor, the greater the charge it can store and the greater the current that can flow. It is also inversely proportional to the frequency ff size 12{f} {}; the greater the frequency, the less time there is to fully charge the capacitor, and so it impedes current less.<\/p>\n<div class=\"example\" id=\"fs-id1169736597928\">\n<div class=\"title\">Calculating Capacitive Reactance and then Current<\/div>\n<p id=\"import-auto-id1169737895572\">(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current if the applied rms voltage is 120 V?<\/p>\n<p id=\"import-auto-id1169737939050\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1169737714880\">The capacitive reactance is found directly from the expression in XC=12\u03c0fCXC=12\u03c0fC size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } } {}. Once<br \/>\nXCXC<\/p>\n<p>has been found at each frequency, Ohm\u2019s law stated as I=V\/XCI=V\/XC size 12{I=V\/X rSub { size 8{C} } } {} can be used to find the current at each frequency.<\/p>\n<p id=\"import-auto-id1169738179315\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738094747\">Entering the frequency and capacitance into XC=12\u03c0fCXC=12\u03c0fC size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } } {} gives<\/p>\n<div class=\"equation\">XC=12\u03c0fC=16.28(60.0\/s)(5.00\u00a0\u03bcF)=531 \u03a9\u00a0at\u00a060 Hz.XC=12\u03c0fC=16.28(60.0\/s)(5.00\u00a0\u03bcF)=531 \u03a9\u00a0at\u00a060 Hz.alignl { stack {<br \/>\nsize 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } } {} #<br \/>\n&#8221; &#8220;= { {1} over {6 &#8220;.&#8221; &#8220;28&#8221; ( &#8220;60&#8221; &#8220;.&#8221; 0\/s ) ( 5 &#8220;.&#8221; &#8220;00&#8221; \u03bcF ) } } =&#8221;531 &#8221; %OMEGA &#8221; at 60 Hz&#8221; {}<br \/>\n} } {}<\/div>\n<p id=\"import-auto-id1169738052092\">Similarly, at 10 kHz,<\/p>\n<div class=\"equation\" id=\"eip-399\">XC=12\u03c0fC=16.28(1.00\u00d7104\/s)(5.00\u00a0\u03bcF)=3.18 \u03a9\u00a0at\u00a010 kHz.XC=12\u03c0fC=16.28(1.00\u00d7104\/s)(5.00\u00a0\u03bcF)=3.18 \u03a9\u00a0at\u00a010 kHz.alignl { stack {<br \/>\nsize 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } = { {1} over {6 &#8220;.&#8221; &#8220;28&#8221; ( 1 &#8220;.&#8221; &#8220;00&#8221; times &#8220;10&#8221; rSup { size 8{4} } \/s ) ( 5 &#8220;.&#8221; &#8220;00&#8221; \u03bcF ) } } } {} #<br \/>\n&#8221; &#8220;=3 &#8220;.&#8221; &#8220;18&#8221; %OMEGA &#8221; at 10 kHz&#8221; {}<br \/>\n} } {}<\/div>\n<p id=\"import-auto-id1169736590721\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169738251850\">The rms current is now found using the version of Ohm\u2019s law in I=V\/XCI=V\/XC size 12{I=V\/X rSub { size 8{C} } } {}, given the applied rms voltage is 120 V. For the first frequency, this yields<\/p>\n<div class=\"equation\">I=VXC=120 V531 \u03a9=0.226 A at 60 Hz.I=VXC=120 V531 \u03a9=0.226 A at 60 Hz. size 12{I= { {V} over {X rSub { size 8{C} } } } = { {&#8220;120&#8243;&#8221; V&#8221;} over {&#8220;531 &#8221; %OMEGA } } =0 &#8220;.&#8221; &#8220;226&#8221;&#8221; A&#8221;} {}<\/div>\n<p id=\"import-auto-id1169736894272\">Similarly, at 10 kHz,<\/p>\n<div class=\"equation\">I=VXC=120 V3.18 \u03a9=37.7 A at 10 kHz.I=VXC=120 V3.18 \u03a9=37.7 A at 10 kHz. size 12{I= { {V} over {X rSub { size 8{C} } } } = { {&#8220;120&#8243;&#8221; V&#8221;} over {3 &#8220;.&#8221; &#8220;18 &#8221; %OMEGA } } =&#8221;37&#8243; &#8220;.&#8221; 7&#8243; A&#8221;} {}<\/div>\n<p id=\"import-auto-id1169737827258\"><strong>Discussion<\/strong><\/p>\n<p>The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum.<\/p>\n<\/div>\n<p id=\"import-auto-id1169738176743\">Although a capacitor is basically an open circuit, there is an rms current in a circuit with an AC voltage applied to a capacitor. This is because the voltage is continually reversing, charging and discharging the capacitor. If the frequency goes to zero (DC), XCXC size 12{X rSub { size 8{C} } } {} tends to infinity, and the current is zero once the capacitor is charged. At very high frequencies, the capacitor\u2019s reactance tends to zero\u2014it has a negligible reactance and does not impede the current (it acts like a simple wire). <em><em>Capacitors have the opposite effect on AC circuits that inductors have<\/em><\/em>.<\/p>\n<\/section>\n<section id=\"fs-id1169737961091\">\n<h1>Resistors in an AC Circuit<\/h1>\n<p id=\"fs-id1169737906325\">Just as a reminder, consider <a href=\"#import-auto-id1169737758127\" class=\"autogenerated-content\">[link]<\/a>, which shows an AC voltage applied to a resistor and a graph of voltage and current versus time. The voltage and current are exactly <em>in <\/em><em><em>phase<\/em><\/em> in a resistor. There is no frequency dependence to the behavior of plain resistance in a circuit:<\/p>\n<figure id=\"import-auto-id1169737758127\"><figcaption>(a) An AC voltage source in series with a resistor. (b) Graph of current and voltage across the resistor as functions of time, showing them to be exactly in phase.<\/figcaption><span id=\"import-auto-id1169737763020\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_03a.jpg\" alt=\"Part a of the diagram shows a resistor R connected across an A C voltage source V. The voltage drop across the resistor R is given by V R.Part b of the diagram shows a graph showing the variation of voltage V R and current I R with time t. the V R and current I R are plotted along Y axis and time t is along the X axis. Both I and V are progressive cosine waves. The amplitude of I wave is more than V wave.\" width=\"400\" \/><\/span><\/p>\n<\/figure>\n<div class=\"note\" id=\"fs-id1169737730746\">\n<div class=\"title\">AC Voltage in a Resistor<\/div>\n<p id=\"import-auto-id1169736702911\">When a sinusoidal voltage is applied to a resistor, the voltage is exactly in phase with the current\u2014they have a 0\u00ba0\u00ba phase angle.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1169737784611\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<ul id=\"fs-id1169738048183\">\n<li id=\"import-auto-id1169737725415\">For inductors in AC circuits, we find that when a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a 90\u00ba90\u00ba phase angle.<\/li>\n<li id=\"import-auto-id1169736844107\">The opposition of an inductor to a change in current is expressed as a type of AC resistance.<\/li>\n<li id=\"import-auto-id1169736583869\">Ohm\u2019s law for an inductor is\n<div class=\"equation\">I=VXL,I=VXL, size 12{I= { {V} over {X rSub { size 8{L} } } } } {}<\/div>\n<p>where VV size 12{V} {} is the rms voltage across the inductor.<\/li>\n<li id=\"import-auto-id1169737882424\">XLXL size 12{X rSub { size 8{L} } } {} is defined to be the inductive reactance, given by\n<div class=\"equation\">XL=2\u03c0fL,XL=2\u03c0fL, size 12{X rSub { size 8{L} } =2\u03c0 ital &#8220;fL&#8221;} {}<\/div>\n<p>with ff size 12{f} {} the frequency of the AC voltage source in hertz.<\/li>\n<li id=\"import-auto-id1169738208811\">Inductive reactance XLXL size 12{X rSub { size 8{L} } } {} has units of ohms and is greatest at high frequencies.<\/li>\n<li id=\"import-auto-id1169736713656\">For capacitors, we find that when a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a 90\u00ba90\u00ba phase angle.<\/li>\n<li id=\"import-auto-id1169738012850\">Since a capacitor can stop current when fully charged, it limits current and offers another form of AC resistance; Ohm\u2019s law for a capacitor is\n<div class=\"equation\">I=VXC,I=VXC, size 12{I= { {V} over {X rSub { size 8{C} } } } } {}<\/div>\n<p>where VV size 12{V} {} is the rms voltage across the capacitor.<\/li>\n<li id=\"import-auto-id1169737821215\">XCXC size 12{X rSub { size 8{C} } } {} is defined to be the capacitive reactance, given by\n<div class=\"equation\">XC=12\u03c0fC.XC=12\u03c0fC. size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } } {}<\/div>\n<\/li>\n<li id=\"import-auto-id1169738016153\">XCXC size 12{X rSub { size 8{C} } } {} has units of ohms and is greatest at low frequencies.<\/li>\n<\/ul>\n<\/section>\n<section id=\"fs-id1169738011145\" class=\"conceptual-questions\">\n<h1>Conceptual Questions<\/h1>\n<div class=\"exercise\" id=\"fs-id1169738010807\">\n<div class=\"problem\" id=\"fs-id1169738250359\">\n<p id=\"import-auto-id1169737050603\">Presbycusis is a hearing loss due to age that progressively affects higher frequencies. A hearing aid amplifier is designed to amplify all frequencies equally. To adjust its output for presbycusis, would you put a capacitor in series or parallel with the hearing aid\u2019s speaker? Explain.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737854523\">\n<div class=\"problem\" id=\"fs-id1169738105398\">\n<p id=\"import-auto-id1169738054334\">Would you use a large inductance or a large capacitance in series with a system to filter out low frequencies, such as the 100 Hz hum in a sound system? Explain.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737930275\">\n<div class=\"problem\" id=\"fs-id1169737968751\">\n<p id=\"import-auto-id1169738211558\">High-frequency noise in AC power can damage computers. Does the plug-in unit designed to prevent this damage use a large inductance or a large capacitance (in series with the computer) to filter out such high frequencies? Explain.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"problem\" id=\"fs-id1169738116308\">\n<p id=\"import-auto-id1169738012609\">Does inductance depend on current, frequency, or both? What about inductive reactance?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737785016\">\n<div class=\"problem\" id=\"fs-id1169737861279\">\n<p id=\"import-auto-id1169738173969\">Explain why the capacitor in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(a) acts as a low-frequency filter between the two circuits, whereas that in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(b) acts as a high-frequency filter.<\/p>\n<\/div>\n<\/div>\n<figure id=\"import-auto-id1169738245387\"><figcaption>Capacitors and inductors. Capacitor with high frequency and low frequency.<\/figcaption><span id=\"import-auto-id1169737793629\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_11_04a.jpg\" alt=\"The figure describes two circuits with two different connections. The first part of the diagram shows circuit one and circuit two connected in series and a capacitor C is connected between them. Both the circuits are shown as grounded. The second part of the diagram shows two circuits circuit one and circuit two connected to each other. At the point of connection one end of the capacitor is connected and the other end of the capacitor is grounded. Both the circuits are shown as grounded.\" width=\"250\" \/><\/span><\/p>\n<\/figure>\n<div class=\"exercise\" id=\"fs-id1169736669333\">\n<div class=\"problem\" id=\"fs-id1169738050827\">\n<p id=\"import-auto-id1169737869873\">If the capacitors in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a> are replaced by inductors, which acts as a low-frequency filter and which as a high-frequency filter?<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1169737726120\" class=\"problems-exercises\">\n<h1>Problems &amp; Exercises<\/h1>\n<div class=\"exercise\" id=\"fs-id1169738117280\">\n<div class=\"problem\" id=\"fs-id1169737793629\">\n<p id=\"import-auto-id1169738200256\">At what frequency will a 30.0 mH inductor have a reactance of<br \/>\n100 \u03a9100 \u03a9?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737917875\">\n<p id=\"import-auto-id1169737728465\">531 Hz<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737786419\">\n<div class=\"problem\" id=\"fs-id1169736656664\">\n<p id=\"import-auto-id1169737812535\">What value of inductance should be used if a<br \/>\n20.0 k\u03a920.0 k\u03a9<br \/>\nreactance is needed at a frequency of 500 Hz?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738077769\">\n<div class=\"problem\" id=\"fs-id1169738072176\">\n<p id=\"import-auto-id1169736606594\">What capacitance should be used to produce a<br \/>\n2.00 M\u03a92.00 M\u03a9 reactance at 60.0 Hz?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737909282\">\n<p id=\"import-auto-id1169738045746\">1.33 nF<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738198876\">\n<div class=\"problem\" id=\"fs-id1169737757642\">\n<p id=\"import-auto-id1169738036322\">At what frequency will an 80.0 mF capacitor have a reactance of 0.250 \u03a90.250 \u03a9?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737729245\">\n<div class=\"problem\" id=\"fs-id1169737754951\">\n<p id=\"import-auto-id1169736825551\">(a) Find the current through a 0.500 H inductor connected to a 60.0 Hz, 480 V AC source. (b) What would the current be at 100 kHz?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738193476\">\n<p id=\"import-auto-id1169737772039\">(a) 2.55 A<\/p>\n<p id=\"import-auto-id1169737826940\">(b) 1.53 mA<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738086496\">\n<div class=\"problem\" id=\"fs-id1169737812738\">\n<p id=\"import-auto-id1169737825035\">(a) What current flows when a 60.0 Hz, 480 V AC source is connected to a<br \/>\n0.250 \u03bcF0.250 \u03bcF<br \/>\ncapacitor? (b) What would the current be at 25.0 kHz?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738198521\">\n<div class=\"problem\" id=\"fs-id1169737848534\">\n<p id=\"import-auto-id1169737972545\">A 20.0 kHz, 16.0 V source connected to an inductor produces a 2.00 A current. What is the inductance?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738077151\">\n<p>63.7 \u00b5H63.7 \u00b5H<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737994347\">\n<div class=\"problem\" id=\"fs-id1169737764442\">\n<p id=\"import-auto-id1169736658894\">A 20.0 Hz, 16.0 V source produces a 2.00 mA current when connected to a capacitor. What is the capacitance?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738250258\">\n<div class=\"problem\" id=\"fs-id1169737802646\">\n<p id=\"import-auto-id1169737734640\">(a) An inductor designed to filter high-frequency noise from power supplied to a personal computer is placed in series with the computer. What minimum inductance should it have to produce a<br \/>\n2.00 k\u03a92.00 k\u03a9 reactance for 15.0 kHz noise? (b) What is its reactance at 60.0 Hz?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736983958\">\n<p id=\"import-auto-id1169737814409\">(a) 21.2 mH<\/p>\n<p id=\"import-auto-id1169737854280\">(b) 8.00 \u03a98.00 \u03a9<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738162931\">\n<div class=\"problem\" id=\"fs-id1169736713656\">\n<p id=\"import-auto-id1169736583393\">The capacitor in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(a) is designed to filter low-frequency signals, impeding their transmission between circuits. (a) What capacitance is needed to produce a<br \/>\n100 k\u03a9100 k\u03a9 reactance at a frequency of 120 Hz? (b) What would its reactance be at 1.00 MHz? (c) Discuss the implications of your answers to (a) and (b).<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736611033\">\n<div class=\"problem\" id=\"fs-id1169736583899\">\n<p id=\"import-auto-id1169738116034\">The capacitor in <a href=\"#import-auto-id1169738245387\" class=\"autogenerated-content\">[link]<\/a>(b) will filter high-frequency signals by shorting them to earth\/ground. (a) What capacitance is needed to produce a reactance of<br \/>\n10.0 m\u03a910.0 m\u03a9 for a 5.00 kHz signal? (b) What would its reactance be at 3.00 Hz? (c) Discuss the implications of your answers to (a) and (b).<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738110286\">\n<p id=\"import-auto-id1169738138113\">(a) 3.18 mF<\/p>\n<p id=\"import-auto-id1169737965361\">(b) 16.7 \u03a916.7 \u03a9<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737802631\">\n<div class=\"problem\" id=\"fs-id1169736756883\">\n<p id=\"import-auto-id1169737819164\"><strong>Unreasonable Results<\/strong><\/p>\n<p>In a recording of voltages due to brain activity (an EEG), a 10.0 mV signal with a 0.500 Hz frequency is applied to a capacitor, producing a current of 100 mA. Resistance is negligible. (a) What is the capacitance? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736919851\">\n<div class=\"problem\" id=\"fs-id1169736620528\">\n<p id=\"import-auto-id1169738131438\"><strong>Construct Your Own Problem<\/strong><\/p>\n<p>Consider the use of an inductor in series with a computer operating on 60 Hz electricity. Construct a problem in which you calculate the relative reduction in voltage of incoming high frequency noise compared to 60 Hz voltage. Among the things to consider are the acceptable series reactance of the inductor for 60 Hz power and the likely frequencies of noise coming through the power lines.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1169738074090\" class=\"definition\">\n<dt>inductive reactance<\/dt>\n<dd>the opposition of an inductor to a change in current; calculated by XL=2\u03c0fLXL=2\u03c0fL size 12{X rSub { size 8{L} } =2\u03c0 ital &#8220;fL&#8221;} {}<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1169736611323\" class=\"definition\">\n<dt>capacitive reactance<\/dt>\n<dd id=\"fs-id1169737724984\">the opposition of a capacitor to a change in current; calculated by XC=12\u03c0fCXC=12\u03c0fC size 12{X rSub { size 8{C} } = { {1} over {2\u03c0 ital &#8220;fC&#8221;} } } {}<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":103,"menu_order":12,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1192","chapter","type-chapter","status-publish","hentry"],"part":1129,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1192","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\/1192\/revisions"}],"predecessor-version":[{"id":1944,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1192\/revisions\/1944"}],"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\/1192\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/media?parent=1192"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapter-type?post=1192"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/contributor?post=1192"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/license?post=1192"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}