Question

In the given circuit find reading of ammeter (in A) when capacitor has attained steady state. (Approximate the answer to the nearest integer)

Answer 1

When the capacitor has attained steady-state, it will behave as an open circuit.
So, the circuit will be as shown

Simplifying the circuit from the combination of $2\,\Omega$ , $1\,\Omega$ and $6\,\Omega$ .
$R=\frac{\left(2 + 1\right) \times 6}{2 + 1 + 6}=2\,\Omega$

This circuit is a balanced wheatstone bridge.
So, the $5.2\,\Omega$ resistor will be as in an open circuit.
The equivalent resistance, $R_{eq}=\frac{\left(2 + 1\right) \times \left(8 + 4\right)}{\left(2 + 1\right) + \left(8 + 4\right)}=2.4\,\Omega$
The reading of the ammeter $I=\frac{10}{2 . 4}\approx4.16\,A$