To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has

Question

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 3.4 V battery connected in series with a switch. When the switch is closed, the battery powers two paths in parallel, one of which has a resistor of resistance R_1 = 100 ohm in series with an inductor of inductance L = 1.3 times 10^-2 H, while the other has a resistor of resistance R_2 = 320 ohm. What is the current supplied by the battery at a time t = 0.35 ms after the switch is closed? What is the current i_1 supplied by the battery a time t = 0.35 ms after the switch is closed? Express your answer in amperes to three significant figures. In the previous part, you found the current supplied by the battery 0.35 ms after the switch is closed. What will be the steady-state current, I_1, supplied by the battery after the switch has been closed for a very long time? Express your answer in amperes to three significant figures.

Explanation / Answer

part D )

at steady state , inductor behaves as short circuit and hence the total resistance of the circuit will be the parallel combination of R1 and R2

Rtotal = R1 R2 / (R1 + R2) = 100 x 320 / (100 + 320) = 76.2 ohm

current = i = V/Rtotal = 3.4/76.2 = 0.045 A

i1 = V/R1 = 3.4 / 100 = 0.034 A

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