A 2.0-μF capacitor is charged to 12 V and then discharged through a 4.0-MΩ resistor.How long will it take for the voltage across the capacitor to drop to 3.0 V?
A)8.0 s
B)11 s
C)22 s
D)24 s

Question

Quizplus · Accepted Answer

The time constant (τ) of the circuit is given by the product of the capacitance (C) and resistance (R) in the circuit.

τ = RC

τ = (4.0 x 10^6 Ω)(2.0 x 10^-6 F)

τ = 8.0 s

After one time constant, the voltage across the capacitor drops to approximately 36.8% of its initial value.

V = V0 e^(-t/τ)

where V0 is the initial voltage (12 V), t is the time elapsed, and e is the mathematical constant (approximately equal to 2.718).

We can solve for t when V = 3.0 V:

3.0 V = 12 V e^(-t/8.0 s)

e^(-t/8.0 s) = 0.25

Taking the natural logarithm of both sides:

-ln(0.25) = -t/8.0 s

t = 11 s

Therefore, it will take 11 seconds for the voltage across the capacitor to drop to 3.0 V.

A 20-μF Capacitor Is Charged to 12 V and Then Discharged