Question

A capacitor of 4 uF is connected as shown in the circuit. The internal resistance of the battery is 0.5 ohm. The amount of charge on the capacitor plates will be​

Answer 1

At steady state the capacitor is open circuited so no current flows through the 10 ohm resistor.

Current flowing through 2 ohm resistor is

i = 2.5/(2+0.5) = 1A

Voltage across capacitor terminal 2.5 - 0.5*1 = 2V

Charge in capacitor plates 4*10^(-6) * 2 = 8 microcoulombs.

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