Charging and Discharging CapacitorsCharging and Discharging Capacitors

Capacitors can be charged and discharged through a resistor. Meters with a data logger can be used to plot a graph of charge, p.d. or current against time.

When a capacitor charges:

Electrons flow onto the plate connected to the negative terminal of the power supply so current flows. As negative charge accumulates, it repels electrons off the plate connected to the positive terminal of the power supply. An equal and opposite charge accumulates on each plate, causing a potential difference.
But as charge builds up, electrostatic repulsion makes it increasingly difficult for electrons to be deposited on the negative plate.
Therefore, the voltage and charge stop increasing so quickly and level off and the current decreases. The current is 0 when the p.d. across the capacitor is equal to the e.m.f.
When a capacitor discharges, the opposite happens. The charge initially falls quickly as electrons are repelled from the negatively charged plate but as the negative charge becomes less electrons are repelled less strongly. Therefore, the charge decreases more and more slowly.

The time constant of a capacitor-resistor circuit is the time taken for the charge stored on

1 a capacitor to fall to of its original value. It is given by:

tau = CR

This is known as the constant ratio property (physical quantities decrease by the same factor in equal time periods).

Equations can be used to describe the exponential graphs associated with capacitor charge and discharge:

$I=frac{V}{R}=frac{Q/C}{R}=frac{Q}{CR}$

$I = -frac{Delta Q}{Delta t}$

$thereforefrac{Delta Q}{Delta t} =-frac{Q}{CR}$

The solution to this equation (for discharging capacitors) is derived by integration:

$frac{1}{Q}Delta Q=-frac{1}{CR}Delta t$

$intlimits_{0}^{Q}{frac{dQ}{left( {{Q}_{0}}-Q right)}}=intlimits_{0}^{t}{frac{dt}{CR}}=frac{1}{CR}$

$left| -ln left( {{Q}_{0}}-Q right) right|_{0}^{Q}=frac{1}{CR}left| t right|_{0}^{t}$

$ln left( {{Q}_{0}}-Q right)+ln {{Q}_{0}}=frac{t}{CR}ln(Q_{0}Q)+lnQ_{0}=CRt$

$ln(frac{Q}{Q_{o}})=-frac{t}{CR}$

$frac{Q}{Q_{o}}=-efrac{t}{CR}$

$Q=Q_{o}e-frac{t}{RC}$

Given $I=-frac{Q}{t} and V=frac{Q}{c},$ it follows that $I=I_{o}e-frac{t}{RC}$ and $V=V_{o}e-frac{t}{RC}$

Analogous equations for charging capacitors can easily be derived:

When charging, $I=I_{o}e-frac{t}{RC}$ in the capacitor and so the resistor. V=IR and R is constant, so $V_{R}=V_{o}efrac{t}{RC}$

$V_{o}=V_{R}+V_{c}$

$V_{c}=V_{o}-V_{R}=V_{o}-V_{o}e-frac{t}{RC}$

$=V_{o}(1-e-frac{t}{RC})$

As above, it follows that $Q=Q_{o}(1-e-frac{t}{RC})$

(Note that current still obeys the $I=I_{o}e-frac{t}{RC}$ equation for charging).

In summary:

Charging capacitors: the charge, p.d. and current on a discharging capacitor decreases exponentially. $x=x_{o}e-frac{t}{CR}$ where can be , I or .

Discharging capacitors: the current in a circuit with a discharging capacitor decreases as above but in the opposite direction. The charge and potential difference across the plates at a given time after a capacitor begins charging are

given by $x-x_{o}(1-efrac{t}{CR})$

where can be or .