Exponential charge and discharge of a capacitor through a resistor, the time constant, and graphical and logarithmic analysis of the decay.

What this dot point is asking

AQA specification point 3.7.4.4 wants you to describe and use the exponential charge and discharge of a capacitor through a resistor, define and use the time constant, and analyse decay graphically, including using log-linear plots to find the time constant.

Discharging a capacitor

When a charged capacitor discharges through a resistor, the rate of loss of charge is proportional to the charge remaining. This is because the discharge current is $I = \dfrac{V}{R} = \dfrac{Q}{RC}$, and a current is a rate of loss of charge, so $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = -\dfrac{Q}{RC}$. Any quantity whose rate of change is proportional to itself decays exponentially.

The voltage and current decay together because $V = \dfrac{Q}{C}$ and $I = \dfrac{V}{R}$ are both proportional to $Q$ at every instant.

The time constant

After one time constant the quantity has fallen to 37 percent; after two, to 13.5 percent; after about five time constants the capacitor is effectively fully discharged (under 1 percent remaining). A larger $R$ or $C$ slows the discharge, because a larger resistance restricts the current and a larger capacitance stores more charge to lose.

Half-life of the decay

This is the time for the charge to halve, directly analogous to radioactive decay, where the activity halves every half-life. The decay constant of the capacitor circuit is $\dfrac{1}{RC}$.

Charging a capacitor

Analysing decay with logarithms

Taking natural logs of the discharge equation gives a straight line:

A graph of $\ln Q$ against $t$ has gradient $-\dfrac{1}{RC}$, so the time constant is found as the negative reciprocal of the gradient. Linearising in this way is the standard exam method because a straight-line fit is more reliable than reading a curve.

Try this

Q1. Define the time constant of a capacitor-resistor circuit. [1 mark]

• Cue. The time for the charge to fall to $\dfrac{1}{e}$ (about 37 percent) of its initial value, $\tau = RC$.

Q2. A capacitor discharges with a time constant of $4.0 \text{ s}$. What fraction of the initial charge remains after $4.0 \text{ s}$? [1 mark]

• Cue. $e^{-1} \approx 0.37$, so about 37 percent remains.

Q3. State how the half-life of a capacitor discharge relates to the time constant. [1 mark]

• Cue. $T_{1/2} = 0.69RC = \ln 2 \times \tau$.