A 6.0-mH inductor and a 3.0-$\Omega$ resistor are wired in series to a 12-V ideal battery. A switch in the circuit is closed at time 0, at which time the current is zero. 2.0 ms later the energy stored in the inductor is: A) 0 B) 2.5 * 10^-² J C) 1.9 * 10^-² J D) 3.8 *10^-² J E) 9.6 * 10^-³ J

Question

Quizplus · Accepted Answer

The energy stored in an inductor is given by W = (1/2)*L*I^2, where L is the inductance and I is the current flowing through the inductor. At time t=0, the current I in the circuit is zero. When the switch is closed, the current starts to flow in the circuit and increases with time. The rate of change of current is determined by the time constant tau = L/R, where R is the resistance in the circuit. In this case, tau = L/R = 6.0 mH / 3.0 kohm = 2.0 ms. After a time t = 2.0 ms, the current in the circuit has reached a steady state value given by I = V/R = 12 V / 3.0 kohm = 4.0 mA. Using these values, we can calculate the energy stored in the inductor: W = (1/2)*L*I^2 = (1/2)*(6.0 mH)*(4.0 mA)^2 = 1.92*10^-5 J Therefore, the answer is C, 1.9 F1F1F1S1F1F1F10 10^-² J.

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