Question

An inductor with an inductance of $3.00H$ and a resistance of $7.00\Omega$ is connected to the terminals of a battery with an emf of $12.0V$ and negligible internal resistance. What is the rate of dissipation of energy in the resistance of the inductor as a function of time?

Answer 1

Given : $R=7\Omega$ $L=3H$ $V=12$ volts
Current flowing as a function of time is given by $I=\frac{V}{R}(1-e^{-\frac{Rt}{L}})$
$\therefore$ $I=\frac{12}{7}(1-e^{-\frac{7t}{3}})$
Or $I=1.7(1-e^{-2.3t})$
Rate of dissipation of energy in resistance $P=I^{2}R$
$\therefore$ $P=\left[1.7\right(1-e^{-2.3t}){]}^2\times 7=20.23(1-e^{-2.3t}{)}^2$ Watt