The charge accumulated on the capacitor connected in the following circuit is ______ μC . (Given C = 150 μ F )

Step-by-step solution

Under the balanced condition, the capacitor will act as an open circuit. The current i 1 through the 1 Ω resistor is given by i 1 = 10 V 1 Ω + 2 Ω = 10 3 A The current i 2 through the 6 Ω resistor is given by i 2 = 10 V 6 Ω + 4 Ω = 1 A From the above diagram, it can be written that V A + i 1 × 1 - i 2 × 6 = V B ⇒ V A - V B = 1 × 6 - 10 3 × 1 = 8 3 V Hence, the charge on the capacitor is given by Q = C V A - V B = 150 μF × 8 3 V = 400 μC