A wire has length 100 cm and diameter 0.38 mm. The metal of the wire has resistivity 4.5 × 10–7 Ω m.

Question 40

(a) A wire has length 100 cm and diameter 0.38 mm. The metal of the wire has resistivity 4.5 × 10^–7 Ω m.

Show that the resistance of the wire is 4.0 Ω. [3]

(b) The ends B and D of the wire in (a) are connected to a cell X, as shown in Fig. 6.1.

The cell X has electromotive force (e.m.f.) 2.0 V and internal resistance 1.0 Ω.

A cell Y of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to the wire at points B and C, as shown in Fig. 6.1.

The point C is distance l from point B. The current in cell Y is zero.

Calculate

(i) the current in cell X, [2]

(ii) the potential difference (p.d.) across the wire BD, [1]

(iii) the distance l. [2]

(c) The connection at C is moved so that l is increased. Explain why the e.m.f. of cell Y is less than its terminal p.d. [2]

Reference: Past Exam Paper – November 2014 Paper 23 Q6

Solution:

(a)

Resistance R = ρl / A

Cross-sectional area, A = [π × (0.38 × 10^–3)2] / 4       (= 0.113 × 10^–6 m²)

Resistance R = (4.5 × 10^–7 × 1.00) / ([π × (0.38 × 10^–3)²] / 4)

R = 4.0 (3.97) Ω

(b)

(i)

Current І = V / R = 2.0 / 5.0

I = 0.4(0) A

{No current flows in cell Y. So, cell Y can be considered to be short circuited and with C at this position, the circuit can be considered to consist of cell X (with internal resistance 1.0 Ω) and the wire BD (having a resistance of 4.0 Ω as calculated in part (a)).

Total resistance = 1.0 + 4.0 = 5.0 Ω}

(ii)

p.d. across BD {= IR} = 4 × 0.4

p.d. = 1.6 V

(iii)

{Since the current in cell Y is zero, the p.d. across BC should be equal to the e.m.f. of cell Y, that is 1.5 V. BC is a distance l.}

p.d. across BC (l) = 1.5 (V)

{For a wire, the resistance is proportional to the length. R = ρl / A. The p.d. across the wire is V = IR, so the p.d. V is proportional to the resistance of the wire, which is itself proportional to the length.

100 cm of wire has a p.d. of 1.6V across it.

l cm of wire has a p.d. of (l / 100) × 1.6 V across it. But, as deduced above, it is equal to 1.5V. So, the distance l can be obtained.}

BC (l) = (1.5 / 1.6) × 100 = 94 (93.75) cm

(c) (Question discounted.)