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 EF will become ineffective. The circuit is now reduced to as below

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

   = \[\left( 3 + 3 \right) \lVert 3 = 2 \Omega\]

Total resistance of the circuit = AD + DF = 2 + 3 = 5 Ω
Total current in the circuit = \[\frac{15}{5} = 3 A\]

The current through AD and BCD will divide in the inverse ratio of the resistance.
∴ Current through AD = \[3 \times \frac{6}{\left( 6 + 3 \right)} = 3 \times \frac{6}{9} = 2 A\]