Question

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

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

Answer 1

Answer: (A)

$17\,W$

Answer 2

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