Question

An $RL$ circuit in which $R=5\Omega$ and $L=4H$ is connected to a $22V$ battery at $t=0$. For this circuit at what rate the energy is being stored in the inductor when the current in circuit is $1A$ ?

• A. $17W$
• B. $22W$
• C. $5W$
• D. Cannot be computed from the given information

Answer 1

Answer: (A)

Energy stored in the inductor when current in circuit is $I$, is given by
$U=\frac{L{I}^2}{2}$
$\therefore \frac{dU}{dt}=LI\frac{dI}{dt}...\left(i\right)$
But, $I=\frac{E}{R}\left[1-e^{-t/\tau }\right]...\left(ii\right)$
$\frac{dI}{dt}=\frac{E}{R}\times \frac{1}{\tau }\times e^{-t/\tau }$
$=\frac{E}{L}\times e^{-t/\tau }\left(\because \tau =\frac{L}{R}\right)$
$\frac{dI}{dt}=\frac{E}{L}\left[1-\frac{IR}{E}\right]...\left(iii\right)$
For $I=1A$
$\frac{dU}{dt}=LI\times \frac{E}{L}\left(1-\frac{IR}{E}\right)$ [ Using $(iii)$]
$=1\times 22[1-\frac{1\times 5}{22}]$
$=17W$