The circuit shown below consists of a 9V battery, three resistors, an ideal Indu

Question

The circuit shown below consists of a 9V battery, three resistors, an ideal Inductor and switch. Assume that the switch has been open long enough that no current flows in the system at any point. The twitch is now closed at t = 0. Immediately afterwards, what is the current I_3 flowing through resistor R_3? 10.2 mA 25.0 mA 33.3 mA S7.8 mA 128.6 mA A very long time after the switch has been closed, what is the current I_1 through R_1? 10 2 mA 25.0 mA 33.3 mA 57.8 mA 128.6 mA The switch is now suddenly opened. How long after opening the switch does it take for the current through the inductor to reach 1/e of Its value Just before the switch is opened? How much current flows through the inductor at time t = 30 mu s after the switch is opened?

Explanation / Answer

2. just after closing,

now current will flow through inductor.

and R1 and R3 are in series.

Req = R1 + R3 = 270 ohm

I = V / Req = 9 /270 = 0.0333 A or 33.3 mA

Ans(C)

3. now R2 and R3 are in parallel,

1/R' = 1/R2 + 1/R3

R' = 85.7 ohm

Req = R1 + R' = 70 + 85.7 = 155.7 ohm

I = 9 / 155.7 = 0.0578 A Or 57.8 mA

Ans(D)

4.when opened,

Req = R2 + R3 = 350 ohm

I = I0 (1 - e^(-t/T))

1/e = 1 - e^(-t/T)

e^(-t/T) = 0.632

t = 0.46T = 0.46 x 0.01 / 350 = 13.1 x 10^-6 s Or 13.1 us

5. I0 = 57.8 x 200 / 350 = 33.03 mA

I = 33.03 [ 1 - e^(-30 / 28.57)] = 21.5 mA

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