In the network shown below, the charge accumulated in the capacitor in steady state will be [2023]

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Solution

Q.
In the network shown below, the charge accumulated in the capacitor in steady state will be [2023]

1 7.2 μC

2 4.8 μC

3 10.3 μC

4 12 μC

Ans.
(1)

No current will flow in the capacitor in steady state. Current flowing in the circuit in steady state is

   $I = \frac{3}{6 + 4} = \frac{3}{10}$

Potential difference across the $6 Ω$ resistance

   $V = 6 \times \frac{3}{10} = 1.8 V$

The capacitor will have the same potential, so the charge is

   $q = C V = (4 μ F) (1.8 V) = 7.2 μ C$

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