Charging and discharging capacitors

What you'll learn

How a capacitor charges and discharges through a resistor.
What the time constant means, and why $τ=CR\tau = CR$ has units of seconds.
How to use exponential equations for charge, potential difference and current.
How to analyse capacitor data using meters, data-loggers, $ln⁡x\ln x$ graphs and spreadsheet models.

The essential starting point

A capacitor stores charge on two conducting plates separated by an insulator. When it is connected in a circuit, charge can move onto or off the plates, but charge does not pass through the insulating gap.

Definition

Capacitance

The capacitance $C$ of a capacitor is the charge stored per unit potential difference across it:

\frac{Q}{V}

where $C$ is capacitance in farads (F), $Q$ is charge in coulombs (C), and $V$ is potential difference in volts (V).

You will also use the circuit ideas:

$V = I R$ for a resistor.
Current is the rate of flow of charge, so $\frac{\Delta Q}{\Delta t}$ for a small time interval.
A resistor controls how quickly charge can move onto or off the capacitor plates.

Charging and discharging through a resistor

When a capacitor is charging, it is connected to a direct current supply through a resistor. At first the capacitor has little or no potential difference across it, so the current is large. As charge builds up, the capacitor potential difference increases, so the current gets smaller.

When a capacitor is discharging, the supply is removed and the charged capacitor is connected across a resistor. The capacitor now acts like a temporary source of energy, driving a current through the resistor. As it loses charge, its potential difference and current both decrease.

Charging and discharging capacitor circuits

Key Idea

Shape of the change

Charging and discharging are not linear processes. The rate of change is greatest at the start, then gradually decreases, giving an exponential curve.

What changes during charging?

For a capacitor charging from zero using a supply of potential difference $V_0$ :

capacitor charge $Q$ increases towards a maximum value $Q_0$ ;
capacitor potential difference $V_C$ increases towards $V_0$ ;
current $I$ starts at its maximum value and decreases towards zero.

What changes during discharging?

For a charged capacitor discharging through a resistor:

charge $Q$ decreases towards zero;
capacitor potential difference $V_C$ decreases towards zero;
current $I$ also decreases towards zero.

Common Mistake

Thinking the current stays constant

In an RC circuit, the current is not constant. The resistor value is constant, but the capacitor potential difference changes, so the current changes with time.

Investigating capacitor charge and discharge

In the laboratory, you can investigate these curves using either meters or a data-logger.

With meters, you usually connect a voltmeter across the capacitor and record $V_C$ at regular time intervals. Since $Q = CV_C$ , the voltage graph has the same shape as the charge graph. You can also measure current directly with an ammeter, but capacitor currents can change quickly, so voltage measurements are often easier.

With a data-logger, a voltage sensor records $V_C$ automatically at short time intervals. This is useful because the first part of the curve changes fastest.

Tip

Choosing component values

Pick $R$ and $C$ so the discharge lasts long enough to measure. A useful rule is that the main change takes about five time constants, so aim for $5 C R$ to be comfortably longer than your reaction time.

For reliable results:

use the correct capacitor polarity for electrolytic capacitors;
discharge the capacitor fully before repeating a charging run;
use a voltmeter or sensor with high resistance so it does not significantly discharge the capacitor;
take more frequent readings near the start of the curve.

The time constant

Definition

Time constant

The time constant $τ\tau$ of a capacitor-resistor circuit is

τ=CR\tau = CR

where $C$ is capacitance in farads (F), $R$ is resistance in ohms (Ω), and $τ\tau$ is time in seconds (s).

The unit works because one ohm-farad is one second:

F=VA×CV=CA=s\Omega \,\text{F} = \frac{\text{V}}{\text{A}} \times \frac{\text{C}}{\text{V}} = \frac{\text{C}}{\text{A}} = \text{s}

After one time constant:

during discharging, $Q$ , $V_C$ or $I$ has fallen to about $0.37$ of its initial value;
during charging, $Q$ or $V_C$ has reached about $0.63$ of its final value.

After about five time constants, the capacitor is usually treated as almost fully charged or almost fully discharged.

Example

Finding the time constant and measurement time

A 220 µF capacitor is connected in series with a 47 kΩ resistor. Estimate the time constant and the time needed for the capacitor to be nearly fully discharged.

Convert the component values into SI units:
$220\,\mu\text{F} = 220 \times 10^{-6}\,\text{F}$ $47\,\text{k}\Omega = 47 \times 10^3\,\Omega$
Use $τ=CR\tau = CR$ :
$s\tau = \left(220 \times 10^{-6}\,\text{F}\right)\left(47 \times 10^3\,\Omega\right) = 10.34\,\text{s}$
Estimate the practical discharge time as about $5τ5\tau$ :
$s5\tau = 5 \times 10.34\,\text{s} = 51.7\,\text{s}$
So the capacitor is nearly discharged after about 52 s.

Exponential equations for RC circuits

OCR uses equations of the form:

x = x_0 e^{-t/CR}

for discharge, and

x_0\left(1 - e^{-t/CR}\right)

for charging.

Here $x$ can represent charge $Q$ , capacitor potential difference $V_C$ , or current $I$ , depending on the situation. The symbol $x_0$ means the starting value for a discharge, or the final maximum value for a charging curve.

For a discharging capacitor, the named equation is:

Q = Q_0 e^{-t/CR}

Capacitor exponential charging and discharging curves

Example

Calculating a charging voltage

A capacitor charges through a resistor from a 6.0 V supply. The circuit has $47\,\text{k}\Omega$ and $220\,\mu\text{F}$ . Find the capacitor potential difference after 15.0 s.

Calculate the time constant:
$\left(47 \times 10^3\,\Omega\right)\left(220 \times 10^{-6}\,\text{F}\right) = 10.34\,\text{s}$
Choose the charging equation because the capacitor potential difference is increasing towards 6.0 V:
$VC=V0(1−e−t/CR)V_C = V_0\left(1 - e^{-t/CR}\right)$
Substitute the values:
$s)V_C = 6.0\,\text{V}\left(1 - e^{-15.0\,\text{s}/10.34\,\text{s}}\right)$ $VV_C = 6.0\,\text{V}\left(1 - e^{-1.45}\right) = 4.6\,\text{V}$

Common Mistake

Using the wrong exponential form

Use $x = x_0e^{-t/CR}$ for a quantity that is decreasing. Use $x = x_0(1 - e^{-t/CR})$ for a quantity that starts at zero and increases towards a maximum.

Using a $ln⁡x\ln x$ against $t$ graph

For a discharging capacitor:

x = x_0 e^{-t/CR}

Taking natural logs gives:

ln⁡x=ln⁡x0−tCR\ln x = \ln x_0 - \frac{t}{CR}

This has the straight-line form $y = m x + c$ :

vertical axis: $ln⁡x\ln x$ ;
horizontal axis: $t$ ;
gradient: $−1CR-\frac{1}{CR}$ ;
intercept: $ln x_0$ .

So if you plot $ln V_C$ against time for a discharge, the gradient lets you determine $C R$ .

Example

Finding CR from a logarithmic graph

A plot of $ln V_C$ against $t$ for a discharging capacitor gives a straight line with gradient $s−1-0.125\,\text{s}^{-1}$ . Find the time constant.

Match the measured gradient to the equation:
$gradient=−1CR\text{gradient} = -\frac{1}{CR}$
Rearrange for $C R$ :
$-\frac{1}{\text{gradient}}$
Substitute the gradient:
$-\frac{1}{-0.125\,\text{s}^{-1}} = 8.00\,\text{s}$
The time constant is 8.00 s.

Tip

Logs and zero readings

You cannot take $ln⁡0\ln 0$ . For a logarithmic graph, ignore readings that have reached zero or are dominated by sensor noise.

Spreadsheet modelling of discharge

For a discharging capacitor, OCR gives:

ΔQΔt=−QCR\frac{\Delta Q}{\Delta t} = -\frac{Q}{CR}

This says the rate of loss of charge is proportional to the charge remaining. The negative sign means $Q$ is decreasing.

In a spreadsheet, use small time steps. For each row:

ΔQ=−QCRΔt\Delta Q = -\frac{Q}{CR}\Delta t

then

Qnext=Q+ΔQQ_{\text{next}} = Q + \Delta Q

This is a numerical model: it estimates the curve step by step. Smaller time steps usually give a better approximation to the true exponential curve.

Example

Modelling one spreadsheet step

A capacitor has $24.0\,\text{mC}$ at the start of a row. The time constant is $10.0\,\text{s}$ and the spreadsheet time step is $s\Delta t = 1.00\,\text{s}$ . Find the next charge value.

Convert the charge into coulombs:
$24.0\,\text{mC} = 24.0 \times 10^{-3}\,\text{C}$
Calculate the change in charge during the time step:
$C\Delta Q = -\frac{24.0 \times 10^{-3}\,\text{C}}{10.0\,\text{s}} \times 1.00\,\text{s} = -2.40 \times 10^{-3}\,\text{C}$
Add the change to the old charge:
$CQ_{\text{next}} = 24.0 \times 10^{-3}\,\text{C} - 2.40 \times 10^{-3}\,\text{C} = 21.6 \times 10^{-3}\,\text{C}$
So the next spreadsheet value is 21.6 mC.

Common Mistake

Time step too large

If $Δt\Delta t$ is too large compared with $C R$ , the spreadsheet model becomes a poor approximation and can even predict unphysical negative charge values.

Constant-ratio property of exponential decay

An exponential decay has a constant-ratio property: in equal time intervals, the quantity is multiplied by the same factor.

For capacitor discharge:

Δtxbefore=e−Δt/CR\frac{x_{\text{after time } \Delta t}}{x_{\text{before}}} = e^{-\Delta t/CR}

So the graph gets flatter, but the percentage decrease over each equal time interval stays the same.

Example

Using the constant-ratio property

A capacitor discharges with time constant 4.0 s. The initial potential difference is 8.0 V. Find the potential difference after each 2.0 s interval for the first 6.0 s.

Calculate the multiplying factor for each 2.0 s interval:
$s=e−0.50=0.607e^{-\Delta t/CR} = e^{-2.0\,\text{s}/4.0\,\text{s}} = e^{-0.50} = 0.607$
Apply the same factor repeatedly:
$VV_1 = 8.0\,\text{V} \times 0.607 = 4.86\,\text{V}$ $VV_2 = 4.86\,\text{V} \times 0.607 = 2.95\,\text{V}$ $VV_3 = 2.95\,\text{V} \times 0.607 = 1.79\,\text{V}$
Link the results to the graph shape: the drops are 3.14 V, then 1.91 V, then 1.16 V, so the absolute decrease gets smaller even though the ratio stays constant.

Exam technique

In the exam

Decide whether the quantity is charging or discharging before choosing the exponential equation.
Calculate $C R$ first, with $R$ in ohms and $C$ in farads.
For $ln⁡x\ln x$ graphs, remember that the gradient is negative and equals $−1CR-\frac{1}{CR}$ .
In practical questions, link better data to short sampling intervals, high-resistance voltmeters, repeated runs and suitable uncertainty treatment.

Self review

Check yourself

Why is the current largest at the very start of charging or discharging?
What fraction of the initial charge remains after one time constant during discharge?
How would you find $C R$ from a straight-line graph of $ln V_C$ against time?

What you'll learn

How a capacitor charges and discharges through a resistor.
What the time constant means, and why $τ=CR\tau = CR$ has units of seconds.
How to use exponential equations for charge, potential difference and current.
How to analyse capacitor data using meters, data-loggers, $ln⁡x\ln x$ graphs and spreadsheet models.

The essential starting point

Definition

Capacitance

The capacitance $C$ of a capacitor is the charge stored per unit potential difference across it:

\frac{Q}{V}

where $C$ is capacitance in farads (F), $Q$ is charge in coulombs (C), and $V$ is potential difference in volts (V).

You will also use the circuit ideas:

$V = I R$ for a resistor.
Current is the rate of flow of charge, so $\frac{\Delta Q}{\Delta t}$ for a small time interval.
A resistor controls how quickly charge can move onto or off the capacitor plates.

Charging and discharging through a resistor

Charging and discharging capacitor circuits

Key Idea

Shape of the change

Charging and discharging are not linear processes. The rate of change is greatest at the start, then gradually decreases, giving an exponential curve.

What changes during charging?

For a capacitor charging from zero using a supply of potential difference $V_0$ :

capacitor charge $Q$ increases towards a maximum value $Q_0$ ;
capacitor potential difference $V_C$ increases towards $V_0$ ;
current $I$ starts at its maximum value and decreases towards zero.

What changes during discharging?

For a charged capacitor discharging through a resistor:

charge $Q$ decreases towards zero;
capacitor potential difference $V_C$ decreases towards zero;
current $I$ also decreases towards zero.

Common Mistake

Thinking the current stays constant

In an RC circuit, the current is not constant. The resistor value is constant, but the capacitor potential difference changes, so the current changes with time.

Investigating capacitor charge and discharge

In the laboratory, you can investigate these curves using either meters or a data-logger.

With a data-logger, a voltage sensor records $V_C$ automatically at short time intervals. This is useful because the first part of the curve changes fastest.

Tip

Choosing component values

For reliable results:

use the correct capacitor polarity for electrolytic capacitors;
discharge the capacitor fully before repeating a charging run;
use a voltmeter or sensor with high resistance so it does not significantly discharge the capacitor;
take more frequent readings near the start of the curve.

The time constant

Definition

Time constant

The time constant $τ\tau$ of a capacitor-resistor circuit is

τ=CR\tau = CR

where $C$ is capacitance in farads (F), $R$ is resistance in ohms (Ω), and $τ\tau$ is time in seconds (s).

The unit works because one ohm-farad is one second:

F=VA×CV=CA=s\Omega \,\text{F} = \frac{\text{V}}{\text{A}} \times \frac{\text{C}}{\text{V}} = \frac{\text{C}}{\text{A}} = \text{s}

After one time constant:

during discharging, $Q$ , $V_C$ or $I$ has fallen to about $0.37$ of its initial value;
during charging, $Q$ or $V_C$ has reached about $0.63$ of its final value.

After about five time constants, the capacitor is usually treated as almost fully charged or almost fully discharged.

Example

Finding the time constant and measurement time

A 220 µF capacitor is connected in series with a 47 kΩ resistor. Estimate the time constant and the time needed for the capacitor to be nearly fully discharged.

Convert the component values into SI units:
$220\,\mu\text{F} = 220 \times 10^{-6}\,\text{F}$ $47\,\text{k}\Omega = 47 \times 10^3\,\Omega$
Use $τ=CR\tau = CR$ :
$s\tau = \left(220 \times 10^{-6}\,\text{F}\right)\left(47 \times 10^3\,\Omega\right) = 10.34\,\text{s}$
Estimate the practical discharge time as about $5τ5\tau$ :
$s5\tau = 5 \times 10.34\,\text{s} = 51.7\,\text{s}$
So the capacitor is nearly discharged after about 52 s.

Exponential equations for RC circuits

OCR uses equations of the form:

x = x_0 e^{-t/CR}

for discharge, and

x_0\left(1 - e^{-t/CR}\right)

for charging.

For a discharging capacitor, the named equation is:

Q = Q_0 e^{-t/CR}

Capacitor exponential charging and discharging curves

Example

Calculating a charging voltage

A capacitor charges through a resistor from a 6.0 V supply. The circuit has $47\,\text{k}\Omega$ and $220\,\mu\text{F}$ . Find the capacitor potential difference after 15.0 s.

Calculate the time constant:
$\left(47 \times 10^3\,\Omega\right)\left(220 \times 10^{-6}\,\text{F}\right) = 10.34\,\text{s}$
Choose the charging equation because the capacitor potential difference is increasing towards 6.0 V:
$VC=V0(1−e−t/CR)V_C = V_0\left(1 - e^{-t/CR}\right)$
Substitute the values:
$s)V_C = 6.0\,\text{V}\left(1 - e^{-15.0\,\text{s}/10.34\,\text{s}}\right)$ $VV_C = 6.0\,\text{V}\left(1 - e^{-1.45}\right) = 4.6\,\text{V}$

Common Mistake

Using the wrong exponential form

Use $x = x_0e^{-t/CR}$ for a quantity that is decreasing. Use $x = x_0(1 - e^{-t/CR})$ for a quantity that starts at zero and increases towards a maximum.

Using a $ln⁡x\ln x$ against $t$ graph

For a discharging capacitor:

x = x_0 e^{-t/CR}

Taking natural logs gives:

ln⁡x=ln⁡x0−tCR\ln x = \ln x_0 - \frac{t}{CR}

This has the straight-line form $y = m x + c$ :

vertical axis: $ln⁡x\ln x$ ;
horizontal axis: $t$ ;
gradient: $−1CR-\frac{1}{CR}$ ;
intercept: $ln x_0$ .

So if you plot $ln V_C$ against time for a discharge, the gradient lets you determine $C R$ .

Example

Finding CR from a logarithmic graph

A plot of $ln V_C$ against $t$ for a discharging capacitor gives a straight line with gradient $s−1-0.125\,\text{s}^{-1}$ . Find the time constant.

Match the measured gradient to the equation:
$gradient=−1CR\text{gradient} = -\frac{1}{CR}$
Rearrange for $C R$ :
$-\frac{1}{\text{gradient}}$
Substitute the gradient:
$-\frac{1}{-0.125\,\text{s}^{-1}} = 8.00\,\text{s}$
The time constant is 8.00 s.

Tip

Logs and zero readings

You cannot take $ln⁡0\ln 0$ . For a logarithmic graph, ignore readings that have reached zero or are dominated by sensor noise.

Spreadsheet modelling of discharge

For a discharging capacitor, OCR gives:

ΔQΔt=−QCR\frac{\Delta Q}{\Delta t} = -\frac{Q}{CR}

This says the rate of loss of charge is proportional to the charge remaining. The negative sign means $Q$ is decreasing.

In a spreadsheet, use small time steps. For each row:

ΔQ=−QCRΔt\Delta Q = -\frac{Q}{CR}\Delta t

then

Qnext=Q+ΔQQ_{\text{next}} = Q + \Delta Q

This is a numerical model: it estimates the curve step by step. Smaller time steps usually give a better approximation to the true exponential curve.

Example

Modelling one spreadsheet step

A capacitor has $24.0\,\text{mC}$ at the start of a row. The time constant is $10.0\,\text{s}$ and the spreadsheet time step is $s\Delta t = 1.00\,\text{s}$ . Find the next charge value.

Convert the charge into coulombs:
$24.0\,\text{mC} = 24.0 \times 10^{-3}\,\text{C}$
Calculate the change in charge during the time step:
$C\Delta Q = -\frac{24.0 \times 10^{-3}\,\text{C}}{10.0\,\text{s}} \times 1.00\,\text{s} = -2.40 \times 10^{-3}\,\text{C}$
Add the change to the old charge:
$CQ_{\text{next}} = 24.0 \times 10^{-3}\,\text{C} - 2.40 \times 10^{-3}\,\text{C} = 21.6 \times 10^{-3}\,\text{C}$
So the next spreadsheet value is 21.6 mC.

Common Mistake

Time step too large

If $Δt\Delta t$ is too large compared with $C R$ , the spreadsheet model becomes a poor approximation and can even predict unphysical negative charge values.

Constant-ratio property of exponential decay

An exponential decay has a constant-ratio property: in equal time intervals, the quantity is multiplied by the same factor.

For capacitor discharge:

Δtxbefore=e−Δt/CR\frac{x_{\text{after time } \Delta t}}{x_{\text{before}}} = e^{-\Delta t/CR}

So the graph gets flatter, but the percentage decrease over each equal time interval stays the same.

Example

Using the constant-ratio property

A capacitor discharges with time constant 4.0 s. The initial potential difference is 8.0 V. Find the potential difference after each 2.0 s interval for the first 6.0 s.

Calculate the multiplying factor for each 2.0 s interval:
$s=e−0.50=0.607e^{-\Delta t/CR} = e^{-2.0\,\text{s}/4.0\,\text{s}} = e^{-0.50} = 0.607$
Apply the same factor repeatedly:
$VV_1 = 8.0\,\text{V} \times 0.607 = 4.86\,\text{V}$ $VV_2 = 4.86\,\text{V} \times 0.607 = 2.95\,\text{V}$ $VV_3 = 2.95\,\text{V} \times 0.607 = 1.79\,\text{V}$
Link the results to the graph shape: the drops are 3.14 V, then 1.91 V, then 1.16 V, so the absolute decrease gets smaller even though the ratio stays constant.

Exam technique

In the exam

Decide whether the quantity is charging or discharging before choosing the exponential equation.
Calculate $C R$ first, with $R$ in ohms and $C$ in farads.
For $ln⁡x\ln x$ graphs, remember that the gradient is negative and equals $−1CR-\frac{1}{CR}$ .
In practical questions, link better data to short sampling intervals, high-resistance voltmeters, repeated runs and suitable uncertainty treatment.

Self review

Check yourself

Why is the current largest at the very start of charging or discharging?
What fraction of the initial charge remains after one time constant during discharge?
How would you find $C R$ from a straight-line graph of $ln V_C$ against time?

What you'll learn

The essential starting point

Charging and discharging through a resistor

What changes during charging?

What changes during discharging?

Investigating capacitor charge and discharge

The time constant

Exponential equations for RC circuits

Using a $ln⁡x\ln x$ against $t$ graph

Spreadsheet modelling of discharge

Constant-ratio property of exponential decay

Review Charging and discharging capacitors by teaching Genie

Remember key concepts with flashcards

Charging and discharging capacitors

What you'll learn

The essential starting point

Charging and discharging through a resistor

What changes during charging?

What changes during discharging?

Investigating capacitor charge and discharge

The time constant

Exponential equations for RC circuits

Using a $ln⁡x\ln x$ against $t$ graph

Spreadsheet modelling of discharge

Constant-ratio property of exponential decay

Review Charging and discharging capacitors by teaching Genie

Remember key concepts with flashcards

Charging and discharging capacitors

What you'll learn

The essential starting point

Charging and discharging through a resistor

What changes during charging?

What changes during discharging?

Investigating capacitor charge and discharge

The time constant

Exponential equations for RC circuits

Using a ln⁡x\ln xlnx against ttt graph

Spreadsheet modelling of discharge

Constant-ratio property of exponential decay

Review Charging and discharging capacitors by teaching Genie

Remember key concepts with flashcards

Charging and discharging capacitors Revision Guide

Charging and discharging capacitors

What you'll learn

The essential starting point

Charging and discharging through a resistor

What changes during charging?

What changes during discharging?

Investigating capacitor charge and discharge

The time constant

Exponential equations for RC circuits

Using a ln⁡x\ln xlnx against ttt graph

Spreadsheet modelling of discharge

Constant-ratio property of exponential decay

Review Charging and discharging capacitors by teaching Genie

Remember key concepts with flashcards

Charging and discharging capacitors Revision Guide

Using a $ln⁡x\ln x$ against $t$ graph

Using a $ln⁡x\ln x$ against $t$ graph