A circuit is constructed with four resistors, one inductor, one battery and a sw

Question

A circuit is constructed with four resistors, one inductor, one battery and a switch as shown. The values for the resistors are: R_1 = R_2 = 56 ohm, R_3 = 45 ohm and R_4 = 91 ohm. The inductance is L = 249 mH and the battery voltage is V = 24 V. The switch has been open for a long time when at time t = 0, the switch is closed. What is I_1 (0), the magnitude of the current through the resistor R-1 just after the switch is closed? What is I_1 (infinity), the magnitude of the current that flows through the resistor R_1 a very long time after the switch has been closed? What is V_L (0), the magnitude of the voltage across the inductor just after the switch is closed? What is I_L (infinity), the magnitude of the current through the inductor after the switch has been closed for a very long time?

Explanation / Answer

1) When the switch was opened for a long time, the current through inductor was zero.

When the switch is closed at t = 0, the current through the inductor would still be zero immediately after t = 0, as the current through indictor cannot change abruptly.

Thus all the resistors would be in series and current through them is same which is given by;

I = 24V/[(56 + 56 + 45 + 91V)] = 9.67X10-2 A

So the current through R1 will be I1(0) = 9.67X10-2 A

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2) A very long time after the switch has been closed, the inductor would be shorted and so R2 and R3 would be shorted and no current would pass through R2 and R3.

So the R1 and R4 would be in series and the current through them would be;

I = 24V/(56 + 91) = 0.163A which is also the required current I1()

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3) When the switch is closed at t = 0, the current through the inductor would still be zero immediately after t = 0, as the current through indictor cannot change abruptly.

So the voltage across inductor would be the same as the voltage across the branch containing R2 and R3.

Current through the branch containing R2 and R3 = 9.67X10-2 A

equivalent resistance of this branch is R = R2 + R3 = (56+45) = 101

So voltage across this branch =  (9.67X10-2 A)(101) = 9.77V which would also be the voltage across indictor.

Thus VL(0) = 9.77V

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4) As we saw in 2) above, a very long time after the switch has been closed, the inductor would be shorted and so R2 and R3 would be shorted and no current would pass through R2 and R3.

So, the current through R1 an R4 would be the current through indcutor as all three are is series;

So, IL() = I1() = 0.163A

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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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