A 25-mH inductor is connected in series with a 20-Ω resistor through a 15-V dc power supply and a switch. If the switch is closed at t = 0 s, what is the current when t = 2.0 ms?

Question

Quizplus · Accepted Answer

First, we need to determine the time constant for the circuit, which is given by:

τ = L/R = 25 mH / 20 Ω = 1.25 ms

When the switch is closed, the current in the circuit will increase exponentially towards its steady-state value of I = V/R = 15 V / 20 Ω = 0.75 A.

The equation for the current as a function of time is:

I(t) = I_final * (1 - e^(-t/τ))

Plugging in t = 2.0 ms and τ = 1.25 ms, we get:

I(2.0 ms) = 0.75 A * (1 - e^(-2.0 ms/1.25 ms)) = 0.60 A

Therefore, the current when t = 2.0 ms is 0.60 A, which corresponds to choice C.

A 25-MH Inductor Is Connected in Series with a 20-Ω