In the given circuit. C 1 = 2 μ F , C 2 = 0 . 2 μ F , C 3 = 2 μ F , C 4 = 4 μ F , C 5 = 2 μ F , C 6 = 2 μ F . The charge stored on capacitor C 4 is ______ μ C .

Step-by-step solution

From the diagram of the circuit, C 3 ,   C 4 ,   C 5 are in series, so the total capacitance is 1 C ' = 1 2 + 1 4 + 1 2 ⇒ C ' = 4 5   μF C '   &   C 2 are parallel, the total capacitance is C ' ' = 4 5 + 0 . 2   μF = 1   μF C 1 ,   C ' ' ,   C 6 are in series. So the equivalent capacitance is, 1 C e q = 1 2 + 1 + 1 2 ⇒ C e q = 0 . 5   μF Using Q = C e q V . The charge through the battery, Q = 10 × 1 2 = 5   μC From