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Practise 19 exam-style questions on Direct and alternating potential difference for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
What is meant by a direct current (d.c.)?
A direct current is a current that always flows in the same direction.
Cells and batteries provide a direct potential difference, so the current they produce always flows the same way round the circuit, from the positive terminal to the negative terminal (conventional current).
a current that flows in one direction only
allow a current that does not change direction
The graph shows how the potential difference of a supply changes with time.
Is the supply a.c. or d.c.? Give a reason for your answer.
The supply is a.c. The potential difference keeps changing from positive to negative and back again, so its direction keeps reversing.
On a graph of potential difference against time, a negative value means the potential difference is in the opposite direction. A d.c. supply would stay on one side of the zero line.
a.c.
allow alternating
the potential difference changes between positive and negative values / reverses direction
allow the graph goes above and below zero; do not accept 'it is a wave' on its own
The graph shows the output of a low-voltage a.c. supply, displayed on an oscilloscope.
The equation linking period and frequency is:
period = 1 ÷ frequency
Show that the supply has the same frequency as the UK mains.
From the graph, one complete cycle takes 20 ms = 0.020 s.
f = 1 ÷ T = 1 ÷ 0.020 = 50 Hz
The UK mains frequency is also 50 Hz, so the frequencies are the same.
Measure the period as the time for one complete cycle, for example from one crest to the next, and convert milliseconds to seconds before using f = 1 ÷ T. Measuring across several cycles and dividing gives a more accurate period.
period read from the graph = 20 ms
allow 0.020 s
f = 1 ÷ 0.020
allow ecf from their period
f = 50 Hz, which is the same as the UK mains frequency
comparison with 50 Hz needed
The graph shows how the potential differences of two a.c. supplies, A and B, vary with time.
The equation linking period and frequency is:
period = 1 ÷ frequency
Compare the peak potential difference and the frequency of supply A with those of supply B.
Supply A has a peak potential difference of 4 V and supply B has a peak of 2 V, so B's peak is half of A's. One cycle of A takes 20 ms, so its frequency is 1 ÷ 0.020 = 50 Hz. One cycle of B takes 10 ms, so its frequency is 1 ÷ 0.010 = 100 Hz. B has twice the frequency of A.
On a potential difference-time graph, the height of the crests gives the peak potential difference and the time for one cycle gives the period. A shorter period means a higher frequency: B completes two cycles in the time A completes one.
peak p.d. of A = 4 V and of B = 2 V
B has half the peak p.d. of A
allow A has twice the peak p.d. of B
period of A = 20 ms (50 Hz) and period of B = 10 ms (100 Hz)
allow either the periods or the frequencies
B has twice the frequency of A
allow A has half the frequency of B
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