A capacitor is connected in series with a resistor and a switch. With the switch open, the capacitor is charged to 9.0 V. When the switch is closed, how long will it take for the voltage across the capacitor to drop to 6.0 V if the time constant of the circuit is 4.0 s?
A) 0.3 s
B) 1.2 s
C) 1.6 s
D) 2.7 s
E) 12 s

Question

Quizplus · Accepted Answer

The time constant of the circuit is given as 4.0 s, which is equal to the product of the resistance and capacitance of the circuit (τ = RC). Since τ is known and the voltage has to drop from 9.0 V to 6.0 V, we can use the formula for voltage across a capacitor in a discharging circuit: V = V0 * e^(-t/τ), where V0 is the initial voltage (9.0 V), V is the final voltage (6.0 V), and t is the time elapsed. Solving for t, we get:
ln(V/V0) = -t/τ
ln(6.0/9.0) = -t/4.0
t ≈ 1.6 s
Therefore, the best choice is C.

A Capacitor Is Connected in Series with a Resistor and a Switch