Revise Discharging a capacitor for Cambridge 9702 (19.3)

When a charged capacitor is connected across a resistor, the stored electrical energy is released. This release drives an electric current through the resistor. Initially, the current is at its maximum, but as the capacitor loses charge, the potential difference across it drops, causing the current to decrease over time. The rate of discharge is always proportional to the amount of charge remaining.

A key characteristic of capacitor discharge is that the charge (Q), voltage (Vc), and current (I) all decrease following an exponential decay pattern. This means they don't drop linearly but rather shed a constant fraction of their remaining value over equal time intervals. This pattern is crucial for predicting circuit behaviour.

The mathematical expressions for exponential decay are:

• $V_c = V_0 e^{-t/\tau}$
• $Q = Q_0 e^{-t/\tau}$
• $I = I_0 e^{-t/\tau}$

Here, $V_0$, $Q_0$, and $I_0$ are the initial values at $t=0$ when discharge begins. The term 'e' is Euler's number (approx. 2.718).

To determine the time constant experimentally from a set of measurements, it's useful to rearrange the decay equation into a linear form. By taking the natural logarithm of the voltage equation, we can plot a straight-line graph.

Starting with $V_c = V_0 e^{-t/\tau}$ Taking natural logs of both sides: $\ln(V_c) = \ln(V_0 e^{-t/\tau})$ $\ln(V_c) = \ln(V_0) + \ln(e^{-t/\tau})$ $\ln(V_c) = \ln(V_0) - \frac{t}{\tau}$ Rearranging into the form $y = mx + c$: $\ln(V_c) = -\left(\frac{1}{\tau}\right)t + \ln(V_0)$

The symbol $\tau$ (tau) represents the time constant of the RC circuit. It's a critical parameter that tells us how quickly the capacitor will discharge. A larger time constant means a slower discharge, while a smaller one means a faster discharge. It's determined by the resistance and capacitance values in the circuit.

The time constant is calculated as: $\tau = RC$

The energy stored in the capacitor is dissipated as heat in the resistor. Since energy is proportional to the square of the voltage ($E = \frac{1}{2}CV^2$), it also decays exponentially, but at a faster rate.

$E = E_0 e^{-2t/\tau}$

Another useful concept is the half-life ($t_{1/2}$), the time it takes for the charge (or voltage) to fall to half its initial value. It's directly related to the time constant.

$t_{1/2} = \tau \ln(2) \approx 0.693\tau$