Home / IGCSE Physics (0625) 4.3.3 Action and use of circuit components Paper 4

IGCSE Physics (0625) 4.3.3 Action and use of circuit components Paper 4

Question

Fig. 8.1 shows a circuit.

                                           

The lamp has a resistance of 3.0 Ω. Line XY represents a uniform resistance wire of resistance 6.0 Ω.

(a) Calculate the reading on the ammeter.
ammeter reading =………………………………………………………………………………………………………….

(b) Fig. 8.2 shows the circuit with a different connection to the resistance wire and an added resistor. The length XY of the whole resistance wire is 2.0 m. The contact is made at Q where the distance XQ is 0.60 m.

 

Calculate the resistance of the circuit.
resistance =……………………………………………………………………………………………………………….

Answer/Explanation

Answer:

(a) {Rs = R1 + R2 + R3 in any form OR (R s )= R1 + R2 + R3 OR (Rs ) = 3 + 2 + 6 (Ω) OR (Rs ) = 11 (Ω)
AND {V= IR in any form OR (I=)V / R OR
(I=) 12 / 11 (A)}
(I=) 1.1 A

(b) uses resistance of wire proportional to length OR (resistance XQ =) 6 Ω 0.6 / 2.0 (Ω) OR 1.8 (Ω) 
1 / Rp = 1 / R1 + 1 / R2 OR (Rp=) R1R2 / (R1 + R2)
1 / Rp = 1 / 1.5 + 1 / (6 × 0.6 / 2)
OR (Rp=) 1.5 × (6 × 0.6 / 2) / (1.5 + 6 × 0.6 / 2)
OR (Rp= 1.5 × 1.8 / {1.5 + 1.8}) = 0.82 (Ω)
(R = 3 + 2 + 0.82 =) 5.8 Ω

Question

 Fig. 7.1 shows a circuit diagram that includes component X.

(a) State the name of component  X

(b) The electromotive force (e.m.f.) of the battery is E. The switch is closed.
The potential difference (p.d.) across the 30 Ω resistor is \(V_{30}\).
The p.d. across the 20 Ω resistor is \(V_{20}\).
The p.d. across component X is VX.
State an equation that relates VX to:
1. \(V_{30}\)
2. E and \(V_{20}\)

(c) The e.m.f. of the battery is 6.0 V and the resistance of component X is 15 Ω.
Calculate:
(i) the total resistance of the circuit
resistance =
(ii) the ammeter reading.
reading =

(d) The temperature of component X increases.
State and explain what happens to the ammeter reading.

Answer/Explanation

Answer:

(a) thermistor c.a.o.

(b) (i) \(V_X = V_{30}\)
(ii) \(V_X = E – V_{20}\) in any form

(c) (i) \(1/ R_1 + 1/ R_2 = 1/ R_{tot}\) OR \((R_{tot} =) R_1 R_2 / ( R_1 + R_2)\) OR \(1/15 + 1/30 = 1/ R_{tot}\)
OR (15 × 30) / (15 + 30)
(ii) I = V / R in any form OR ( I =)V / R OR 6.0 / 30
0.20 A

(d) resistance of X decreases ammeter reading / it increases and (total) resistance (of circuit) decreases / more voltage across 20 Ω resistor

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