# Capacitor Charge and Discharge

## Capacitor Charge and Discharge Revision

**Capacitor Charge and Discharge**

For this unit it is important to be able to read and interpret the shapes of** charging** and **discharging **graphs for capacitors. For each we need to know the graphs of** current**, **potential difference** and **charge against time**.

**Charging Graphs**

As previously mentioned, **work is done** on the **electrons** in the circuit to overcome the **electrostatic forces **present in a **capacitor**. At the **positive plate**, **electrons** are attracted back towards the plate but the **potential difference** of the supply overcomes this force. Similarly at the** negative plate**, **electrons** from the circuit have to overcome the **repulsive forces** between the like charges.

As seen in the **current-time graph**, as the capacitor **charges**, the **current **decreases exponentially until it reaches zero. This is due to the forces acting within the** capacitor** increasing over time until they prevent electron flow.

The **potential difference** needs to increase over time exponentially as does **charge**. This is because of the build-up of **electrons** on the negative plate and the removal of **electrons **on the positive plate over time. The **potential difference** increases until it reaches an equal potential difference as the power supply. The **maximum charge** is determined by the rating of the capacitor.

**Discharging graphs**

When a **capacitor discharges**, it always discharges through a **resistor** when disconnected from the power supply (or the power supply is switched off).

As soon as the power supply is switched off and the **capacitor **is connected to the **resistor**, it rapidly **discharges **causing the electrons on the negative plate to return to the positive plate until the p.d reaches zero.

All three graphs for **discharging** are identical exponential decay curves.

As the **electrons** stop flowing from the negative plate to the positive plate, the **current**, **potential difference **and **charge** all return to zero. The **capacitor **is ready to be charged again by connecting back to the power supply.

To increase the** rate of discharge**, the **resistance **of the circuit should be reduced. This would be represented by a steeper gradient on the decay curve.

**The Time Constant**

The **time constant** of a **discharging capacitor **is the time taken for the **current**, **charge **or **potential difference **to decrease to 37 \% of the original amount. It can also be calculated for a charging capacitor to reach 63 \% of its **maximum charge** or **potential difference**.

The **time constant** \left(\tau\right) is proportional to the **resistance** and the **capacitance **of the capacitor. This can be represented in the equation:

\tau = RC

- \tau is the time constant in seconds \left(\text{s}\right)
- R is the resistance of the resistor connected to the capacitor in ohms \Omega
- C is the capacitance in the capacitor \left(\text{F}\right)

**Example:** Calculate the time constant for a 10 \: \text{nF} capacitor discharging through a 500 \: \text{k} \Omega resistor.

**[1 mark]**

\tau = RC

\tau = 500 \times 10^3 \times 10 \times 10^{-9} = 5 \times 10^{-3} \: \text{s} \: \text{or} \: 5 \: \text{ms}

Alternatively, the time constant can be found using a **voltage-time graph** for a **capacitor **by reading the time from the graph when the voltage of a **charging capacitor **reaches 63 \: \% or the voltage of a **discharging capacitor** reaches 37 \: \% .

As both charging and discharging graphs are** exponential graphs**, the 63 \: % (or 0.63) is the exponential function \left(e\right) whilst the 37 \: \% for a discharging capacitor is equal to \dfrac{1}{e}.

It is also useful to note that the time to **charge**/**discharge** to **half** the amount of **charg**e is given by:

T_{1/2} = 0.69 RC

**Charging and Discharging equations**

Using the** time constant** \left(\tau\right), we can calculate the amount of **current**, **charge **or** p.d** left after an amount of time **discharging**. The equations we use are:

I = I_0 e^{-t/RC}

- I is the
**current**in amps \left(\text{A}\right), I_0 is the**initial current**, t is the**time**and RC is the**time constant**.

V = V_0 e^{-t/RC}

- V is the voltage in volts \left(\text{V}\right), V_0 is the
**initial p.d**, t is the**time**and RC is the**time constant**.

Q = Q_0 e^{-t/RC}

- Q is the charge in coulombs \left(\text{C}\right), Q_0 is the
**initial charge**, t is the**time**and RC is the**time constant**.

For a capacitor charging, we can use the following equation:

Q = Q_0 \left(1-e^{-t/RC}\right)

where the elements of the equation have the same meaning, however Q_0 is the **maximum charge** of the capacitor.

**Example:** Calculate the current after 0.2 seconds for a 0.8 \: \text{mF} capacitor discharges from an initial current of 0.5 \: \text{A} through a 500 \: \Omega resistor.

**[3 mark]**

I = I_0 e^{-t/RC}

\text{Time Constant} \left(RC\right) = 500 \times 0.8 \times 10^{-3} = 0.4 \: \text{s}

I = 0.5 \times e^{-0.2/0.4}

I = 0.3 \: \text{A}

**Required Practical 9**

**Investigating Charging and Discharging Capacitors.**

This experiment will involve **charging** and **discharging **a capacitor, and using the data recorded to calculate the **capacitance **of the **capacitor**. It’s important to note that a large resistance resistor (such as a 10 \: \text{kΩ} resistor) is used to allow the **discharge** to be slow enough to measure readings at suitable time intervals.

We will measure the time taken for a capacitor to **discharge** in this method. A similar experiment could be conducted where you measure the time taken for the capacitor to **charge** instead.

**Doing the experiment:**

- Set up the circuit as shown, however ensure the switch is open (not connected to X or Y). You will also need a stopwatch for this experiment.
- The
**capacitor**at this stage should be fully discharged as no current has yet passed through the capacitor. - Set the power supply to 10 \: \text{V}.
- Move the switch to position X, which will begin charging the
**capacitor**. You can tell when the capacitor is fully charged when the voltmeter reading reads 10 \: \text{V}. - Once fully charged, the switch should be moved to position Y and the capacitor will begin
**discharging**. - Record the voltage on the voltmeter every 10 seconds until the capacitor has fully
**discharged**\left(0 \: \text{V}\right).

**Analysing the results:**

We know from earlier in this page that the equation for a **capacitor discharging** in terms of voltage is:

V = V_0 e^{-t/RC}

If we want to calculate the **capacitance **C using our results in the experiment, we need to rearrange this equation.

First if we take the natural log of both sides:

\text{ln}\left(\dfrac{V}{V_0}\right) = -\dfrac{t}{RC}

Using the log rules this can be written as:

\text{ln}\left(V\right) - \text{ln}\left(V_0\right) = -\dfrac{t}{RC}

\text{ln} \left(V\right) = -\dfrac{t}{RC} + \text{ln}\left(V_0\right)

If we compare this to the straight line equation:

\begin{aligned} \textcolor{2730e9}{y} &= \textcolor{f21cc2}{m}\textcolor{00d865}{x} + \textcolor{d11149}{c} \\ \textcolor{2730e9}{\text{ln} \left(V\right)} &= \textcolor{f21cc2}{-\dfrac{1}{RC}} \times \textcolor{00d865}{t} + \textcolor{d11149}{\text{ln}\left(V_0\right)} \end{aligned}So to calculate **capacitance**, you need to begin by plotting a graph of \text{ln}\left(V\right) against t as shown on the right.

\text{ln} \left(V\right) must be plotted on the y-axis and t must be plotted on the x-axis. This then means the **gradient **will be equal to \textcolor{f21cc2}{-\dfrac{1}{RC}}.

So then to calculate the **capacitance** of the **capacitor**, we simply rearrange for C using our calculated gradient:

\text{gradient} = \textcolor{f21cc2}{-\dfrac{1}{RC}}

C = -\dfrac{1}{R \times \text{gradient}}

## Capacitor Charge and Discharge Example Questions

**Question 1:** Describe the charging curves for current, charge and p.d of a capacitor over time.

**[3 marks]**

All three curves are **exponential** curves.

The **current-time graph shows an exponential decrease over time until zero current**.

The **charge-time and p.d-time graphs both show an exponential increase over time until the plateau**.

**Question 2:** What is the time constant of a discharging capacitor?

**[1 mark]**

Either:

**The time taken for the current, charge or p.d to decrease to 37\% of the original value.**-
**The time taken for the charge to reach 63 \% of its maximum charge or p.d.**

**Question 3:** Calculate the charge after 0.4 seconds for a 2.0 \: \text{mF} capacitor discharges from an initial charge of 0.8 \: \text{C} through a 1 \: \text{kΩ} resistor.

**[3 marks]**