Potential difference is the energy transferred per unit charge as electric charge moves around an electric circuit. A power supply (cell or battery) does work on a charge and transfers energy to the charge to increase its potential difference. When a charge passes through a component, it does work against the resistance of the component. When the charge does work in a component, it goes through a potential difference fall (it loses some of its potential difference) and transfers energy to the component. The potential difference between two points is equal to the work done divided by the charge (work done can also be viewed as energy transferred, so the potential difference between two points is equal to the energy transferred divided by the charge).
V is potential difference in volts (V), E is energy transferred or work done in joules (J) and Q is charge in coulombs (C). 1 volt is 1 joule per coulomb.
A greater fall in potential difference means that more energy is transferred per coulomb of charge. Therefore, a battery that produces a larger potential difference will supply more energy to the circuit per coulomb of charge.
A greater fall in potential difference means that more energy is transferred per coulomb of charge. Therefore, a battery that produces a larger potential difference will supply more energy to the circuit per coulomb of charge.
Remember This
An easier way to remember this formula is to have energy transferred as the subject. When we are working out energy transferred, we multiply the charge flow by the potential difference.
An easier way to remember this formula is to have energy transferred as the subject. When we are working out energy transferred, we multiply the charge flow by the potential difference.
E is energy transferred in joules (J), Q is charge in coulombs (C) and V is potential difference in volts (V).
Example 1
A motor in an electric fan is powered by a 3 V battery. How much energy is transferred if 320 C of charge passes through the motor in the fan?
We work out the energy transferred by multiplying the charge by the potential difference.
A motor in an electric fan is powered by a 3 V battery. How much energy is transferred if 320 C of charge passes through the motor in the fan?
We work out the energy transferred by multiplying the charge by the potential difference.
The question tells us that the charge is 320 C and the potential difference is 3 V. We sub these values into the calculation.
The energy transferred is 960 joules.
Example 2
The potential difference across a component is 12 V and the energy transferred is 1,500 J. What is the charge flow?
We can work out the charge flow by dividing the energy transferred by the potential difference.
The potential difference across a component is 12 V and the energy transferred is 1,500 J. What is the charge flow?
We can work out the charge flow by dividing the energy transferred by the potential difference.
The energy transferred is 1,500 J and the potential difference is 12 V.
The charge flow is 125 C.
Example 3
The next example is going to be a bit more complex because we are going to use a formula that we looked at in one of the first sections of this whole electricity topic. The formula that we need is the one that works out the charge flow. We can find the charge flow by multiplying the current by the time. The formula for this is shown below.
The next example is going to be a bit more complex because we are going to use a formula that we looked at in one of the first sections of this whole electricity topic. The formula that we need is the one that works out the charge flow. We can find the charge flow by multiplying the current by the time. The formula for this is shown below.
Q is charge flow measured in coulombs (C), I is current measured in amps (A) and t is time measured in seconds (s). Let’s now have the question.
Question
A 4 V battery passes a current of 1.2 A through a lamp for 2 minutes. Find the energy transferred to the lamp.
The first step in answering this question is to work out the charge flow, which we do by multiplying the current by the time.
Question
A 4 V battery passes a current of 1.2 A through a lamp for 2 minutes. Find the energy transferred to the lamp.
The first step in answering this question is to work out the charge flow, which we do by multiplying the current by the time.
The question tells us that the current is 1.2 A, which is in the correct units. The question also tells us that the lamp is on for 2 minutes, which is not in the correct units as time needs to be in seconds and not in minutes; there are 60 seconds in 1 minute, so we convert minutes to seconds by multiplying by 60.
The charge flow is 144 C.
We can now work out the energy transferred by multiplying the charge flow (144 C) by the potential difference (4 V; we were told this in the question).
We can now work out the energy transferred by multiplying the charge flow (144 C) by the potential difference (4 V; we were told this in the question).
The energy transferred to the lamp is 576 J.