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 inductor increases linearly with time when a constant voltage is applied. The rate of increase is determined by the inductance and the voltage. In this case, the current will increase by:```ΔI = (V/L) * t```where:* ΔI is the change in current* V is the voltage* L is the inductance* t is the timePlugging in the values given in the question, we get:```ΔI = (15 V / 45 mH) * 7.0 ms = 225 mA```Therefore, the current 7.0 ms after the switch is closed will be:```I = ΔI + 0 A = 225 mA```which is closest to choice A, 250 mA.

A 45-MH Ideal Inductor Is Connected in Series with a 60-Ω