{"id":1187,"date":"2017-04-10T15:10:10","date_gmt":"2017-04-10T19:10:10","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/?post_type=chapter&#038;p=1187"},"modified":"2017-04-10T15:10:10","modified_gmt":"2017-04-10T19:10:10","slug":"23-10-rl-circuits","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/chapter\/23-10-rl-circuits\/","title":{"raw":"23.10 RL Circuits","rendered":"23.10 RL Circuits"},"content":{"raw":"<div>RL Circuits<\/div>\n<div>\n<ul><li>Calculate the current in an RL circuit after a specified number of characteristic time steps.<\/li>\n \t<li>Calculate the characteristic time of an RL circuit.<\/li>\n \t<li>Sketch the current in an RL circuit over time.<\/li>\n<\/ul><\/div>\n<p id=\"import-auto-id1169738093650\">We know that the current through an inductor LL size 12{L} {} cannot be turned on or off instantaneously. The change in current changes flux, inducing an emf opposing the change (Lenz\u2019s law). How long does the opposition last? Current <em><em>will<\/em><\/em> flow and <em><em>can<\/em><\/em> be turned off, but how long does it take? <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a> shows a switching circuit that can be used to examine current through an inductor as a function of time.<\/p>\n\n<figure id=\"import-auto-id1169737969272\"><figcaption>(a) An <em>RL<\/em> circuit with a switch to turn current on and off. When in position 1, the battery, resistor, and inductor are in series and a current is established. In position 2, the battery is removed and the current eventually stops because of energy loss in the resistor. (b) A graph of current growth versus time when the switch is moved to position 1. (c) A graph of current decay when the switch is moved to position 2.<\/figcaption><span id=\"import-auto-id1169737758284\">\n<img src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_10_01.jpg\" alt=\"Part a of the figure shows an inductor connected in series with a resistor. The arrangement is connected across a cell by an on and off switch with two positions. When in position one, the battery, resistor, and inductor are in series and a current is established. In position two, the battery is removed and the current stops eventually because of energy loss in the resistor. Part b of the diagram shows the graph when the switch is in position one. It shows a graph for current growth verses time. The current is along the Y axis and the time is along the X axis. The graph shows a smooth rise from origin to a maximum value I zero corresponding to Y axis and value four tau on X axis. Part c of the diagram shows the graph when the switch is in position two. It shows a graph for current decay verses time is shown. The current is along the Y axis and the time is along the X axis. The graph is decreasing curve from a value I zero on Y axis, touching the X axis at a point where value of time equals four tau.\" width=\"550\"\/><\/span>\n\n<\/figure><p id=\"import-auto-id1169737757267\">When the switch is first moved to position 1 (at <em>t=0t=0 size 12{t=0} {}<\/em>), the current is zero and it eventually rises to I0=V\/RI0=V\/R size 12{I rSub { size 8{0} } = ital \"V\/R\"} {}, where RR is the total resistance of the circuit. The opposition of the inductor LL size 12{L} {} is greatest at the beginning, because the amount of change is greatest. The opposition it poses is in the form of an induced emf, which decreases to zero as the current approaches its final value. The opposing emf is proportional to the amount of change left. This is the hallmark of an exponential behavior, and it can be shown with calculus that<\/p>\n\n<div class=\"equation\" id=\"eip-314\">I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on),I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on), size 12{I=I rSub { size 8{0} } ( 1 - e rSup { size 8{ - t\/\u03c4} } ) } {}<\/div>\n<p id=\"import-auto-id1169737811419\">is the current in an <em>RL<\/em> circuit when switched on (Note the similarity to the exponential behavior of the voltage on a charging capacitor). The initial current is zero and approaches I0=V\/RI0=V\/R size 12{I rSub { size 8{0} } = ital \"V\/R\"} {} with a <span id=\"import-auto-id1169737897754\">characteristic time constant<\/span>\n\u03c4\u03c4\nfor an <em>RL<\/em> circuit, given by<\/p>\n\n<div class=\"equation\">\u03c4=LR,\u03c4=LR, size 12{\u03c4= { {L} over {R} } } {}<\/div>\n<p id=\"import-auto-id1169737803195\">where \u03c4\u03c4 size 12{\u03c4} {} has units of seconds, since\n1H=1\u03a9\u00b7s1H=1\u03a9\u00b7s.\nIn the first period of time \u03c4\u03c4 size 12{\u03c4} {}, the current rises from zero to 0.632I00.632I0 size 12{0 \".\" \"632\"I rSub { size 8{0} } } {}, since I=I0(1\u2212e\u22121)=I0(1\u22120.368)=0.632I0I=I0(1\u2212e\u22121)=I0(1\u22120.368)=0.632I0 size 12{I=I rSub { size 8{0} } ( 1 - e rSup { size 8{ - 1} } ) =I rSub { size 8{0} } ( 1 - 0 \".\" \"368\" ) =0 \".\" \"632\"I rSub { size 8{0} } } {}. The current will go 0.632 of the remainder in the next time \u03c4\u03c4 size 12{\u03c4} {}. A well-known property of the exponential is that the final value is never exactly reached, but 0.632 of the remainder to that value is achieved in every characteristic time \u03c4\u03c4 size 12{\u03c4} {}. In just a few multiples of the time \u03c4\u03c4 size 12{\u03c4} {}, the final value is very nearly achieved, as the graph in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(b) illustrates.<\/p>\n<p id=\"import-auto-id1169735754248\">The characteristic time \u03c4\u03c4 size 12{\u03c4} {} depends on only two factors, the inductance LL size 12{L} {} and the resistance RR size 12{R} {}. The greater the inductance LL size 12{L} {}, the greater \u03c4\u03c4 size 12{\u03c4} {} is, which makes sense since a large inductance is very effective in opposing change. The smaller the resistance RR size 12{R} {}, the greater \u03c4\u03c4 size 12{\u03c4} {} is. Again this makes sense, since a small resistance means a large final current and a greater change to get there. In both cases\u2014large LL size 12{L} {} and small RR size 12{R} {} \u2014more energy is stored in the inductor and more time is required to get it in and out.<\/p>\n<p id=\"import-auto-id1169737882691\">When the switch in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(a) is moved to position 2 and cuts the battery out of the circuit, the current drops because of energy dissipation by the resistor. But this is also not instantaneous, since the inductor opposes the decrease in current by inducing an emf in the same direction as the battery that drove the current. Furthermore, there is a certain amount of energy, (1\/2)LI02(1\/2)LI02 size 12{ ( \"1\/2\" ) ital \"LI\" rSub { size 8{0} } rSup { size 8{2} } } {}, stored in the inductor, and it is dissipated at a finite rate. As the current approaches zero, the rate of decrease slows, since the energy dissipation rate is I2RI2R size 12{ I rSup { size 8{2} } R} {}. Once again the behavior is exponential, and\nII\nis found to be<\/p>\n\n<div class=\"equation\">I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off).I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off). size 12{I=I rSub { size 8{0} } e rSup { size 8{ - t\/\u03c4} } } {}<\/div>\n<p id=\"import-auto-id1169737711668\">(See <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(c).) In the first period of time \u03c4=L\/R\u03c4=L\/R size 12{\u03c4=L\/R} {} after the switch is closed, the current falls to 0.368 of its initial value, since I=I0e\u22121=0.368I0I=I0e\u22121=0.368I0 size 12{I=I rSub { size 8{0} } e rSup { size 8{ - 1} } =0 \".\" \"368\"I rSub { size 8{0} } } {}. In each successive time \u03c4\u03c4 size 12{\u03c4} {}, the current falls to 0.368 of the preceding value, and in a few multiples of \u03c4\u03c4 size 12{\u03c4} {}, the current becomes very close to zero, as seen in the graph in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(c).<\/p>\n\n<div class=\"example\" id=\"fs-id1169738136126\">\n<div class=\"title\">Calculating Characteristic Time and Current in an <em>RL<\/em> Circuit<\/div>\n<p id=\"import-auto-id1169737045564\">(a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 \u03a93.00 \u03a9 resistor? (b) Find the current 5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A.<\/p>\n<p id=\"import-auto-id1169737804462\"><strong>Strategy for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169737787249\">The time constant for an <em>RL<\/em> circuit is defined by \u03c4=L\/R\u03c4=L\/R size 12{\u03c4=L\/R} {}.<\/p>\n<p id=\"import-auto-id1169737713276\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738036454\">Entering known values into the expression for \u03c4\u03c4 size 12{\u03c4} {} given in \u03c4=L\/R\u03c4=L\/R size 12{\u03c4=L\/R} {} yields<\/p>\n\n<div class=\"equation\">\u03c4=LR=7.50 mH3.00\u03a9=2.50 ms.\u03c4=LR=7.50 mH3.00\u03a9=2.50 ms. size 12{\u03c4= { {L} over {R} } = { {7 \".\" \"50\"\" mH\"} over {3 \".\" \"00 \" %OMEGA } } =2 \".\" \"50\"\" ms\"} {}<\/div>\n<p id=\"import-auto-id1169738164481\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169737723042\">This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms.<\/p>\n<p id=\"import-auto-id1169736758585\"><strong>Strategy for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169738010703\">We can find the current by using I=I0e\u2212t\/\u03c4I=I0e\u2212t\/\u03c4 size 12{I=I rSub { size 8{0} } e rSup { size 8{ - t\/\u03c4} } } {}, or by considering the decline in steps. Since the time is twice the characteristic time, we consider the process in steps.<\/p>\n<p id=\"import-auto-id1169736614228\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169738189000\">In the first 2.50 ms, the current declines to 0.368 of its initial value, which is<\/p>\n\n<div class=\"equation\">I=0.368I0=(0.368)(10.0 A)=3.68 A\u00a0at\u00a0t=2.50\u00a0ms.I=0.368I0=(0.368)(10.0 A)=3.68 A\u00a0at\u00a0t=2.50\u00a0ms.<\/div>\n<p id=\"import-auto-id1169738055826\">After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,<\/p>\n\n<div class=\"equation\">I\u2032=0.368I=(0.368)(3.68 A)=1.35\u00a0A\u00a0at\u00a0t=5.00\u00a0ms.I\u2032=0.368I=(0.368)(3.68 A)=1.35\u00a0A\u00a0at\u00a0t=5.00\u00a0ms.alignl { stack {\nsize 12{ { {I}} sup { ' }=0 \".\" \"368\"I= ( 0 \".\" \"368\" ) ( 3 \".\" \"68\"\" A\" ) } {} #\nsize 12{\" \"=1 \".\" \"35\"\" A at \"t=5 \".\" \"00\"\" ms\"} {}\n} } {}<\/div>\n<p id=\"import-auto-id1169737972496\"><strong>Discussion for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737896000\">After another 5.00 ms has passed, the current will be 0.183 A (see <a href=\"#fs-id1169738227350\" class=\"autogenerated-content\">[link]<\/a>); so, although it does die out, the current certainly does not go to zero instantaneously.<\/p>\n\n<\/div>\n<p id=\"import-auto-id1169737717434\">In summary, when the voltage applied to an inductor is changed, the current also changes, <em><em>but the change in current lags the change in voltage in an RL circuit<\/em><\/em>. In <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4\">Reactance, Inductive and Capacitive<\/a>, we explore how an <em>RL<\/em> circuit behaves when a sinusoidal AC voltage is applied.<\/p>\n\n<section id=\"fs-id1169737941056\" class=\"section-summary\"><h1>Section Summary<\/h1>\n<ul id=\"fs-id1169736611192\"><li id=\"import-auto-id1169737712209\">When a series connection of a resistor and an inductor\u2014an <em>RL<\/em> circuit\u2014is connected to a voltage source, the time variation of the current is\n<div class=\"equation\">I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on).I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on). size 12{I=I rSub { size 8{0} } ( 1 - e rSup { size 8{ - t\/\u03c4} } ) } {}<\/div>\nwhere I0=V\/RI0=V\/R size 12{I rSub { size 8{0} } =V\/R} {} is the final current.<\/li>\n \t<li id=\"import-auto-id1169737936637\">The characteristic time constant \u03c4\u03c4 size 12{\u03c4} {} is \u03c4=LR\u03c4=LR size 12{\u03c4= { {L} over {R} } } {} , where LL is the inductance and RR is the resistance.<\/li>\n \t<li id=\"import-auto-id1169738035430\">In the first time constant \u03c4\u03c4 size 12{\u03c4} {}, the current rises from zero to 0.632I00.632I0 size 12{0 \".\" \"632\"I rSub { size 8{0} } } {}, and 0.632 of the remainder in every subsequent time interval \u03c4\u03c4 size 12{\u03c4} {}.<\/li>\n \t<li id=\"import-auto-id1169738048163\">When the inductor is shorted through a resistor, current decreases as\n<div class=\"equation\">I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off).I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off). size 12{I=I rSub { size 8{0} } e rSup { size 8{ - t\/\u03c4} } } {}<\/div>\nHere I0I0 size 12{I rSub { size 8{0} } } {} is the initial current.<\/li>\n \t<li id=\"import-auto-id1169737060067\">Current falls to 0.368I00.368I0 size 12{0 \".\" \"368\"I rSub { size 8{0} } } {} in the first time interval \u03c4\u03c4 size 12{\u03c4} {}, and 0.368 of the remainder toward zero in each subsequent time \u03c4\u03c4 size 12{\u03c4} {}.<\/li>\n<\/ul><\/section><section id=\"fs-id1169738010703\" class=\"problems-exercises\"><h1>Problem Exercises<\/h1>\n<div class=\"exercise\" id=\"fs-id1169738227350\">\n<div class=\"problem\" id=\"fs-id1169738245588\">\n<p id=\"import-auto-id1169738210359\">If you want a characteristic <em>RL<\/em> time constant of 1.00 s, and you have a 500 \u03a9500 \u03a9 resistor, what value of self-inductance is needed?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737933054\">\n<p id=\"import-auto-id1169738164414\">500 H<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737723038\">\n<div class=\"problem\" id=\"fs-id1169736814918\">\n<p id=\"import-auto-id1169738117421\">Your <em>RL<\/em> circuit has a characteristic time constant of 20.0 ns, and a resistance of 5.00 M\u03a95.00 M\u03a9. (a) What is the inductance of the circuit? (b) What resistance would give you a 1.00 ns time constant, perhaps needed for quick response in an oscilloscope?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737788008\">\n<div class=\"problem\" id=\"fs-id1169736620644\">\n\nA large superconducting magnet, used for magnetic resonance imaging, has a 50.0 H inductance. If you want current through it to be adjustable with a 1.00 s characteristic time constant, what is the minimum resistance of system?\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738094791\">\n<p id=\"import-auto-id1169736659279\">50.0 \u03a950.0 \u03a9<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736581925\">\n<div class=\"problem\" id=\"fs-id1169736620836\">\n<p id=\"import-auto-id1169738110369\">Verify that after a time of 10.0 ms, the current for the situation considered in <a href=\"#fs-id1169738136126\" class=\"autogenerated-content\">[link]<\/a> will be 0.183 A as stated.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736596063\">\n<div class=\"problem\" id=\"fs-id1169736590714\">\n<p id=\"import-auto-id1169737770657\">Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and resistors ranging from\n0.100 \u03a90.100 \u03a9 to\n1.00 M\u03a91.00 M\u03a9. What is the range of characteristic <em>RL<\/em> time constants you can produce by connecting a single resistor to a single inductor?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738045475\">\n\n1.00\u00d710\u201318 s1.00\u00d710\u201318 s size 12{1 \".\" \"00\" times \"10\" rSup { size 8{\"-15\"} } \" s\"} {} to 0.100 s\n\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"problem\" id=\"fs-id1169737812307\">\n<p id=\"import-auto-id1169738113687\">(a) What is the characteristic time constant of a 25.0 mH inductor that has a resistance of\n4.00 \u03a94.00 \u03a9? (b) If it is connected to a 12.0 V battery, what is the current after 12.5 ms?<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737723042\">\n<div class=\"problem\" id=\"fs-id1169737785691\">\n<p id=\"import-auto-id1169738075685\">What percentage of the final current\nI0I0\nflows through an inductor LL size 12{L} {} in series with a resistor RR size 12{R} {}, three time constants after the circuit is completed?<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736758585\">\n<p id=\"import-auto-id1169738251076\">95.0%<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738248980\">\n<div class=\"problem\" id=\"fs-id1169738223614\">\n<p id=\"import-auto-id1169737898346\">The 5.00 A current through a 1.50 H inductor is dissipated by a 2.00 \u03a92.00 \u03a9 resistor in a circuit like that in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a> with the switch in position 2. (a) What is the initial energy in the inductor? (b) How long will it take the current to decline to 5.00% of its initial value? (c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737787416\">\n<div class=\"problem\" id=\"fs-id1169736624109\">\n<p id=\"import-auto-id1169738250359\">(a) Use the exact exponential treatment to find how much time is required to bring the current through an 80.0 mH inductor in series with a\n15.0 \u03a915.0 \u03a9\nresistor to 99.0% of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of \u03c4\u03c4 size 12{\u03c4} {}. (c) Discuss how significant the difference is.<\/p>\n\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737768546\">\n<p id=\"import-auto-id1169738036121\">(a) 24.6 ms<\/p>\n<p id=\"import-auto-id1169737895428\">(b) 26.7 ms<\/p>\n<p id=\"import-auto-id1169738051862\">(c) 9% difference, which is greater than the inherent uncertainty in the given parameters.<\/p>\n\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737851564\">\n<div class=\"problem\" id=\"fs-id1169738011205\">\n<p id=\"import-auto-id1169737961873\">(a) Using the exact exponential treatment, find the time required for the current through a 2.00 H inductor in series with a\n0.500 \u03a90.500 \u03a9\nresistor to be reduced to 0.100% of its original value. (b) Compare your answer to the approximate treatment using integral numbers of \u03c4\u03c4 size 12{\u03c4} {}. (c) Discuss how significant the difference is.<\/p>\n\n<\/div>\n<\/div>\n<\/section><div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1169738071203\" class=\"definition\"><dt>characteristic time constant<\/dt>\n \t<dd id=\"fs-id1169736581702\">denoted by \u03c4\u03c4 size 12{\u03c4} {}, of a particular series <em>RL<\/em> circuit is calculated by \u03c4=LR\u03c4=LR size 12{\u03c4= { {L} over {R} } } {}, where LL size 12{L} {} is the inductance and RR is the resistance<\/dd>\n<\/dl><\/div>","rendered":"<div>RL Circuits<\/div>\n<div>\n<ul>\n<li>Calculate the current in an RL circuit after a specified number of characteristic time steps.<\/li>\n<li>Calculate the characteristic time of an RL circuit.<\/li>\n<li>Sketch the current in an RL circuit over time.<\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id1169738093650\">We know that the current through an inductor LL size 12{L} {} cannot be turned on or off instantaneously. The change in current changes flux, inducing an emf opposing the change (Lenz\u2019s law). How long does the opposition last? Current <em><em>will<\/em><\/em> flow and <em><em>can<\/em><\/em> be turned off, but how long does it take? <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a> shows a switching circuit that can be used to examine current through an inductor as a function of time.<\/p>\n<figure id=\"import-auto-id1169737969272\"><figcaption>(a) An <em>RL<\/em> circuit with a switch to turn current on and off. When in position 1, the battery, resistor, and inductor are in series and a current is established. In position 2, the battery is removed and the current eventually stops because of energy loss in the resistor. (b) A graph of current growth versus time when the switch is moved to position 1. (c) A graph of current decay when the switch is moved to position 2.<\/figcaption><span id=\"import-auto-id1169737758284\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-content\/uploads\/sites\/113\/2017\/02\/Figure_24_10_01.jpg\" alt=\"Part a of the figure shows an inductor connected in series with a resistor. The arrangement is connected across a cell by an on and off switch with two positions. When in position one, the battery, resistor, and inductor are in series and a current is established. In position two, the battery is removed and the current stops eventually because of energy loss in the resistor. Part b of the diagram shows the graph when the switch is in position one. It shows a graph for current growth verses time. The current is along the Y axis and the time is along the X axis. The graph shows a smooth rise from origin to a maximum value I zero corresponding to Y axis and value four tau on X axis. Part c of the diagram shows the graph when the switch is in position two. It shows a graph for current decay verses time is shown. The current is along the Y axis and the time is along the X axis. The graph is decreasing curve from a value I zero on Y axis, touching the X axis at a point where value of time equals four tau.\" width=\"550\" \/><\/span><\/p>\n<\/figure>\n<p id=\"import-auto-id1169737757267\">When the switch is first moved to position 1 (at <em>t=0t=0 size 12{t=0} {}<\/em>), the current is zero and it eventually rises to I0=V\/RI0=V\/R size 12{I rSub { size 8{0} } = ital &#8220;V\/R&#8221;} {}, where RR is the total resistance of the circuit. The opposition of the inductor LL size 12{L} {} is greatest at the beginning, because the amount of change is greatest. The opposition it poses is in the form of an induced emf, which decreases to zero as the current approaches its final value. The opposing emf is proportional to the amount of change left. This is the hallmark of an exponential behavior, and it can be shown with calculus that<\/p>\n<div class=\"equation\" id=\"eip-314\">I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on),I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on), size 12{I=I rSub { size 8{0} } ( 1 &#8211; e rSup { size 8{ &#8211; t\/\u03c4} } ) } {}<\/div>\n<p id=\"import-auto-id1169737811419\">is the current in an <em>RL<\/em> circuit when switched on (Note the similarity to the exponential behavior of the voltage on a charging capacitor). The initial current is zero and approaches I0=V\/RI0=V\/R size 12{I rSub { size 8{0} } = ital &#8220;V\/R&#8221;} {} with a <span id=\"import-auto-id1169737897754\">characteristic time constant<\/span><br \/>\n\u03c4\u03c4<br \/>\nfor an <em>RL<\/em> circuit, given by<\/p>\n<div class=\"equation\">\u03c4=LR,\u03c4=LR, size 12{\u03c4= { {L} over {R} } } {}<\/div>\n<p id=\"import-auto-id1169737803195\">where \u03c4\u03c4 size 12{\u03c4} {} has units of seconds, since<br \/>\n1H=1\u03a9\u00b7s1H=1\u03a9\u00b7s.<br \/>\nIn the first period of time \u03c4\u03c4 size 12{\u03c4} {}, the current rises from zero to 0.632I00.632I0 size 12{0 &#8220;.&#8221; &#8220;632&#8221;I rSub { size 8{0} } } {}, since I=I0(1\u2212e\u22121)=I0(1\u22120.368)=0.632I0I=I0(1\u2212e\u22121)=I0(1\u22120.368)=0.632I0 size 12{I=I rSub { size 8{0} } ( 1 &#8211; e rSup { size 8{ &#8211; 1} } ) =I rSub { size 8{0} } ( 1 &#8211; 0 &#8220;.&#8221; &#8220;368&#8221; ) =0 &#8220;.&#8221; &#8220;632&#8221;I rSub { size 8{0} } } {}. The current will go 0.632 of the remainder in the next time \u03c4\u03c4 size 12{\u03c4} {}. A well-known property of the exponential is that the final value is never exactly reached, but 0.632 of the remainder to that value is achieved in every characteristic time \u03c4\u03c4 size 12{\u03c4} {}. In just a few multiples of the time \u03c4\u03c4 size 12{\u03c4} {}, the final value is very nearly achieved, as the graph in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(b) illustrates.<\/p>\n<p id=\"import-auto-id1169735754248\">The characteristic time \u03c4\u03c4 size 12{\u03c4} {} depends on only two factors, the inductance LL size 12{L} {} and the resistance RR size 12{R} {}. The greater the inductance LL size 12{L} {}, the greater \u03c4\u03c4 size 12{\u03c4} {} is, which makes sense since a large inductance is very effective in opposing change. The smaller the resistance RR size 12{R} {}, the greater \u03c4\u03c4 size 12{\u03c4} {} is. Again this makes sense, since a small resistance means a large final current and a greater change to get there. In both cases\u2014large LL size 12{L} {} and small RR size 12{R} {} \u2014more energy is stored in the inductor and more time is required to get it in and out.<\/p>\n<p id=\"import-auto-id1169737882691\">When the switch in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(a) is moved to position 2 and cuts the battery out of the circuit, the current drops because of energy dissipation by the resistor. But this is also not instantaneous, since the inductor opposes the decrease in current by inducing an emf in the same direction as the battery that drove the current. Furthermore, there is a certain amount of energy, (1\/2)LI02(1\/2)LI02 size 12{ ( &#8220;1\/2&#8221; ) ital &#8220;LI&#8221; rSub { size 8{0} } rSup { size 8{2} } } {}, stored in the inductor, and it is dissipated at a finite rate. As the current approaches zero, the rate of decrease slows, since the energy dissipation rate is I2RI2R size 12{ I rSup { size 8{2} } R} {}. Once again the behavior is exponential, and<br \/>\nII<br \/>\nis found to be<\/p>\n<div class=\"equation\">I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off).I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off). size 12{I=I rSub { size 8{0} } e rSup { size 8{ &#8211; t\/\u03c4} } } {}<\/div>\n<p id=\"import-auto-id1169737711668\">(See <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(c).) In the first period of time \u03c4=L\/R\u03c4=L\/R size 12{\u03c4=L\/R} {} after the switch is closed, the current falls to 0.368 of its initial value, since I=I0e\u22121=0.368I0I=I0e\u22121=0.368I0 size 12{I=I rSub { size 8{0} } e rSup { size 8{ &#8211; 1} } =0 &#8220;.&#8221; &#8220;368&#8221;I rSub { size 8{0} } } {}. In each successive time \u03c4\u03c4 size 12{\u03c4} {}, the current falls to 0.368 of the preceding value, and in a few multiples of \u03c4\u03c4 size 12{\u03c4} {}, the current becomes very close to zero, as seen in the graph in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a>(c).<\/p>\n<div class=\"example\" id=\"fs-id1169738136126\">\n<div class=\"title\">Calculating Characteristic Time and Current in an <em>RL<\/em> Circuit<\/div>\n<p id=\"import-auto-id1169737045564\">(a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 \u03a93.00 \u03a9 resistor? (b) Find the current 5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A.<\/p>\n<p id=\"import-auto-id1169737804462\"><strong>Strategy for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169737787249\">The time constant for an <em>RL<\/em> circuit is defined by \u03c4=L\/R\u03c4=L\/R size 12{\u03c4=L\/R} {}.<\/p>\n<p id=\"import-auto-id1169737713276\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169738036454\">Entering known values into the expression for \u03c4\u03c4 size 12{\u03c4} {} given in \u03c4=L\/R\u03c4=L\/R size 12{\u03c4=L\/R} {} yields<\/p>\n<div class=\"equation\">\u03c4=LR=7.50 mH3.00\u03a9=2.50 ms.\u03c4=LR=7.50 mH3.00\u03a9=2.50 ms. size 12{\u03c4= { {L} over {R} } = { {7 &#8220;.&#8221; &#8220;50&#8221;&#8221; mH&#8221;} over {3 &#8220;.&#8221; &#8220;00 &#8221; %OMEGA } } =2 &#8220;.&#8221; &#8220;50&#8221;&#8221; ms&#8221;} {}<\/div>\n<p id=\"import-auto-id1169738164481\"><strong>Discussion for (a)<\/strong><\/p>\n<p id=\"import-auto-id1169737723042\">This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms.<\/p>\n<p id=\"import-auto-id1169736758585\"><strong>Strategy for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169738010703\">We can find the current by using I=I0e\u2212t\/\u03c4I=I0e\u2212t\/\u03c4 size 12{I=I rSub { size 8{0} } e rSup { size 8{ &#8211; t\/\u03c4} } } {}, or by considering the decline in steps. Since the time is twice the characteristic time, we consider the process in steps.<\/p>\n<p id=\"import-auto-id1169736614228\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169738189000\">In the first 2.50 ms, the current declines to 0.368 of its initial value, which is<\/p>\n<div class=\"equation\">I=0.368I0=(0.368)(10.0 A)=3.68 A\u00a0at\u00a0t=2.50\u00a0ms.I=0.368I0=(0.368)(10.0 A)=3.68 A\u00a0at\u00a0t=2.50\u00a0ms.<\/div>\n<p id=\"import-auto-id1169738055826\">After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,<\/p>\n<div class=\"equation\">I\u2032=0.368I=(0.368)(3.68 A)=1.35\u00a0A\u00a0at\u00a0t=5.00\u00a0ms.I\u2032=0.368I=(0.368)(3.68 A)=1.35\u00a0A\u00a0at\u00a0t=5.00\u00a0ms.alignl { stack {<br \/>\nsize 12{ { {I}} sup { &#8216; }=0 &#8220;.&#8221; &#8220;368&#8221;I= ( 0 &#8220;.&#8221; &#8220;368&#8221; ) ( 3 &#8220;.&#8221; &#8220;68&#8221;&#8221; A&#8221; ) } {} #<br \/>\nsize 12{&#8221; &#8220;=1 &#8220;.&#8221; &#8220;35&#8221;&#8221; A at &#8220;t=5 &#8220;.&#8221; &#8220;00&#8221;&#8221; ms&#8221;} {}<br \/>\n} } {}<\/div>\n<p id=\"import-auto-id1169737972496\"><strong>Discussion for (b)<\/strong><\/p>\n<p id=\"import-auto-id1169737896000\">After another 5.00 ms has passed, the current will be 0.183 A (see <a href=\"#fs-id1169738227350\" class=\"autogenerated-content\">[link]<\/a>); so, although it does die out, the current certainly does not go to zero instantaneously.<\/p>\n<\/div>\n<p id=\"import-auto-id1169737717434\">In summary, when the voltage applied to an inductor is changed, the current also changes, <em><em>but the change in current lags the change in voltage in an RL circuit<\/em><\/em>. In <a href=\"\/contents\/663b7dc1-df6f-4a85-bae8-10d20a358d01@4\">Reactance, Inductive and Capacitive<\/a>, we explore how an <em>RL<\/em> circuit behaves when a sinusoidal AC voltage is applied.<\/p>\n<section id=\"fs-id1169737941056\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<ul id=\"fs-id1169736611192\">\n<li id=\"import-auto-id1169737712209\">When a series connection of a resistor and an inductor\u2014an <em>RL<\/em> circuit\u2014is connected to a voltage source, the time variation of the current is\n<div class=\"equation\">I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on).I=I0(1\u2212e\u2212t\/\u03c4)\u00a0\u00a0\u00a0\u00a0(turning on). size 12{I=I rSub { size 8{0} } ( 1 &#8211; e rSup { size 8{ &#8211; t\/\u03c4} } ) } {}<\/div>\n<p>where I0=V\/RI0=V\/R size 12{I rSub { size 8{0} } =V\/R} {} is the final current.<\/li>\n<li id=\"import-auto-id1169737936637\">The characteristic time constant \u03c4\u03c4 size 12{\u03c4} {} is \u03c4=LR\u03c4=LR size 12{\u03c4= { {L} over {R} } } {} , where LL is the inductance and RR is the resistance.<\/li>\n<li id=\"import-auto-id1169738035430\">In the first time constant \u03c4\u03c4 size 12{\u03c4} {}, the current rises from zero to 0.632I00.632I0 size 12{0 &#8220;.&#8221; &#8220;632&#8221;I rSub { size 8{0} } } {}, and 0.632 of the remainder in every subsequent time interval \u03c4\u03c4 size 12{\u03c4} {}.<\/li>\n<li id=\"import-auto-id1169738048163\">When the inductor is shorted through a resistor, current decreases as\n<div class=\"equation\">I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off).I=I0e\u2212t\/\u03c4\u00a0\u00a0\u00a0\u00a0(turning off). size 12{I=I rSub { size 8{0} } e rSup { size 8{ &#8211; t\/\u03c4} } } {}<\/div>\n<p>Here I0I0 size 12{I rSub { size 8{0} } } {} is the initial current.<\/li>\n<li id=\"import-auto-id1169737060067\">Current falls to 0.368I00.368I0 size 12{0 &#8220;.&#8221; &#8220;368&#8221;I rSub { size 8{0} } } {} in the first time interval \u03c4\u03c4 size 12{\u03c4} {}, and 0.368 of the remainder toward zero in each subsequent time \u03c4\u03c4 size 12{\u03c4} {}.<\/li>\n<\/ul>\n<\/section>\n<section id=\"fs-id1169738010703\" class=\"problems-exercises\">\n<h1>Problem Exercises<\/h1>\n<div class=\"exercise\" id=\"fs-id1169738227350\">\n<div class=\"problem\" id=\"fs-id1169738245588\">\n<p id=\"import-auto-id1169738210359\">If you want a characteristic <em>RL<\/em> time constant of 1.00 s, and you have a 500 \u03a9500 \u03a9 resistor, what value of self-inductance is needed?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737933054\">\n<p id=\"import-auto-id1169738164414\">500 H<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737723038\">\n<div class=\"problem\" id=\"fs-id1169736814918\">\n<p id=\"import-auto-id1169738117421\">Your <em>RL<\/em> circuit has a characteristic time constant of 20.0 ns, and a resistance of 5.00 M\u03a95.00 M\u03a9. (a) What is the inductance of the circuit? (b) What resistance would give you a 1.00 ns time constant, perhaps needed for quick response in an oscilloscope?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737788008\">\n<div class=\"problem\" id=\"fs-id1169736620644\">\n<p>A large superconducting magnet, used for magnetic resonance imaging, has a 50.0 H inductance. If you want current through it to be adjustable with a 1.00 s characteristic time constant, what is the minimum resistance of system?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738094791\">\n<p id=\"import-auto-id1169736659279\">50.0 \u03a950.0 \u03a9<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736581925\">\n<div class=\"problem\" id=\"fs-id1169736620836\">\n<p id=\"import-auto-id1169738110369\">Verify that after a time of 10.0 ms, the current for the situation considered in <a href=\"#fs-id1169738136126\" class=\"autogenerated-content\">[link]<\/a> will be 0.183 A as stated.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169736596063\">\n<div class=\"problem\" id=\"fs-id1169736590714\">\n<p id=\"import-auto-id1169737770657\">Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and resistors ranging from<br \/>\n0.100 \u03a90.100 \u03a9 to<br \/>\n1.00 M\u03a91.00 M\u03a9. What is the range of characteristic <em>RL<\/em> time constants you can produce by connecting a single resistor to a single inductor?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169738045475\">\n<p>1.00\u00d710\u201318 s1.00\u00d710\u201318 s size 12{1 &#8220;.&#8221; &#8220;00&#8221; times &#8220;10&#8221; rSup { size 8{&#8220;-15&#8243;} } &#8221; s&#8221;} {} to 0.100 s<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"problem\" id=\"fs-id1169737812307\">\n<p id=\"import-auto-id1169738113687\">(a) What is the characteristic time constant of a 25.0 mH inductor that has a resistance of<br \/>\n4.00 \u03a94.00 \u03a9? (b) If it is connected to a 12.0 V battery, what is the current after 12.5 ms?<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737723042\">\n<div class=\"problem\" id=\"fs-id1169737785691\">\n<p id=\"import-auto-id1169738075685\">What percentage of the final current<br \/>\nI0I0<br \/>\nflows through an inductor LL size 12{L} {} in series with a resistor RR size 12{R} {}, three time constants after the circuit is completed?<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169736758585\">\n<p id=\"import-auto-id1169738251076\">95.0%<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169738248980\">\n<div class=\"problem\" id=\"fs-id1169738223614\">\n<p id=\"import-auto-id1169737898346\">The 5.00 A current through a 1.50 H inductor is dissipated by a 2.00 \u03a92.00 \u03a9 resistor in a circuit like that in <a href=\"#import-auto-id1169737969272\" class=\"autogenerated-content\">[link]<\/a> with the switch in position 2. (a) What is the initial energy in the inductor? (b) How long will it take the current to decline to 5.00% of its initial value? (c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737787416\">\n<div class=\"problem\" id=\"fs-id1169736624109\">\n<p id=\"import-auto-id1169738250359\">(a) Use the exact exponential treatment to find how much time is required to bring the current through an 80.0 mH inductor in series with a<br \/>\n15.0 \u03a915.0 \u03a9<br \/>\nresistor to 99.0% of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of \u03c4\u03c4 size 12{\u03c4} {}. (c) Discuss how significant the difference is.<\/p>\n<\/div>\n<div class=\"solution\" id=\"fs-id1169737768546\">\n<p id=\"import-auto-id1169738036121\">(a) 24.6 ms<\/p>\n<p id=\"import-auto-id1169737895428\">(b) 26.7 ms<\/p>\n<p id=\"import-auto-id1169738051862\">(c) 9% difference, which is greater than the inherent uncertainty in the given parameters.<\/p>\n<\/div>\n<\/div>\n<div class=\"exercise\" id=\"fs-id1169737851564\">\n<div class=\"problem\" id=\"fs-id1169738011205\">\n<p id=\"import-auto-id1169737961873\">(a) Using the exact exponential treatment, find the time required for the current through a 2.00 H inductor in series with a<br \/>\n0.500 \u03a90.500 \u03a9<br \/>\nresistor to be reduced to 0.100% of its original value. (b) Compare your answer to the approximate treatment using integral numbers of \u03c4\u03c4 size 12{\u03c4} {}. (c) Discuss how significant the difference is.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1169738071203\" class=\"definition\">\n<dt>characteristic time constant<\/dt>\n<dd id=\"fs-id1169736581702\">denoted by \u03c4\u03c4 size 12{\u03c4} {}, of a particular series <em>RL<\/em> circuit is calculated by \u03c4=LR\u03c4=LR size 12{\u03c4= { {L} over {R} } } {}, where LL size 12{L} {} is the inductance and RR is the resistance<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":103,"menu_order":11,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1187","chapter","type-chapter","status-publish","hentry"],"part":1129,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1187","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\/1187\/revisions"}],"predecessor-version":[{"id":1943,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapters\/1187\/revisions\/1943"}],"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\/1187\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/media?parent=1187"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/pressbooks\/v2\/chapter-type?post=1187"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/contributor?post=1187"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/physics0312chooge\/wp-json\/wp\/v2\/license?post=1187"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}