example lr circuit In the circuit drawn below, epsilon = 60.0 V, L = 3.00 H, R_1
ID: 1561640 • Letter: E
Question
example lr circuit
In the circuit drawn below, epsilon = 60.0 V, L = 3.00 H, R_1 = 20.0 Ohm, and R_2 = 5.00 Ohm. What is the value of i) the current in each resistor, ii) the voltage across the inductor, and iii) the magnetic potential energy stored in the inductor: a) Immediately after the switch is closed? b) A very long time after the switch has been closed? c) Immediately after the switch is opened? d) A very long lime after the switch has been opened? e) What happened to the magnetic energy stored in the inductor? f) What could happen if the 20.0-Ohm resistor failed immediately after the switch was opened?Explanation / Answer
a) Immediately after the switch is closed, inductor will act as open circuit. So there would be no current through the indictor and hence no current through the 5 resistor.
Current through the 20 resistor is I20 = 60V/20 = 3A.
Voltage across the 20 resistor is V20 = 3A(20) = 60V
The voltage across the inductor will be same as the voltage across the 20 resistor as there is no current through the inductor. So the voltage through the inductor will be 60V.
Magnetic potential energy stored in the inductor = 1/2LI2 = 0 as the current I through the inductor is zero.
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b) A very long time afer switch has been closed, the inductor would be shor circuit and voltage throught inductor would be zero.
The equuvalent resistance of the circuit would be Req = (20)(5)/(20 + 5) = 4
So current given by the battery would be I = 60V/Req = 15A
So current throught the 20 resistor obtained by current division is I20 = (I)(5)/(20 + 5) = 3A
So current throught the 5 resistor wpuld be I5 = 15A - 3A = 12A
So the current through inductor would also be 12A
So Magnetic potential energy stored in the inductor = 1/2LI2 = (0.5)(3H)(12A)2 = 216J
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c) Immediatelty after the switch is opened the current through inductor would be 12A, as current through indictor cannot change abruptly.
So the Magnetic potential energy stored in the inductor = 1/2LI2 = (0.5)(3H)(12A)2 = 216J
Also the same current 12 A will flow through both the resistors as they are all in series now.
Voltage across the inductor will be = 12A(5 + 20) = 300V
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d) A very long time after the switch is opned, current through the ciruit will be zero. Voltage across the inductor would be zero. Also the magnetic potential energy stored in the indictor would be zero.
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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige....
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