Charging and discharging capacitors

In a nutshell

A charging capacitor charges very quickly initially and then slows down as it gets near to full. Similar to the potential difference. The current starts off initially high but then slowly decreases until it is fully charged where no current flows.

A discharging capacitor loses its charge and potential difference very quickly and then the rate slows down as the capacitor loses its charge and potential difference. Initially the current is high as and then slowly decreases until no more potential difference remains across the plates.

Equations

Description	Equation
Discharging equations for a capacitor	$I=I_0 e^{-\frac t{CR}} \newline \\[0.1in]V=V_0 e^{-\frac t{CR}} \newline \\[0.1in]Q=Q_0 e^{-\frac t{CR}} \newline \\[0.1in]$
Charging equations for a capacitor	$I=I_0 e^{-\frac t{CR}} \newline \\[0.1in]V=V_0 (1-e^{-\frac t{CR}}) \newline \\[0.1in]Q=Q_0 (1-e^{-\frac t{CR}}) \newline \\[0.1in]$
Time constant	$\tau = CR$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$maximum \ value$	$I_0 \\ V_0 \\ Q_0$	$A \\ V \\ C$	$A \\ kg \, m^2\, s^{-3}\, A^{-1} \\ A\, s$
$value \ at \ time \ t$	$I \\ V \\ Q$	$A \\ V \\ C$	$A \\ kg \, m^2\, s^{-3}\, A^{-1} \\ A\, s$
$time \ elapsed$	$t$	$s$	$s$
$capacitance$	$C$	$F$	$kg^{-1} \, m^{-2}\, s^{4}\, A^{2}$
$resistance$	$R$	$\Omega$	$kg \, m^2\, s^{-3}\, A^{-2}$
$time \ constant$	$\tau$	$s$	$s$

How capacitors charge

Consider the following circuit:

Physics; Capacitance; KS5 Year 12; Charging and discharging capacitors

When the switch is in position P, the capacitor is connected to the battery and will be charging. When the capacitor first starts charging there are an equal amount of positive and negative charges on both of the plates.

Electrons will leave the negative terminal of the battery and flow clockwise onto plate X. Each electron gained by plate X will repel an electron from plate Y which is then attracted to the positive terminal of the battery, leaving plate Y slightly positive.

Each electron deposited onto plate X makes it more negative, which means it becomes exponentially more difficult for electrons to be deposited.

When the potential difference across the plates is equal to the emf of the battery, the capacitor is fully charged and no more electrons can be deposited.

Note: Capacitors can only charge up until the potential difference across the capacitor is equal to the value of the emf of the power supply or battery which is charging them.

When the switch is moved to position Q the capacitor discharges, as the electrons from plate X are attracted to the positive plate Y. The electrons can't pass through the dielectric between the plates.

The only pathway available for the electrons is through the fixed resistor and onto plate Y.

Initially there is a large potential difference which means the current is large initially. As plate X and Y lose their negative and positive charges respectively, the attractiveness gets less and current starts to decrease until it falls to zero when the plates are no longer charged.

Capacitor graphs

Charging capacitors

For a charging capacitor, the graphs for current, potential difference and charge are the following:

	Initially the current is high, as there is no charge on the capacitor plates. Therefore, it is easy for electrons to be deposited and lots of electrons flow.
	The potential difference is initially zero as both of the plates are not charged. As more electrons are added the potential difference increases. Initially this is quick, due to the ease of being able to add the electrons to a neutral plate, but then slows down as the plates become more charged.
	The charge at the start is initially zero and increases quickly as the electrons are deposited on the plates rapidly at the start. It then starts to slow down as it becomes more difficult for the electrons to be deposited on the already negative plate.

The equations for a charging capacitor are:

$I=I_0 e^{-\frac t{CR}} \newline \\[0.1in]V=V_0 (1-e^{-\frac t{CR}}) \newline \\[0.1in]Q=Q_0 (1-e^{-\frac t{CR}}) \newline \\[0.1in]$

Discharging capacitors

	Initially the plates are very charged and there is a large attraction between the positive plate and the electrons. This means lots of the electrons travel very quickly. and therefore the current is very large. As the attractiveness decreases, the electrons travel less and more slowly. The rate at which the current decreases slows down until the plates are equally charged and no more current flows.
	Initially the potential difference across the plates is at a maximum. It decreases very rapidly due to the large current at the start and lots of electrons moving away from the negative plate. The rate of electrons leaving the plates will decrease, which means the rate of potential difference loss will also decrease.
	Initially the charge on the plates is at a maximum and decreases very rapidly due to the large amount of electrons moving away from the negative plate. When the potential difference falls to zero, there will be no more current and the plates will no longer be charged.

The equations for a discharging capacitor are:

$I=I_0 e^{-\frac t{CR}} \newline \\[0.1in]V=V_0 e^{-\frac t{CR}} \newline \\[0.1in]Q=Q_0 e^{-\frac t{CR}} \newline \\[0.1in]$

Note: All of the equations for a discharging capacitor are the same as initially they are all at maximum and exponentially decay.

Example

A $460 \, mF$ capacitor is charged using a $15\, V$ battery. It is then discharged through a $680 \, \Omega$ resistor. What is the potential difference of the capacitor after $5\, minutes$ .

Firstly, write down the known values:

$C=460 \, mF = 460 \times 10^{-3} \, F \newline \\[0.1in] V_0=15 \, V\newline \\[0.1in] R=680 \, \Omega \newline \\[0.1in] t=5\, min=300 \, s$

Next, write down the equations needed and rearrange if necessary:

$V=V_0 e^{-\frac t{CR}}$

Then, substitute the values into the equation:

$V=15e^{-\frac {300}{460\times 10^{-3}\times 680}} \newline \\[0.1in]V=5.74868...$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$potential \ difference , \ V=5.7\, V$

The potential difference across the capacitor after $5 \, minutes$ is $\underline{5.7\, V}.$ 

Time constant

The time taken to charge or discharge a capacitor depends on two factors, the capacitance of the capacitor and the resistance of the circuit.

The time constant is given as:

$\tau =CR$

Substituting the time constant for time in the exponential equations for charging current:

$I=I_0 e^{-\frac t{CR}} \newline \\[0.1in]I=I_0 e^{-\frac {\tau}{CR}} \newline \\[0.1in]I=I_0 e^{-\frac {CR}{CR}} \newline \\[0.1in]I=I_0 e^{-1}$

Dividing by $I_0$ :

$\\[0.1in]\frac I{I_0}=e^{-1} \newline \\[0.1in]\frac I{I_0}=\frac 1 e$

Substituting in the value for $e$ :

$\newline \\[0.1in]\frac I{I_0}=\frac 1 {2.71828...} \newline \\[0.1in]\frac I{I_0}\approx 0.37 \newline \\[0.1in]$

Therefore, the time constant is the time taken for the current on a charging or discharging capacitor to fall to $37 \%$ of its initial value.

This is the same for the all of the discharging quantities, current, potential difference and charge.

For the potential difference and charge for a charging capacitor, the time constant is the time taken for the capacitor to charge to $63\%$ of its maximum value.

Note: In practice, it takes approximately $5\tau$ for a capacitor to fully charge or discharge.