# Exam-style Questions: Current, Charge and Voltage

1. Fig. 1.1 shows an electrical circuit. a) Name the component marked X.

(1 Mark)

b) Suggest one reason why component X may be useful in the electrical circuit.

(1 Mark)

c) On Fig. 1.1 indicate the direction of flow of electrons.

(1 Mark)

d) State the energy changes taking place in the component Y.

(1 Mark)

(Marks available: 4)

2. The current I through a wire id given by the expression I = nAve.

a) Explain the meaning of n.

(1 Mark)

b) Calculate the value of n for a semiconductor in which I = 3.6 mA, v = 80 m s-1 and A = 8.2 x 10-6 m2.

n = ........... m-3

(2 Marks)

c) The value of n for copper is about 109 times greater than that for the semiconductor. Briefly explain why in terms of band theory.

(2 Marks)

(Marks available: 5)

3. Fig. 4.1 shows an electrical circuit. The battery has negligible internal resistance.

a) Show that the current I is 25 mA.

(2 Marks)

b) Calculate the potential difference (p.d.) across the resistor of resistance 120Ω.

p.d. = ...................... V

(1 Mark)

c) Explain why a voltmeter connected between points A and B will read 0 V.

(2 Marks)

(Marks available: 5)

(Marks available: 14)

Answer outline and marking scheme for question:

1. a) Variable resistor / rheostat

(1 Mark)

b) To change (allow 'control') current/ ammter reading / resistance (of the circuit) / brightness (of lamp) / p.d. (across) lamp / X

(1 Mark)

c) Clockwise

(1 Mark)

d) Electrical to heat / light

(1 Mark)

(Marks available: 4)

2. a) n is the number of free electrons / charge carriers per unit volume / m3

(1 Mark)

b) n = l/Ave

= 0.0036/(8.2 x 10-6 x 80 x 1.6 x 10-19) = 3.4 x 1019 m-3

(2 Marks)

c) In a metal the conduction band (of energy levels) is oermanently occupied by electrons, so many are available for conduction; In a semiconductor electrons must be promoted from the valence to the conduction band by thermal energy and few are available at normal temperatures.

(2 Marks)

(Marks available: 5)

3. a) R = R1 + R2 / R = 200 + 120 / R = 320 current = 2.5 x 10-2 (A)

(2 Marks)

b) v = 25 x 10-3 x 120 / v = 3.0 (V)

(1 Mark)

c) p.d. across the 60 (Ω) resistor = p.d. across the 120 (Ω) resistor / There is no current between A and B / in the voltmeter (Allow 'A & B have same voltage' - BOD)

The p.d.calculated across 360 Ω resistor is shown to be 3.0 V / The ratio of the resistances of the resistors is shown to be the same.

(2 Marks)

(Marks available: 5) 