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Circuits - Series and Parallel

Q4.

(a) A student is given three resistors of resistance 3.0 , 4.0 and 6.0 respectively.

(i) Draw the arrangement, using all three resistors, which will give the largest resistance.

The largest resistance will be the three in series.

(ii) Calculate the resistance of the arrangement you have drawn.

R = 3.0 + 4.0 + 6.0 = 13

(iii) Draw the arrangement, using all three resistors, which will give the smallest resistance.

The smallest resistance will be the three in parallel.

(iv) Calculate the resistance of the arrangement you have drawn.

1/R = 1/3.0 + 1/4.0 + 1/6.0 = (4.0 + 3.0 + 2.0)/12 = 9.0/12 = 0.75

R = 1/0.75 = 1.3

(5)

(b) The three resistors are now connected to a battery that provides a potential difference of 12V, as shown in the diagram below.

(i) Calculate the total resistance in the circuit.

First we have to replace the parallel arrangement with a single resistor.

1/R = 1/3.0 + 1/6.0 = 1/2.0

Rparallel = 2.0

We now have a series arrangement

RTotal = 2.0 + 4.0 = 6.0

(ii) Calculate the voltage across the 6.0 resistor.

The 6.0 resistor is part of the parallel arrangement that has a resistance of 2.0 . It will therefore get that share of the voltage.

The 12V from the battery is shared out across 6.0 . Each ohm gets 2.0 V - the parallel arrangement will therefore get 4.0 V and the voltage across the 6.0 resistor will be 4.0 V

(4)

(Total 9 marks)