GCSE Standard Questions: Electric Circuits
Q15.
(a) Lex investigated how the current in a series circuit varied with the resistance of a variable resistor. The diagram below shows the circuit he used:
![](q15circuit1.png)
This graph shows his results:
![](15graphANS.png)
(i) The battery had a power output of 230 mW when the resistance of the variable box resistor was 36 Ω.
Determine the potential difference across the battery.
![](../../../../graphics/Bulletpoints/FuzzleGREEN.png)
P = I V
V = IR
I = V/R
P = V2/R ![](../../../../graphics/symbols/nuclides/ticksmall.png)
230 x 10-3
= V2/ 36
V2 = 230 x 10-3 x 36
V2 = 8.28 ![](../../../../graphics/symbols/nuclides/ticksmall.png)
V = 2.88 volts ![](../../../../graphics/symbols/nuclides/ticksmall.png)
P = IV
I = 0.08 A - from the graph ![](../../../../graphics/symbols/nuclides/ticksmall.png)
0.230
= 0.08 × V
V = 0.230 /0.08 ![](../../../../graphics/symbols/nuclides/ticksmall.png)
V = 2.88 volts ![](../../../../graphics/symbols/nuclides/ticksmall.png)
[4 marks]
(ii) Lex concluded: 'the current in the circuit was inversely proportional to the resistance of the variable resistor.'
Explain how the results on his graph show that he is correct.
![](../../../../graphics/symbols/nuclides/ticksmall.png)
![](../../../../graphics/symbols/nuclides/ticksmall.png)
[2 marks]
(b) The diagram below shows a circuit with a switch connected incorrectly.
![](q15circuit2.png)
Explain how closing the switch would affect the current in the variable resistor.
When you close the switch you cause a short circuit.
The current would be almost zero in the variable resistor because the switch has effectively zero resistance
, therefore most of the current would flow through the path with the switch in it not the one with the resistor in it. ![](../../../../graphics/symbols/nuclides/ticksmall.png)
[3 marks]
(Total 9 marks)