Question

Considering the given circuit, find out the values of current given by the battery

(i) just after closing the key
(ii) long time after closing the key

• A. (i) $2A$ (ii) $2.5A$
• B. (i) $2.5A$ (ii) $2A$
• C. (i) $4A$ (ii) $3A$
• D. none of these

Answer 1

Answer: (A)

JUST AFTER CLOSING THE KEY:
Let $t=0$ , be the time when we need to find the current, i.e, when the switch is just turned on. The inductor will oppose the flow of current and behaves as open circuit so, the current will flow only in the $2\Omega$ and $3\Omega$ resistors only.
The equivalent resistance of the circuit is
$R_s=2+3=5\Omega$
so, $i=\frac{V}{R}=\frac{10}{5}=2A$
LONG TIME AFTER THE CLOSING:
After a long time, the inductor will not oppose the flow of current and hence will acts as short circuit.
Now the equivalent resistance of the circuit is
$\Rightarrow \frac{1}{R_P}=\frac{1}{6}+\frac{1}{3}$
$\Rightarrow \frac{1}{R_P}=\frac{1+2}{6}=\frac{3}{6}$
$\Rightarrow R_P=\frac{6}{3}=2\Omega$ $\Rightarrow R_{eq}=2+2=4\Omega$
so, $i=\frac{10}{4}=2.5A$