Question

The total heat produced in resistor R in an RL circuit when the current in the inductor decreases from $I_0$ to 0 is

• A. $L{I}_0^2$
• B. $\frac{1}{2}L{I}_0^2$
• C. $\frac{3}{2}L{I}_0^2$
• D. $\frac{1}{3}L{I}_0^2$

Answer 1

Answer: (B)

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
$W={\int }_0^{\infty }{I}^{2}Rdt$
The current in an RL circuit is $I=I_0{e}^{-(R/L)t}$
$W={\int }_0^{\infty }{I}_0^2{e}^{-(2R/I)t}dt=\frac{I_0^{2}R}{-2R/L}{\left[e^{-\frac{2R}{L}t}\right]}_0^{infty}$
= $\frac{1}{2}L{I}_0^2$
We can integrate by substituting
Note that the total heat produced equals the energy $(1/2)L{I}_0^2$ .
Originally stored in the conductor.