Question

In the $L-R$ circuit shown, what is the variation of the current $I$ as a function of time $t$, if the switch is closed at time $t=0\text{s}$.

• A. $\frac{V}{R}(1-e^{\frac{Rt}{L}})$
• B. $\frac{V}{R}{e}^{-\frac{Rt}{L}}$
• C. $-\frac{V}{R}{e}^{-\frac{Rt}{L}}$
• D. None

Answer 1

The correct option is C $-\frac{V}{R}{e}^{-\frac{Rt}{L}}$


Applying $\text{KCL}$ at junction $\text{C}:$

$i_1+i_2=I.........\left(1\right)$

Applying $\text{KVL}$ in loop $\text{CDEFC}:$

$-2V+V-i_{1}R=0$

$i_1=-\frac{V}{R}........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow i_2=I+\frac{V}{R}.....\left(3\right)$

Applying $\text{KVL}$ in loop $\text{ABFGA:}$

$-L\frac{d{i}_2}{dt}+i_{1}R-V+3V-i_{2}R=0$

Using equation $\left(3\right)$,

$-L\frac{dI}{dt}-\frac{V}{R}\times R-V+3V-(I+\frac{V}{R})R=0[\because \frac{V}{R}\text{is constant}]$

$\Rightarrow \frac{dI}{dt}=-\frac{IR}{L}$

$\Rightarrow \frac{dI}{I}=\frac{-R}{L}dt$

Integrating both side for the given limit,

${\int }_{I_0}^I\frac{dI}{I}={\int }_0^t-\frac{R}{L}dt$

At $t=0$, when the switch is open, the current in the circuit,

$\because I_0=-\frac{V}{R}$

$[\ln I{]}_{-V/R}^I=\frac{-Rt}{L}$

$\ln \left(\frac{I}{-V/R}\right)=-\frac{R}{L}t$

$\therefore I=-\frac{V}{R}{e}^{-\frac{R}{L}t}$

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

Why this question ? Caution: This is a case of discharging $L-R$ circuit, so don't apply the condition of charging current. Tip: Such type of question is frequently asked in the competitive exams. It enhances the concept of Kirchhoff's law and charging/ discharging current in an inductor.