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Question
In the given circuit in the steady state, obtain the expressions for (a) the potential drop (b) the charge and (c) the energy stored in the capacitor, C.
Solution
(a)
In the steady state, the capacitor behaves as an open circuit. When the steady state is reached, there is no current through arm BE. The potential difference across the two plates of the capacitor is equal to the potential difference across EF.
Applying Kirchhoff's voltage law in the loop ACDF
\[- 2V + 2RI + RI + V = 0\]
\[V = 3RI\]
\[ \Rightarrow I = \frac{V}{3R}\]
\[V_E - V_B = 2V - \left( \frac{V}{3R} \right) \times 2R = \frac{4V}{3}\]
Since there is no current through the battery in branch BE, therefore,
\[V_E - V_B = \left( V_E - V_P \right) + \left( V_P - V_B \right)\]
\[\frac{4V}{3} = V_E - V_P + V\]
\[ \Rightarrow V_E - V_P = \frac{V}{3}\]
\[\left( b \right) \text { The charge stored in the capacitor }, Q = CV = \frac{CV}{3}\]
\[\left( c \right) \text { The energy stored in the capacitor, E } = \frac{1}{2}C \left( V_E - V_P \right)^2 = \frac{1}{2}C \left( \frac{V}{3} \right)^2 = \frac{1}{18}C V^2\]
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