Voltage & Potential Difference (Oxford AQA IGCSE Physics)
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Voltage
The voltage of a source is the energy supplied by a source in driving charges around a complete circuit
It is also called e.m.f. or the source of potential difference
It is measured in volts, V
Voltage is supplied by:
A cell
Batteries (multiple cells)
Electrical generator
Potential Difference
As charge flows around a circuit, energy is transferred to or from the charge
The potential difference (p.d. or voltage) across a component measures the amount of energy that is transferred by the charge flowing in the circuit to the component
Potential difference is also measured in volts (V)
The potential difference in a circuit
Measuring potential difference
Potential difference (or voltage) is measured using a voltmeter
A voltmeter is always set up in parallel to the component you are measuring the potential difference across
Voltmeter in parallel
Exam Tip
In your exam, questions will be asked about potential difference. In your answer you can use either term; potential difference or voltage.
Potential Difference, Energy & Charge
The potential difference (voltage) across a component measures the energy transfer by charges
Charge is measured in coulombs, C
The unit of voltage, the volt (V), is the same as a joule per coulomb (J/C)
1 V = 1 J/C
So, for example:
If a bulb has a potential difference of 3 V, every coulomb of charge passing through the bulb will lose 3 J of energy
Potential difference is calculated using the equation:
Where:
V is the potential difference measured in volts, V
E is energy transferred measured in joules, J
Q is charge flow measured in coulombs, C
Worked Example
The normal operating voltage for a lamp is 6 V.
Calculate how much energy the lamp transfers when 4200 C of charge flows through it.
Answer:
Step 1: List the known quantities
Voltage, V = 6 V
Charge, Q = 4200 C
Step 2: State the equation linking potential difference, energy and charge
The equation linking potential difference, energy and charge is:
Step 3: Rearrange the equation to make E the subject
Step 4: Substitute the known values and calculate the energy transferred
Therefore, 25 200 J of energy is transferred in the lamp
Exam Tip
Think of the potential difference as being the energy per coulomb.
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