Question

In the given circuit, find the ratio of $i_1$ and $i_2$. $i_1$ is the initial (at $t=0$) current, and $i_2$ is the steady state (at $t=\infty )$ current through the battery:

• A. $2:5$
• B. $4:5$
• C. $5:2$
• D. $3:2$

Answer 1

The correct option is B $4:5$

At $t=0$, the inductor will behave as open terminals, so the equivalent circuit can be drawn as:


The equivalent resistance of the circuit is:

$R_{eq}=6+4=10\Omega$

$\therefore$ Current in the circuit at $t=0$ is,

$i_1=\frac{10}{10}=1\text{A}$

At $t=\infty$, the inductor will behave as a short circuit. The equivalent circuit can be drawn as:


The equivalent resistance of the circuit is,

$R_{eq}^{\prime }=6+\frac{4\times 4}{4+4}=8\Omega$

$\therefore$ Current in the circuit in steady state is,

$i_2=\frac{10}{8}\text{A}$

So, the ratio of $i_1$ and $i_2$ will be

$i_1:i_2=1:\frac{10}{8}$

$\therefore i_1:i_2=8:10=4:5$

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

Why this question ? Concept: As we know that inductor opposes the change in the circuit, so at $t=0$, it creates high resistance $(R=\infty )$, and at time $t=\infty$, it behaves as a ideal conducting wire $(R=0)$.