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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:

This graph shows his results:

(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.

You can do this simply by calculation:

P = I V

V = IR

I = V/R

P = V2/R

230 x 10-3 = V2/ 36

V2 = 230 x 10-3 x 36

V2 = 8.28

V = 2.88 volts

OR you can use the graph:

P = IV

I = 0.08 A - from the graph

0.230 = 0.08 × V

V = 0.230 /0.08

V = 2.88 volts

 

[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.

the product of any current and resistance from the graph = the constant 2.88

You should then prove this by using three or more pairs of values to show this is always the case

OR you could get just one mark by saying that doubling one quantity (R or I) halves the other quantity.

[2 marks]

(b) The diagram below shows a circuit with a switch connected incorrectly.

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.

[3 marks]

(Total 9 marks)