Question

Two resistors of $10\Omega$ and $20\Omega$, and an ideal inductor of $10\text{H}$ are connected to a $2V$ battery as shown. The key $K$ is closed at time $t=0$. At the initial $(t=0)$ and final $(t\to \infty )$, the currents through the battery are

• A. $\frac{1}{15}\text{A},\frac{1}{10}\text{A}$
• B. $\frac{1}{10}\text{A},\frac{1}{15}\text{A}$
• C. $\frac{1}{10}\text{A},\frac{1}{10}\text{A}$
• D. $\frac{1}{15}\text{A},\frac{3}{10}\text{A}$

Answer 1

The correct option is A $\frac{1}{15}\text{A},\frac{1}{10}\text{A}$

At $t=0$ the inductor will behave like an open circuit.

The equivalent circuit diagram can be drawn as,


The total resistance of the circuit is,

$R_{eq}=10+30=30\Omega$

$\therefore$ Current flowing through the battery is,

$i=\frac{2}{30}=\frac{1}{15}\text{A}$

At $t=\infty$ the inductor will behave as a short circuit and the equivalent circuit diagram is,


The resistance of the circuit is,

$R_{eq}=20\Omega$

$\therefore$ Current flowing through the battery is,

$i=\frac{2}{20}=\frac{1}{10}\text{A}$

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