A 20-μF capacitor is charged to 200 V and is then connected across a 1000-Ω resistor. The amount of charge on the capacitor after 10 ms is approximately

Question

Quizplus · Accepted Answer

The charge on a capacitor Q is given by Q = CV, where C is the capacitance and V is the voltage across the capacitor.

When the capacitor is connected across the resistor, it will discharge through the resistor, and the voltage across the capacitor will decrease over time.

The time constant of the circuit is given by τ = RC, where R is the resistance and C is the capacitance. In this case, τ = (1000 Ω)(20 μF) = 20 ms.

After one time constant, the voltage across the capacitor will have decreased to approximately 36.8% of its initial value. After two time constants, it will have decreased to approximately 13.5% of its initial value, and so on.

After 10 ms, the capacitor will have discharged approximately halfway to its final voltage. Therefore, the voltage across the capacitor will be approximately 100 V.

The charge on the capacitor is then Q = CV = (20 μF)(100 V) = 2 mC.

Therefore, the answer is B) 2.4 mC.

A 20-μF Capacitor Is Charged to 200 V and Is