Example: Charging RC

Batteries supply EMF of 2.8V
Capacitor has capacitance of 1.0F
Light bulb has resistance of 6.0Ω
If the capacitor is initially uncharged, what is the current, charge, potential difference, and stored energy of the capacitor after 15s?
How much time is required for ΔV to reach 1.0V?

Solution

2: Circuit diagram
3: ℰ = 2.8V, C = 1.0F, R = 6.0Ω
4: Capacitor, and charging RC circuit
5: Q = C ΔV ← capacitor PE = 1/2 C (ΔV)²
ΔV = ℰ (1 - e^(-t/τ)) ← charging RC circuit
τ = R C
6/7: τ = 6.0s
Initially, ΔV = 0, Q = 0, PE = 0, and I = ℰ / R = 0.47A.
I = 2.8V / 6.0Ω × e^(-15s/6s) = 0.038A
ΔV = 2.8V (1-e^(-15s/6.0s)) = 2.57V
Q = ΔV / C = 2.57C
t = -τ ln(1-ΔV / ℰ) = -6.0s * ln(.64) = 2.7s
PE = 1/2 C (ΔV)² = 3.3J
8: Units (charge = C, potential difference = V, PE = J)
15s = 2.5 time constants. After 1 time constant, the capacitor will be 63% (1-1/e) charged. So, we expect the charge to be nearly its maximum (ΔV = 2.8V, Q = ΔV/C = 2.8C) and the current to be relatively low compared to initial (I ≪ 0.47A).
Note: A charge of 2.57C would be extraordinary, but remember that is not the net charge of the capacitor (0C), but the amount of charge that has been separated.

Problem: discharging RC (worksheet?)

A 24μF capacitor is charged to 72μC.
It is then connected to a 120kΩ resistor (without a power supply).
What is the initial potential difference of the capacitor, and the potential difference after 4.0s?
What is the initial current through the bulb, and the current after 3.6s?
How much energy is initially on the capacitor, and how much after 2.4s?

Solution

τ = 2.88s
at 0s ΔV = 3V, at 4s ΔV = 0.75V
at 0s I = 25μA, at 3.6s I = 7.2μA
at 0s PE = 0.11mJ, at 2.4s PE = 6.7μJ