A 15-mH inductor is connected in series with a battery and a 12-$\varOmega$ resistor,in a circuit with an open switch.0.50 ms after the switch is closed,the current in the circuit is 1.2A. What is the EMF supplied by the battery?
A)44 V
B)21 V
C)3.6 V
D)1.8 V

Question

Quizplus · Accepted Answer

Using the formula for the current in an RL circuit with a battery,
I = (EMF/R) * (1 - e^(-Rt/L))

We have:
L = 15 mH = 0.015 H
R = 12 ohms
t = 0.50 ms = 0.00050 s
I = 1.2 A

Solving for EMF:
EMF = I * (R + L/T * ln(1 + R/L * I))

EMF = 1.2 A * (12 ohms + 0.015 H / 0.00050 s * ln(1 + 12 ohms / 0.015 H * 1.2 A))

EMF = 44 V (rounded to nearest whole number)

Therefore, the best choice is (A) 44V, as it is the only answer that matches the calculated value of EMF.

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