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Potential Difference Definition
Potential difference (also referred to as voltage) can be a confusing concept that many students have difficulty grasping at first. But don't let that dissuade you - we can use a hydraulic analogy to more easily understand what potential difference is, and how it affects the behaviour of an electrical circuit.
Let's consider a system of pumps and reservoirs like the one shown in the diagram below. When a pump raises water to an elevated reservoir, it adds energy to the system by raising the gravitational potential energy of the water (1). The gravity acting on the elevated water means the water at the bottom of the reservoir is at high pressure (2). If we open a valve allowing the water to descend through a pipe and drive a turbine, it flows at a rate dependent on the pressure of the water and the resistance of the turbine (3). Finally, by driving the turbine, the gravitational potential energy stored in the water is converted into kinetic energy, which is removed from the system (4). The water at the bottom of the pipe has no more stored energy - until we can use a pump to add potential energy back into the water and return it to the top of the system (1).
Diagram showing the hydraulic analogy of an electrical circuit. [Left] - a system of pumps and reservoirs which stores energy as gravitational potential and releases it as kinetic energy. [Right] - an electrical circuit that stores energy as potential difference in charge carriers, which is released as work when they pass through a resistor. StudySmarter Originals.
This system is a useful analogy of a simple electrical circuit consisting of a battery and resistor, like that shown in the diagram. The battery provides a potential difference to the charge carriers (1), which is analogous to the gravitational potential energy added to each particle of water. After passing through the battery, each electron is carrying electrical energy due to the potential difference provided by the battery (2). The charge carriers flow around the circuit which can be measured by the current \(I\) dependent on the potential difference \(V\) and the resistance of the circuit \(R\) (3), as defined by Ohm's law:
$$V=IR$$
or in words,
$$\mathrm{Potential}\;\mathrm{difference}\;=\;\mathrm{current}\;\times\;\mathrm{resistance}$$
As the charge carriers pass through the resistor their electrical energy is converted into work, using up their stored energy (4). As the charge carriers are no longer storing electrical energy, their potential difference is now zero as they flow back to the battery. When they pass through the battery again, they gain potential difference and the process repeats (1).
Hopefully this analogy helps you understand that potential difference is a measure of the amount of potential electrical energy a charge carrier holds between two points in a circuit. Potential difference is measured in the SI unit of volts \(\mathrm{V}\).
The potential difference between two points in a circuit is equal to the difference in the amount of potential energy that charge carriers have per unit of charge between those two points. In other words, the energy transferred (in Joules) by charge carriers per coulomb of charge between two points in a circuit is equal to the potential difference (or voltage) between the points.
The hydraulic analogy can also be used to help understand how current, voltage, and resistance behave in series and parallel circuits.
You will be familiar with the Joule \(\mathrm{J}\) as the primary unit of energy we use in physics. However, a different unit often used in nuclear and atomic physics is the electron-volt \(\mathrm{eV}\)! This is the quantity of energy a single electron gains when it moves through a potential difference of \(1\;\mathrm{V}\). \(1\;\mathrm{eV}\) is equivalent to \(1.602\times10^{-19}\;\mathrm{J}\).
Potential Difference Formula
As we just established, potential difference is a measure of the difference in potential energy each unit of charge contains between two points in an electrical circuit. This means that if we know a quantity of energy transferred \(E\)between two points in a circuit and the amount of charge \(Q\) that flowed to transfer the energy, we can calculate the potential difference \(V\) between the points using the formula below:
$$V=\frac EQ\;$$
or in words,
$$\mathrm{Potential}\;\mathrm{difference}\;=\;\frac{\mathrm{energy}\;\mathrm{transferred}}{\mathrm{charge}\;\mathrm{flow}}$$
| Quantity | Symbol | SI unit |
| Potential difference | V | V (Volts) |
| Energy transferred | E | J (Joule) |
| Charge | Q | C (Coulomb) |
Potential difference unit
By studying the formula \(V=\frac EQ\), we can see that a potential difference of \(1\;\mathrm{V}\) is equal to \(1\;\mathrm{J}\) of energy per \(1\;\mathrm{C}\) of charge.
The SI unit of potential difference is the Volt \(\mathrm{V}\), equivalent to one Joule-per-coulomb \((\mathrm{J}/\mathrm{C})\).
Potential difference equation and symbol
Another set of equations which deal with potential difference are those defined in Ohm's law, which describe how the current \(I\) passing through an Ohmic electrical component is determined by the potential difference \(V\) across the component and the component's electrical resistance \(R\). We referred to this law and its formula earlier in the article.
Linking potential difference to electrical power
Recalling that current is the rate at which charge flows in a circuit, measured in coulombs-per-second, we can derive an equation for the power (energy-per-second) of an electrical circuit.
If we know that a certain quantity of energy \(Q\) was transferred between two points in a circuit by a certain amount of charge \(C\), we can calculate the potential difference \(V\) of the charge between these points:
$$V=\frac EQ$$
If we also know that the energy transfer took a duration of \(t\) seconds, then we can determine the current \(I\):
$$I=\frac Qt$$
Rearranging both these equations for \(Q\), we get:
$$\begin{array}{rcl}Q&=&\textstyle\frac EV\\Q&=&It\end{array}$$
Finally, we can combine these equations to find an equation for power \(P\):
$$\begin{array}{rcl}\textstyle\frac EV&=&It\\E&=&IVt\\\textstyle\frac Et&=&IV\\P&=&IV\end{array}$$
or in words,
$$\mathrm{Power}\;=\;\mathrm{current}\;\times\;\mathrm{potential}\;\mathrm{difference}$$
Potential difference in a battery
A battery adds electric potential to charge carriers as they pass through it. The battery can also be considered as the source of current in a circuit - As the positive terminal of the battery has a higher electric potential than the negative terminal, positive charges are repelled by the positive battery terminal and attracted to the negative terminal, setting up a current as they flow around the circuit. This attraction/repulsion is equivalent to gravity in the hydraulic example.
Batteries in a circuit transfer stored chemical energy into charge carrying electrons, raising their potential difference as they flow through it. Flickr.
In a real electric circuit, the charge carriers are electrons - which are actually negatively charged! This means the charge carriers in a circuit move from the negative terminal to the positive terminal. When electricity was first discovered this was not yet know, and conventional current was defined as moving from positive to negative. This is the opposite of electron current, which flows from negative to positive. This difference is important to be aware of when performing circuit analysis, but it is outside the specification of the GCSE exams.
- When the power source of an electric circuit transfers one Joule of energy to each coulomb of charge, we say that the power source provides one volt of potential difference.
- To measure the potential difference between the two terminals of a cell, battery or an alternative power source we use a voltmeter. Voltmeters can be used to measure the potential difference across any electric circuit component or between any two points.
Voltmeters can be used to measure the potential difference between two points in a circuit. If there is no energy being transferred between the two points the voltmeter measures between, the potential difference reading will be 0 volts. StudySmarter Originals.
Potential difference examples
Lets look at some example problems that will allow us to apply our newfound knowledge to different scenarios.
If a battery in a circuit transfers a total of \(100\;\mathrm{J}\) of energy to \(25\;\mathrm{C}\) of charge, what potential difference (voltage) would be measured across the terminals of the battery?
- Determine what information we get from the question. We know the quantity of energy \(E=100\;\mathrm{J}\). We also know that the amount of charge \(Q=25\;\mathrm{C}\). The question asks us to find the potential difference, \(\mathrm{V}\).
- Select the correct equation to use. As the question involves the quantities \(E, Q, V\), we can use \(V=\frac EQ \).
- Plug in values to find the answer:
$$V=\;\frac{100\;\mathrm J}{25\;\mathrm C}=4\;\mathrm J/\mathrm C=\mathbf4\boldsymbol\;\mathbf V$$
The same battery as in the previous question takes 20 seconds to transfer 240 joules of energy. What current do charge carriers/electrons flow through the battery?
- Determine what information we get from the question. We already know that the potential difference of the battery is\(V=4\;\mathrm{V}\). The amount of energy transferred \(E=240\;\mathrm{J}\) and the time taken\(t=20\;\mathrm{s}\) .
- The first step is to determine the amount of charge that flowed through the battery. We can rearrange the equationto \(V=E/Q\) to find this: \(Q=240\;\mathrm{J}/4\;\mathrm{V}=60\;\mathrm{C}\).
- Now that we know the quantity of charge, we divide by the time take to find the charge-per-second, the current:
#
A \(1000\;\mathrm{W}\) space heater draws \(4\;\mathrm{A}\) of current from its mains power supply. Determine what potential difference the appliance operates at.
- Establish what information is given. The question tells us the power of the appliance \(P=1000\;\mathrm{W}\), and that it draws a current of \(I=4\;\mathrm{A}\). This tells us that the power is \(1000\;\mathrm{joules-per-second}\) and the current is \(4\;\mathrm{coulombs-per-second}\) respectively.
- We can determine the potential difference from the amount of energy transferred per coulomb of charge. In one second \(4\;\mathrm{C}\) transfers \(1000\;\mathrm{J}\) of energy, meaning \(1\;\mathrm{C}\) transfers \(250\;\mathrm{J}\) .
- The potential difference can be found using \(V=\frac EQ\):
$$V\;=\;\frac EQ=\frac{1000\;\mathrm J}{4\;\mathrm C}=\mathbf{250}\boldsymbol\;\mathbf V$$
Potential Difference - Key takeaways
- The potential difference is the amount of energy transferred per unit of charge between two points in an electric circuit.
- A current will flow across a resistor if there is a potential difference or voltage across the component.
- The electric potential difference between any two points can be calculated using the following formula:
$$V=\frac EQ$$
- The SI unit for potential difference is the Volt \(\mathrm{V}\).
- The relationship between current, potential difference and resistance in an Ohmic electrical component is defined by Ohm's law:
$$V=IR$$
or in words,
$$\mathrm{Power}\;=\;\mathrm{current}\;\times\;\mathrm{potential}\;\mathrm{difference}$$
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Frequently Asked Questions about Potential Difference
What is potential difference?
The potential difference is equal to the amount of energy transferred (in joules) per unit of charge (in coulombs).
The equation for potential difference is:
V = E/Q
- V is the potential difference between two points in a circuit in Volts.
- Q is the amount of charge passing between the points in Coulombs.
- E is the energy transferred by the charge flow in Joules.
How to calculate potential difference?
The equation for potential difference is:
V = E/Q
- V is the potential difference between two points in a circuit in Volts.
- Q is the amount of charge passing between the points in Coulombs.
- E is the energy transferred by the charge flow in Joules.
What is the formula and examples of potential difference?
The equation for potential difference is:
V = E/Q
- V is the potential difference between two points in a circuit in Volts.
- Q is the amount of charge passing between the points in Coulombs.
- E is the energy transferred by the charge flow in Joules.
Example: A 4V battery powers a lamp. 8J of energy is supplied to the lamp. How much charge provided this energy to the lamp?
Answer: We rearrange the formula V = E/Q to make Q the subject:
VQ = E
Q = E/V
Therefore, the charge Q is calculated by dividing the energy transferred E by the potential difference V:
Q = 8J/4V = 2C
Therefore the answer is 2 Coulombs of charge.
What is difference between electric potential and potential difference?
Electric potential at a a certain point in an electric field equals the amount of work required to bring a unit positive charge from infinity to that point.
Potential difference defines the amount of energy a charge carrier transfers (in Joules) per unit of charge between two points in a circuit. This is equal to the difference in electric potential between these two points or the work done to these charges in moving through this potential difference per unit of charge.
What is potential difference measured in?
The standard SI unit of potential difference is the Volt (V). 1 Volt is equivalent to 1 Joule of energy being transferred per Coulomb of charge.
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