Have you ever thought about why you can hold a small battery between your fingers and feel nothing, but if you're struck by lightning you'd be lucky to survive? The answer, among other factors, is the voltage (or potential difference) of these two different sources of electricity. A typical battery may have a potential difference of 1.5 Volts, while a lightning strike can be as much as 150 million volts! This article explains what potential difference actually is, which will help you better understand why circuits behave the way they do, and why not to get struck by lightning!
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Jetzt kostenlos anmeldenHave you ever thought about why you can hold a small battery between your fingers and feel nothing, but if you're struck by lightning you'd be lucky to survive? The answer, among other factors, is the voltage (or potential difference) of these two different sources of electricity. A typical battery may have a potential difference of 1.5 Volts, while a lightning strike can be as much as 150 million volts! This article explains what potential difference actually is, which will help you better understand why circuits behave the way they do, and why not to get struck by lightning!
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).
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 \(1.602\times10^{-19}\;\mathrm{J}\).
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) |
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 \((\mathrm{J}/\mathrm{C})\).
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.
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,
$$\mathrm{Power}\;=\;\mathrm{current}\;\times\;\mathrm{potential}\;\mathrm{difference}$$
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.
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?
$$V=\;\frac{100\;\mathrm J}{25\;\mathrm C}=4\;\mathrm J/\mathrm C=\mathbf4\boldsymbol\;\mathbf V$$
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?
#
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.
$$V\;=\;\frac EQ=\frac{1000\;\mathrm J}{4\;\mathrm C}=\mathbf{250}\boldsymbol\;\mathbf V$$
$$V=\frac EQ$$
$$V=IR$$
or in words,
$$\mathrm{Power}\;=\;\mathrm{current}\;\times\;\mathrm{potential}\;\mathrm{difference}$$
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
The equation for potential difference is:
V = E/Q
The equation for potential difference is:
V = E/Q
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.
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.
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.
Why does a lightening strike transfer a large amount of energy, even though its duration is very short?
Lightening strikes have an extremely high potential difference, as much as 150 million volts. This means that they can transfer a large amount of energy in a small amount of charge. In addition, the high voltage will create a high current due to Ohm's law - meaning more charge is transferred in a short period.
In the hydraulic analogy of an electric circuit, what does the water pressure represent?
The water pressure is representative of the potential difference of the electrical circuit.
In the hydraulic analogy of an electric circuit, what does the flow rate of water represent?
The rate at which the water flows represents the current in an electrical circuit.
The energy delivered to charge carriers in a circuit is measured by the potential difference (or voltage) of a supply. Is it true or false?
True
The potential difference is a measurement of how much electrical work a power supply does, per unit of charge that moves through the circuit. Is it rue or false?
Somewhat true - in an theoretical scenario this is true, however in reality there will always be inefficiencies that mean the power supply does extra work.
What is the SI unit for the electric potential difference?
Volts (V)
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