Question

In the given circuit, with steady current, calculate the potential drop across the capacitor and the charge stored in it.

Answer 1

 In steady state, no current will flow from the branch containing capacitor.

Applying KVL in the loop ACDFA
Let current flows anticlockwise in this loop

\[- 8 + I_1 \times 4 + I_1 \times 2 + 4 = 0\]

\[ - 8 + 6 I_1 + 4 = 0\]

\[ - 4 + 6 I_1 = 0\]

\[ I_1 = \frac{4}{6} = \frac{2}{3} A\]

Let the potential drop in capactor be

\[V_c\]

Applying KVL in the loop BEDCB, moving anticlockwise in the loop starting from E

\[V_c + 4 - 8 + 4 \times \frac{2}{3} = 0\]

\[ V_c - 4 + \frac{8}{3} = 0\]

\[ V_c - \frac{4}{3} = 0\]

\[ V_c = \frac{4}{3} = 1 . 3 V\]

Charge on the capacitor, Q = CV =

\[10 \mu F \times \frac{4}{3} V = \frac{40}{3} \mu C = 13 . 3 \times {10}^{- 6} C\]