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Question
In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm AD.
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Solution
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 = \[3 \times \frac{6}{\left( 6 + 3 \right)} = 3 \times \frac{6}{9} = 2 A\]
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