Question

A battery of emf $E$ and of negligible internal resistance is connected in a series $L-R$ circuit. The inductor has a piece of soft iron inside it. When steady state is reached, the piece of soft iron is abruptly pulled out so that the inductance of the inductor decreases to $nL$, where $n<1$ with the battery remaining connected. Find the current in the circuit as a function of time if at $t=0$ is the instant when the soft iron piece is pulled out-

• A. $\frac{E}{R}[1-\frac{1}{n}]e^{(-\frac{Rt}{nL})}$
• B. $\frac{E}{R}+\frac{E}{R}[1-\frac{1}{n}]e^{(-\frac{Rt}{nL})}$
• C. $\frac{E}{R}$
• D. $\frac{E}{R}-\frac{E}{R}[1-\frac{1}{n}]e^{(-\frac{Rt}{nL})}$

Answer 1

The correct option is D $\frac{E}{R}-\frac{E}{R}[1-\frac{1}{n}]e^{(-\frac{Rt}{nL})}$

At $t=0,$ steady state current in the circuit is,

$i_0=\frac{E}{R}.......\left(1\right)$

Now, when the soft iron is suddenly pulled out from the coil its inductance $L$ reduces to $nL(n<1).$

Now, the flux linked with the coil is, $\varphi =Li$

To keep the flux constant, the current in the circuit at time $t=0$ will increase to

$i=\frac{i_0}{n}=\frac{E}{nR}.......\left(2\right)$



Let the current in the circuit at time $t$ be $i.$

Applying $KVL$ to the given loop,

$E-nL\left(\frac{di}{dt}\right)-iR=0$

$\frac{di}{E-iR}=\frac{dt}{nL}$

Integrating the above expression with proper limits we get,

${\int }_{\frac{i_0}{n}}^i\frac{di}{E-iR}={\int }_0^t\frac{dt}{nL}$

$\Rightarrow {\left[\frac{\ln (E-iR)}{-R}\right]}_{\frac{i_0}{n}}^i=\frac{t}{nL}$

$\Rightarrow \ln \left[\frac{E-iR}{E-\frac{i_0}{n}R}\right]=-\frac{Rt}{nL}$

Taking $\text{Antilog}$ on both sides, we get,

$E-iR=[E-\frac{i_0}{n}R]\times e^{\left(\frac{-Rt}{nL}\right)}.......\left(3\right)$

From $\left(1\right)$ and $\left(3\right)$ we get,

$E-iR=[E-\frac{E}{R}\times \frac{R}{n}]\times e^{\left(\frac{-Rt}{nL}\right)}$

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

Hence, option $\left(\text{D}\right)$ is the correct answer.