Question

In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm $AD$.

Answer 1

Since the capacitor will act as an open circuit here, the arm CE and CF will become ineffective. The circuit is now reduced to as below

Effective resistance across AD = (BC + CD) ||AD

= ( 3 + 3 ) || 3 = 2$\Omega$

Total resistance of the circuit = AD + DF = 2 + 3 = $5\Omega$

Total current in the circuit = $\frac{15}{5}$ = $3A$

The current through AD and BCD will divide in the inverse ratio of the resistance.

$\therefore$ Current through AD = $3\times \frac{6}{6+3}=3\times \frac{6}{9}=2A$