An LR circuit having `L = 4.0 H, R = 1.0 Omega and epsilon = 6.0 V` is switched on at `t = 0`. find the power dissipated in joule heating at `t = 4.0 s`.

Explanation

The time constant of the circuit is `tau = L/R = 4
0 H/ 1
0 Omega = 4
0 s`
The current at t= 4
0 s is, therfore, ` I = (epsilon)/(R ) (1-e ^(-t/tau) = (6A) ((1 - (1)/(e )) = (6 A)X(0
63) = 3
8 A
the power dissipated in Joule heating = i^2 R = ((3
8 A)^2 X 10 Omega ~~ 140 Omega