Question

At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 40.6 ms, at what time (in ms) is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?

Answer 1

power dissapated by resistor: P=(I^2)R

Power stored by inductor: P=IL(di/dt)

Current in an RL cuircuit: I=V/R(1-exp(-t/T))

where T is the time constant tau = (L/R)

just set them equal, then simplified

(I^2)R = IL(di/dt)

I=(L/R)(di/dt)

replace L/R with Tau and Current with above equation

V/R(1-exp(-t/T)) = T (di/dt)

take derivative of current

V/R(1-exp(-t/T)) = T (V/R)(1/T)*exp(-t/T)

cancelations

1-exp(-t/T)=exp(-t/T)

add exp(-t/T) to both sides and take natural logs

Ln(1/2) = -t/T

so t = -T*Ln(1/2) = T*Ln2 = 40.6*ln2 = 28.14 ms

Reference https://www.physicsforums.com/threads/power-in-rl-circuits.354475/