Capacitors in d.c. circuits

A capacitor is a gap in a with space for on the 'plates' shown as the horizontal lines.

When a capacitor is charged, electrons on the lower plate repel from the upper plate, which then move to the positive terminal of the supply.

The applied $$V_{c}$$ to charge the capacitor (circuit 1 below) is measured with a and charge accumulated $$Q$$ is measured by removing the charged capacitor from circuit 1 and connecting it to a coulomb meter (circuit 2).

By varying the charging voltage and measuring the associated charge$$Q$$ a graph can be drawn.

The gradient of this graph is equal to the capacitance of the capacitor.

$C= \frac{Q}{V}$

And the area under the graph is the energy stored by the capacitor.

$E= \frac{1}{2}QV$

This can be combined with the equation for capacitance above to give two alternative equations for energy stored.

$$E= \frac{1}{2}\frac{Q^2}{C}$$ and

$E= \frac{1}{2}C{V^2}$

Question

A $$2200 \mu F$$ capacitor is charged up with a $$1.5 V$$ cell.

(a) What charge is stored?

(b) What energy is stored?

$C= 2200\mu F$

$=2200 \times {10^{ - 6}}F$

$V = 1.5 V$

$Q=?$

a) $$Q=CV$$

$=2200 \times {10^{ - 6}} \times 1.5$

$=0.0033C$

b) $$E= \frac{1}{2}C{V^2}$$

$= \frac{1}{2} \times 2200 \times {10^{-6}} \times {1.5^2}$

$= 0.00248J$

• The S.I. unit of charge is the Coulomb, $$C$$
• The S.I. unit of capacitance is the Farad, $$F$$
• A capacitor of capacitance 1 Farad will store 1 Coulomb of charge with the potenial difference across it is 1 volt