Question

A 8.2-V battery is connected in series with a 35-mH inductor, a 150-Ω resistor, and an open switch.

Part A What is the current in the circuit 0.130 ms after the switch is closed? (in mA)

Part B How much energy is stored in the inductor at this time? (in uJ)

Answer 1

First assume that the battery has negligeable internal resistance, so that the total resistance (R) in the circuit is 150 ohms. Then use this equation for current (I) at time t in terms of inductance

(L): I(t) = (V/R)*(1 - exp(-t*R/L))

a) at t = 0.13, I = 95.4mA

b) energy is 0.5*L*I^2 = 15.92uJ