Question

In the circuit shown, the key ($K$) is closed at $t=0$, the current through the key at the instant $t={10}^{-3}\ln \left(2\right)$, is :

• A. $2A$
• B. $8A$
• C. $4A$
• D. zero

Answer 1

The correct option is A $2A$

We know that

Time constant given by

${\tau }_L=\frac{L}{R}=\frac{0.01}{10}={10}^{-3}s$
${\tau }_C=CR=(0.1\times {10}^{-3})\left(10\right)={10}^{-3}s$
${\left(i_0\right)}_L=\frac{20}{10}=2A$
${\left(i_0\right)}_C=\frac{20}{10}=2A$
The given time is the half-life time of both the circuits
$\therefore$ $i_L=i_C=\frac{2}{2}=1A$
or total current is $2A$