# How a Capacitor Discharge

### How does a capacitor work?

When a capacitor is connected across a power source, it starts charging up as current flows in the circuit.

Initially when the capacitor has not stored any charge, its potential difference is 0 V. As current flows, charges start storing across its plates, and the potential difference increases. The current also starts to decrease because the capacitor reduces the overall e.m.f of the circuit. When the potential difference of the capacitor equals to the e.m.f. of the power source, current stops flowing.

After the capacitor is fully charged, the switches are set to the figure above. The capacitor starts discharging and a current flows in the left part of the circuit. The potential difference across the capacitor follows an exponential curve.

### Wolfram Demonstration

Click this Wolfram Demonstration Project link to access the simulation. This simulation shows how the potential difference across the capacitor decreases during a discharge. The rate of discharge is dependent on the capacitance of the capacitor. A larger capacitor stores more charge. Although the maximum voltage across it is dependent on the power source, a large capacitor produces a larger current.

Procedure:

1. Adjust all the controls to the left such that d = 0.4, L = 1.5 and time = 0.
1. Observe that the capacitor discharges almost all its stored charge after about 50 s.
2. The capacitor has a capacitance of $0.481\mu \text{F}$ and its maximum stored charge is $21.6 \mu \text{C}$
2. Increase the separation d to 0.7.
1. Observe that the capacitance is now $0.275\mu \text{F}$ and maximum charge $12.4 \mu \text{C}$. This means that increasing the distance between the plates decreases the capacitance, and correspondingly the charge stored, since $Q=CV$ .
2. Lesser stored charge means that the current stops flowing after about 30 s.
3. Increase the length of side to 3.0.
1. Observe that the capacitance is now $1.1 \mu \text{F}$ and that the charge stored is now $49.4 \mu \text{C}$ . Increasing the cross-sectional area increases the capacitance and the stored charge.
2. Since the stored charge is larger, the capacitor takes a longer time to discharge to 0 V.

### Summary

1. Capacitance depends on the physical property. The capacitance is larger is the distance between the plates is small and the cross-sectional is large.
2. A large capacitor stores a larger charge, and takes a longer time to discharge. Conversely, it should take a larger time to charge a large capacitor to its maximum voltage.

# Capacitance

### What is a capacitor?

A capacitor is an electrical component that stores charge. It is usually made by having two parallel plates thinly separated by an insulating material. This material is also known as a dielectric. Current flows through an insulator if the voltage across it is sufficient. In a capacitor, the conducting plates are close enough for the current to flow, even through it is separated by the insulating dielectric. However, because of the dielectric, some charges are stored on the plates as the current flows, resulting in higher potential difference as time passes. When the potential difference across the capacitor equals to the e.m.f of the circuit, current stops flowing.

### Physics of capacitance

Capacitance is defined as

Capacitance is the ratio of the charge stored to potential difference across a capacitor.

Mathematically,

\begin{aligned}C=\frac{Q}{V} \end{aligned}

The SI  unit of capacitance is farad (F).

A capacitor of 1 farad has a stored charge of 1 C when the potential difference across it is 1 V.

### Combined Capacitance

Capacitors can be connected either in series or parallel.

In a series connection, the combined capacitance can be found from the relationship

\begin{aligned} V_\text{total} &= V_1 + V_2 + ...\\\frac{Q}{C_\text{total}} &= \frac{Q}{C_1}+\frac{Q}{C_2}+...\end{aligned}

Since the charge $Q$ stored in each capacitance is the same,

\begin{aligned}\frac{1}{C_\text{total}} &= \frac{1}{C_1}+\frac{1}{C_2}+...\end{aligned}

In a parallel connection, the combined resistance is derived as follow:

\begin{aligned} Q_\text{total}&=Q_1+Q_2+...\\C_\text{total}V&=C_1V+C_2V+...\\\end{aligned}

Since the potential difference $V$ is the same in a parallel circuit,

\begin{aligned} C_\text{total}=C_1+C_2+... \end{aligned}

## Summary

1. Capacitance is the ratio of the charge stored to the potential difference across the component.
2. SI unit of capacitance is farad (F).
3. The total capacitance of series connected capacitors is \begin{aligned} \frac{1}{C_\text{total}}=\frac{1}{C_1}+\frac{1}{C_2}+... \end{aligned}
4. The total capacitance of parallel connected capacitors is \begin{aligned} C_\text{total} = C_1+C_2+... \end{aligned}

## Review

### Question 1

The figure shows three capacitors connected in series with a cell of e.m.f. 3.0 V.

Calculate the p.d. across each capacitor.

### Solution

Since the charge stored in each capacitor is the same as the charge stored across all the capacitors, we shall go ahead and find the charge stored in the combined capacitors and then use it to calculate the potential difference across each capacitor.

\begin{aligned} \frac{1}{C_\text{total}} &= \frac{1}{100 \times 10^{-3}}+\frac{1}{200 \times10^{-3}} + \frac{1}{400 \times 10^{-3}}\\C_\text{total}&= 0.0571\text{ F}\\ Q_\text{total}&=3.0 \times 0.0571\\&=0.1713\text{ C}\end{aligned}

Hence,

\begin{aligned} V_1&=\frac{0.1713}{100 \times 10^{-3}}\\&=1.713\text{ V}\\V_2&=\frac{0.1713}{200 \times 10^{-3}}\\&=0.857\text{ V} \\V_3&=\frac{0.1713}{400 \times 10^{-3}}\\&=0.428\text{ V}\end{aligned}

You can observe that the sum of all the three p.d. is the e.m.f. of the cell.

# Work done in a Capacitor

Capacitor is an electrical component that stores charge. As charges are stored, potential energy in the capacitor also increases. In this post, I would like to explain how to calculate the energy stored in a capacitor.

##### Question

The switch is closed at time t = 0 s. During the time interval from t = 0.0 s to t = 4.0 s, 15 mC charge passes through the resistor.

Calculate the energy transferred by the battery during this 4.0 seconds and the energy stored in the capacitor after 4.0 s.

#### Definition of Potential Difference

In many textbooks, we learnt that the potential difference across a component is the energy released for every unit charge passing through it.

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

If we are to calculate the energy released by the battery, it would be

$W=V \times Q = 9.0 \times 15 \times 10^{-3} =0.135 \text{J}$

#### Energy stored in the capacitor

However, the energy stored in the capacitor cannot be 0.135 J. This is because a capacitor’s potential difference changes as it stores charge. The equation above requires that the p.d. remains constant. Hence, we need to use another formula.

Since capacitance is defined as the charge stored for every unit potential difference,

$Q=VC$

i.e. the charge stored is directly proportional to the potential difference across it. Since the work done is the area under a charge-voltage graph, we will use the formula for a triangle.

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

At time t = 4.0 s,

$V=\frac{15 \times 10^{-3}}{2000 \times 10^{-6}}=7.5 \text{V}$

Hence,

$W=\frac{1}{2} \times 15 \times 10^{-3} \times 7.5 = 0.05625 \text{J}$

#### Lost energy?

The energy stored in the capacitor(0.05625 J) is not the same as the energy transferred from the battery(0.135 J) because heat is lost through the resistor as current flows.