An 8 μF capacitor is connected across the 10 Ω resistor in a series circuit containing 10 Ω and 15 Ω resistors with a 5 V DC source. Find the steady-state charge on the capacitor. (The source paper prints the charge unit as μF.)
Physics · Capacitor · RC circuit
JEE Mains 2025 23 Jan Evening Shift

Question

An 8 μF capacitor is connected across the 10 Ω resistor in a series circuit containing 10 Ω and 15 Ω resistors with a 5 V DC source. Find the steady-state charge on the capacitor. (The source paper prints the charge unit as μF.)

Physics · Capacitor · RC circuit

JEE Mains 2025 23 Jan Evening Shift

G1 Physics · Accepted Answer

At steady state the capacitor is open, so current through the 10 Ω and 15 Ω series resistors is I=5/(10+15)=0.2 A. Voltage across the 10 Ω resistor is V=0.2×10=2 V. Thus Q=CV=8 μF×2 V=16 μC. The numerical answer is 16; the source paper prints μF for the charge unit.