LR circuits: A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
A) 250 mA
B) 650 mA
C) 550 mA
D) 280 mA
E) 850 mA

Question

Quizplus · Accepted Answer

The current in an LR circuit is given by:$$i(t) = \frac{V}{R} (1 - e^{-t/	au})$$where:* V is the voltage of the power supply* R is the resistance of the resistor* L is the inductance of the inductor* $	au$ is the time constant of the circuit, which is equal to L/RIn this problem, we have:* V = 15 V* R = 60 Ω* L = 45 mH = 0.045 H* t = 7.0 ms = 0.007 sSo, the time constant is:$$	au = \frac{L}{R} = \frac{0.045 H}{60 \Omega} = 0.00075 s$$And the current at t = 7.0 ms is:$$i(0.007 s) = \frac{15 V}{60 \Omega} (1 - e^{-0.007 s/0.00075 s}) = 0.25 A = 250 mA$$Therefore, the correct answer is A.

LR Circuits: a 45-MH Ideal Inductor Is Connected in Series