Question

In the circuit shown in the figure, time constant and steady state current will be

• A. $0.25\text{s},0.25\text{A}$
• B. $0.75\text{s},0.75\text{A}$
• C. $0.25\text{s},0.75\text{A}$
• D. $0.5\text{s},0.75\text{A}$

Answer 1

The correct option is C $0.25\text{s},0.75\text{A}$

Time constant for $LR$ circuit is given by
$\tau =\frac{L}{R_{eq}}$


From circuit diagram it is clear that both $8\Omega$ resistors are in parallel which is further in series with $4\Omega$ resistor.

$\therefore R_{eq}=4+\frac{8\times 8}{16}=8\Omega$

So the time constant will be

$\tau =\frac{L}{R_{eq}}=\frac{2}{8}=\frac{1}{4}\text{s}$

At steady state, inductor will act like a perfect conductor

$\therefore$ circuit will be


$\therefore i_{\text{steady}}=\frac{V}{R_{eq}}=\frac{6}{8}=0.75\text{A}$

$\therefore \tau =0.25\text{s},i_{\text{steady}}=0.75\text{A}$

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