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Circuits - series and parallel

Q5. The circuit shown below shows an arrangement of resistors, W, X, Y, Z, connected to a battery of negligible internal resistance. The emf of the battery is 10V and the reading on the ammeter is 2.0 A.

(a)

(i) Calculate the total resistance of the circuit.

'Negligible internal resistance' means that we don't have to use the EMF version of the circuit.

V = IR

R = V/I = 10/2.0

R = 5.0 Ω

(1 mark)

(ii) The resistors W, X, Y, and Z all have the same resistance. Show that your answer to part (a) (i) is consistent with the resistance of each resistor being 3.0 Ω.

W and X in series:

RWX = 3.0 + 3.0 = 6.0 Ω

WX and Y in parallel:

1/RWXY= 1/6.0 + 1/3.0 = 3/6.0 = 1/2.0

RWXY = 2.0Ω

This parallel arrangement is in series with Z

RTOTAL = 2.0 + 3.0 = 5.0Ω QED

(3 marks)

(b)

(i) Calculate the current through resistor Y

voltage across Z = IR = 2.0 × 3.0 = 6.0 V

voltage across Y = cell voltage - voltage across Z

10.0 – 6.0 = 4.0 V

current in Y = V/R = 4.0/3.0 = 1.3 A

(2 marks)

(ii) Calculate the pd across resistor W.

Current through the parallel arrangement is 1.3A It will split so that double the current goes through W than goes through XY (as it has double the resistance!).

current through W = 1.3 x 2/3 = 0.67 A

voltage = IR = 0.67 × 3.0 = 2.0 V

(2 marks)

(Total 8 marks)