In the circuit shown in the figure (Figure 1) , the switch has been open for a l

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.

A) What do the ammeter read just after S is closed?

B) What do the voltmeter read just after S is closed?

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

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

E) What do the ammeter read 0.110 ms after S is closed?

F) What do the voltmeter read 0.110 ms after S is closed?

Explanation / Answer

Part A)-

As we know that Voltmeter has high resistance and Ammeter has low resistance.

And inductor we be act as open circiut just after closing the switch. then-

As per circuit, the circuit is look like open circuit just after closing the switch then Ammeter will read the Zero reading. (i = 0)

Part B)-

As the voltmeter connected parallel in battery then-

V=20 v

Part C)-

After long time-

Inductor will be act like short circuit then 50 ohm and 25 ohm are in series then Req = 75 ohm.

i = V/Req

i= 20/75

i= 0.266 A

Part D)-

-25(i) +20-50(i) = V

V = 20 V ( Because Voltmeter is short circuited by the inductors so it will show the voltage approx equal to the battery)

Thank you !!! Have fun

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