The charge stored by the capacitor C in the given circuit in the steady state is __________ µC.

Solution

Physics sir · 5 steps

Find which diode conducts, then the capacitor just reads the voltage across that live branch.

1. Battery direction. The long plate (positive terminal) of the 2.5 V cell faces the left wire. So the left wire is at a higher potential than the right wire, and current flows through the lower branches from left to right.
2. Check each diode. The diode with the 3 Ω resistor points right to left, so it is reverse biased and carries no current. The diode with the 4 Ω resistor points left to right, so it is forward biased and acts as an ideal closed switch.
3. Steady-state current. In steady state the capacitor branch carries no current. The only loop is the cell, the 1 Ω resistor and the 4 Ω resistor, so I = 2.5/(1+4) = 0.5 A.
4. Capacitor voltage. The potential difference between the two side wires is 0.5 × 4 = 2 V. No current flows through the 2 Ω resistor, so it drops nothing and the capacitor gets the full 2 V.

   Q = CV = 5 μF × 2 V = 10 μC

5. So. The charge stored is 10 μC, option (c).

The idea to keep. In DC steady state a capacitor branch carries no current, so its resistor drops no voltage. The capacitor takes the terminal voltage across its branch, set by whichever diodes are forward biased.

Answer from the NTA final answer key. Solution by PrepWiser.