Thank you for reporting, we will resolve it shortly
Q.
The total heat produced in resistor R in an RL circuit when the current in the inductor decreases from $I_0$ to 0 is
Electromagnetic Induction
Solution:
The power dissipated in the resistor ,
$P = \frac{dW}{dt} = I^2 R$
Since the current through resistor varies with time, we must integrate.
The totoal energy produced as heat in the resistor
$W = \int^{\infty}_{0} \, I^2 \, Rdt$
The current in an RL circuit is $I = I_0 e^{-(R/L)t}$
$W = \int^{\infty}_{0} \, I^2_0e^{-(2R/I)t} dt = \frac{I^2_0 R}{- 2R / L} \left[ e^{-\frac{2R}{L} t} \right]^{infty}_{0} $
= $\frac{1}{2} LI^2_0$
We can integrate by substituting
Note that the total heat produced equals the energy $(1/2)LI^2_0$ .
Originally stored in the conductor.