Question

A 1.53 μF capacitor is charged through a 129 Ω resistor and then discharged through the same resistor by short-circuiting the battery.

A) While the capacitor is being charged, find the time for the charge on its plates to reach 1/e of its maximum value.
B) While the capacitor is being charged, find the current in the circuit at the time when the charge on its plates has reached 1/e of its maximum value.
C) During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1/e of its initial value.
D) Find the time for the current in the circuit to decrease to 1/e of its initial value.

Answer 1

The time for the charge on the capacitor to reach 1/e of its maximum value during charging and to decrease to 1/e of its initial value during discharge is one RC time constant, which is 197.37 μs. The current will reach 1/e of its maximum value at the same time.

Step-by-step explanation:

When a capacitor is being charged or discharged through a resistor, the time it takes for the capacitor to reach a certain fraction of its maximum charge (or decrease to a certain fraction of its initial charge) can be described using the RC time constant (τ), where τ = R•C. The charge on a capacitor (Q) as a function of time (t) during charging is Q(t) = Qmax(1 - e-t/τ), and during discharge, it's Q(t) = Qinitiale-t/τ. For a capacitor with capacitance (C) of 1.53 μF charged through a resistor (R) of 129 Ω:

1. The time constant τ = R•C = (129 Ω)(1.53 μF) = 197.37 μs.
2. To reach 1/e of its maximum charge during charging, it takes times equal to one RC time constant, so t = 197.37 μs.
3. The current (I) at any time t during charging is I(t) = (V/R)e-t/τ, and at t = τ, I(t) will also be 1/e of its maximum value.
4. The same RC time constant applies for discharge, so the time to reach 1/e of the initial value will be t = 197.37 μs.
5. For the current during discharge, it will also reach 1/e of its initial value at one time constant, t = 197.37 μs.