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 at which the energy of the magnetic field in the inductor is increasing, 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 $I=1.7(1-e^{-2.3t})$
Differentiating it w.r.t. time, we get $\frac{dI}{dt}=1.7\times -e^{-2.3t}\times -2.3=3.9e^{-2.3t}$
Energy stored in the inductor $U=\frac{1}{2}L{I}^2$
Rate of increase in energy $\frac{dU}{dt}=LI\frac{dI}{dt}$
$⟹\frac{dU}{dt}=3\times 1.7(1-e^{-2.3t})\times 3.9\times e^{-2.3t}=2(e^{-2.3t}-e^{-4.6t})$ Watts