Discharging a capacitor — practice questions

Convert units: $C = 2200 \mu F = 2200 \times 10^{-6} F = 2.2 \times 10^{-3} F$ $R = 1.5 k\Omega = 1.5 \times 10^3 \Omega$

Calculate the time constant ($\tau$): $\tau = RC$ $\tau = (1.5 \times 10^3 \Omega) \times (2.2 \times 10^{-3} F)$ $\tau = 3.3 s$

Calculate voltage after 3.0 seconds: Use the exponential decay formula: $V_c = V_0 e^{-t/\tau}$ $V_0 = 12 V$ $t = 3.0 s$ $\tau = 3.3 s$ $V_c = 12 \times e^{-3.0/3.3}$ $V_c = 12 \times e^{-0.909...}$ $V_c = 12 \times 0.4027...$ $V_c \approx 4.83 V$

The time constant is 3.3 s. After 3.0 seconds, the voltage across the capacitor is approximately 4.83 V.

Determine the gradient: Plotting ln(V) against t gives a straight line. We can find the gradient of this line. Let's use the first and last data points for the calculation. Gradient, $m = \frac{\Delta \ln(V)}{\Delta t}$ $m = \frac{0.18 - 2.20}{40.0 - 0.0}$ $m = \frac{-2.02}{40.0} = -0.0505 \text{ s}^{-1}$

Calculate the time constant (τ): The gradient of the ln(V) vs t graph is equal to $-1/\tau$. $m = -\frac{1}{\tau}$ $-0.0505 = -\frac{1}{\tau}$ $\tau = \frac{1}{0.0505}$ $\tau \approx 19.8 \text{ s}$

The time constant for the circuit is approximately 19.8 s.