Capacitance and the charge stored, the energy stored in a capacitor, and the charging and discharging of a capacitor through a resistor.

What this key area is asking

The SQA wants you to define capacitance and calculate the charge stored, calculate the energy stored in a capacitor, and describe and explain how the current and the voltage change as a capacitor charges and discharges through a resistor.

Capacitance and charge

For a given capacitor the charge stored is directly proportional to the voltage across it, so a graph of $Q$ against $V$ is a straight line through the origin with gradient equal to the capacitance $C$. Real capacitors used in the lab are usually in the microfarad ($\mu\text{F} = 10^{-6}$ F) range, so always convert the prefix before substituting.

Energy stored

The energy stored equals the area under a charge-voltage graph. Because the voltage rises from zero as charge accumulates, the average voltage during charging is $\tfrac{1}{2}V$, and the work done in moving the charge is $\tfrac{1}{2}QV$. This is the reason the factor of one half appears, and it is also why charging a capacitor through a resistor wastes exactly half of the energy taken from the supply as heat in the resistor.

Charging and discharging through a resistor

While charging through a resistor, the current starts at a maximum and falls towards zero, while the capacitor voltage rises from zero towards the supply voltage. While discharging, the current and the capacitor voltage both fall from their maximum towards zero (and the current flows in the opposite direction). A larger resistance or capacitance makes both processes take longer, because the larger resistance limits the charging current and the larger capacitance needs more charge to reach a given voltage.

The instantaneous charging current is set by the net driving voltage across the resistor, $I = \frac{V_{\text{supply}} - V_C}{R}$, where $V_C$ is the voltage already across the capacitor. At the start $V_C = 0$ so $I_{\max} = \frac{V_{\text{supply}}}{R}$; as $V_C$ rises towards the supply value the current falls towards zero.

Examples in context

A camera flash stores energy in a large capacitor charged slowly from the battery, then dumps it through the flash tube in a few milliseconds, delivering a high power burst the battery alone could not. A smoothing capacitor in a power supply charges on the peaks of the rectified AC and discharges between peaks to hold the output voltage steady. A defibrillator charges a capacitor to a few kilovolts so that the stored energy $E = \tfrac{1}{2}CV^2$ can be released through the patient in a controlled pulse. In each case the time to charge or discharge is governed by the resistance in series with the capacitor, which is why an RC combination is used as a timing element.

Try this

Q1. A $100\ \mu\text{F}$ capacitor is charged to $12\ \text{V}$. Calculate the charge stored. [2 marks]

• Cue. $Q = CV = 100 \times 10^{-6} \times 12 = 1.2 \times 10^{-3}$ C.

Q2. State how the current changes while a capacitor charges through a resistor. [1 mark]

• Cue. It starts at a maximum and decreases towards zero.

Q3. A $2200\ \mu\text{F}$ capacitor stores $0.10\ \text{J}$. Calculate the voltage across it. [3 marks]

• Cue. Rearrange $E = \tfrac{1}{2}CV^2$ to give $V = \sqrt{\frac{2E}{C}} = \sqrt{\frac{2 \times 0.10}{2200 \times 10^{-6}}} = 9.5$ V.