A power supply of electromotive force (e.m.f.) 8.7 V and negligible internal resistance is connected by two identical wires to three filament lamps, as shown in Fig. 5.1.

Question 23

(a) (i) State what is meant by an electric current. [1]

(ii) Define electric potential difference (p.d.). [1]

(b) A power supply of electromotive force (e.m.f.) 8.7 V and negligible internal resistance is connected by two identical wires to three filament lamps, as shown in Fig. 5.1.

Fig. 5.1 (not to scale)

The power supply provides a current of 0.30 A to the circuit.

The filament lamps are identical. The I–V characteristic for one of the lamps is shown in

Fig. 5.2.

Fig. 5.2

(i) Show that the resistance of each connecting wire is 2.0 Ω. [2]

(ii) The resistivity of the metal of the connecting wires does not vary with temperature.

On Fig. 5.2, sketch the I–V characteristic for one of the connecting wires. [2]

(iii) Calculate the power loss in one of the connecting wires. [2]

(iv) Some data for the connecting wires are given below.

cross-sectional area = 0.40 mm²

resistivity = 1.7 × 10^-8 Ω m

number density of free electrons = 8.5 × 10²⁸ m^-3

Calculate

1. the length of one of the connecting wires, [2]

2. the drift speed of a free electron in the connecting wires. [2]

[Total: 12]

Reference: Past Exam Paper – March 2016 Paper 22 Q5

Solution:

(a) (i) Electric current is the movement / flow of charge carriers.

(ii) Electric potential difference is defined as the energy transferred from electrical to other forms per unit charge passing between two points.

(b)

(i)

{From the graph, when I = 0.30 A, p.d. = 2.5 V}

p.d. across one lamp = 2.5 V

{p.d. across 1 lamp = 2.5 V

p.d. across 3 lamps = 3 × 2.5 = 7.5 V since the lamps are identical

So, p.d. across the 2 wires = (8.7 – 7.5) V

p.d. across 1 wire = ((8.7 – 7.5) / 2) V = 0.6 V

R = V / I}

Resistance = [(8.7 – 7.5) / 0.3] / 2 = 2.0 (Ω)

(ii)

straight line through the origin

with gradient of 0.5

{Since the resistivity of the metal does not change with temperature, its resistance also does not change. So, the metal obeys Ohm’s law. The I-V graph is a straight line through the origin.

Resistance of the wire = V / I = 2

Gradient of graph = I / V

This represents the reciprocal of the resistance.

So, gradient = I / V = 1 / R = ½ = 0.5

Equation of line: y = 0.5x + c = 0.5x  since y-intercept is zero (passes through origin)

When y = 0.40 V, 0.40 = 0.5x             giving x = 0.4 / 0.5 = 0.8 A

So, the line passes through the point (0.8, 0.4).}

(iii)

{As calculated above,

p.d. V across 1 wire = 0.6 V

Current through wire = 0.3 A

Resistance of 1 wire = 2.0 Ω}

P = I²R                        or P = VI and V= IR                or P = V² / R and V= IR

P = 0.30² × 2.0                    = 0.60 × 0.30                           = 0.60² / 2.0

P = 0.18 (W)

(iv)

1.

R = ρl / A

{l = RA / ρ                   where l is the length

A = 0.40 mm² = 0.40 × 10^-6 m² }

l = (2.0 × 0.40 × 10^–6) / 1.7 × 10^–8

I = 47 (m)

2.

I = Anvq

{v = I / Anq                 

where I is the current              and q is the charge of an electron}

v = 0.30 / (0.40×10^–6 × 8.5×10²⁸ × 1.6×10^–19)

v = 5.5 × 10^–5 (m s^–1)