Question

The figure shows a circuit consisting of an ideal cell of emf $E=10\text{V}$, an inductor of inductance $L=2\text{H}$ and a resistor of resistance $R=5\Omega$ connected in series. The switch $\text{S}$ is closed at $t=0.$ Suppose at $t=0$, current in the inductor is $i_0=1\text{A}$, then find out the equation of current as a function of time.

• A. $2-e^{(-\frac{5t}{2})}$
• B. $2-2e^{(-\frac{5t}{2})}$
• C. $e^{(-\frac{5t}{2})}$
• D. $2-3e^{(-\frac{5t}{2})}$

Answer 1

The correct option is A $2-e^{(-\frac{5t}{2})}$


Let at any instant $t$ current in the circuit is $i$ which is increasing at the rate of $\frac{di}{dt}$

Using $KVL$ to the given circuit, we get,

$E-\frac{Ldi}{dt}-iR=0$

$\Rightarrow \frac{Ldi}{dt}=E-iR$

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

On integrating both sides, with proper limit we get,

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

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

$\Rightarrow \ln {\left[E-iR\right]}_{i_0}^i=-\frac{Rt}{L}$

$\Rightarrow \ln \left[\frac{E-iR}{E-i_{0}R}\right]=-\frac{Rt}{L}$

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

$\left[\frac{E-iR}{E-i_{0}R}\right]=e^{(-\frac{Rt}{L})}$

$E-iR=(E-i_{0}R)e^{(-\frac{Rt}{L})}$

$i=\frac{E-(E-i_{0}R)e^{(-\frac{Rt}{L})}}{R}$

Here,
$L=2\text{H};R=5\Omega E=10\text{V};i_0=1\text{A}$

$i=\frac{10-(10-1\times 5)e^{(-\frac{5t}{2})}}{5}$

$\therefore i=2-e^{(-\frac{5t}{2})}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.

Why this question ? Caution: Pay attention on the direction of current while applying $\text{KVL}$ in the circuit and rate of current in the inductor.