In the circuit shown in the figure (Figure 1), the switch has been open for a long time and is suddenly closed. Neither the battery nor the inductors have any appreciable resistance.

Question

Question

In the circuit shown in the figure (Figure 1), the switch has been open for a long time and is suddenly closed. Neither the battery nor the inductors have any appreciable resistance.

Part A

What do the ammeter read just after S is closed?

Part B

What do the voltmeter read just after S is closed?

Part C

What do the ammeter read after S has been closed a very long time?

Part D

What do the voltmeter read after S has been closed a very long time?

Part E

What do the ammeter read 0.160 ms after S is closed?

Part F

What do the voltmeter read 0.160 ms after S is closed?

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Answer 1

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Answer 2

Answer #2

For Part F, the solution is much more simple than the image above. V-I*R_eq=V_L. Once you've found I, it's just 20V-0.185*75=V_L. Kirchoff's Voltage Law is all that is necessary to solve F once E is completed.

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answered by: Student1

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Answer 3

Answer #3

Hope this helps...

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Answer 4

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In the circuit shown in the figure (Figure 1), the switch has been open for a long time and is suddenly closed. Neither the battery nor the inductors have any appreciable resistance.

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Homework Answers

Add Answer to: In the circuit shown in the figure (Figure 1), the switch has been open for a long time and is suddenly closed. Neither the battery nor the inductors have any appreciable resistance.

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In the circuit shown in the figure (Figure 1), the switch has been open for a long time and is suddenly closed. Neither the battery nor the inductors have any appreciable resistance.