Question

An inductor of $L=400\text{mH}$ and a resistor of $R=2\Omega$ are connected to a battery of EMF $12\text{V}$ in series. The internal resistance of the battery is negligible. The switch is closed at $t=0$. The magnitude of potential difference $\left(\text{in V}\right)$ across the inductor, as a function of time $\left(t\right)$, is

• A. $12e^{-5t}$
• B. $24e^{-5t}$
• C. $1.2e^{-5t}$
• D. $2.4e^{-5t}$

Answer 1

The correct option is A $12e^{-5t}$

Growth in current in the circuit, after switch is closed, is given by,

$i=\frac{E}{R}[1-e^{\left(\frac{-Rt}{L}\right)}]$

$\Rightarrow \frac{di}{dt}=\frac{E}{R}[0-e^{\left(\frac{-Rt}{L}\right)}.\frac{-R}{L}]=\frac{E}{L}.e^{\left(\frac{-Rt}{L}\right)}$

$\Rightarrow \frac{di}{dt}=\frac{12}{0.4}\times e^{\left(\frac{-2t}{0.4}\right)}=30e^{-5t}$

Now, potential difference across the inductor is,

$V=L\frac{di}{dt}$

$\Rightarrow V=0.4\times 30e^{-5t}=12e^{-5t}$

Hence, option $\left(A\right)$ is the correct answer.