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Questions on Characteristic Curves - Filament Lamp

Q7. The sketch graph below shows the current–voltage (I–V) characteristics for a resistor and a filament lamp.

(a) Explain, in terms of electron motion, why the I–V characteristic for the filament lamp is a curve.

Current is due to the movement of 'free' electrons in the wire of the filament. The movement is impeded by interaction with the atom lattice.

An increase in p.d. produces an increase in current through the filament. This leads to more heat being generated and an increase in temperature.

This causes an increase in the vibration of the lattice/ions/atoms

and therefore an increase in the rate of collisions between the lattice and the electrons

Initially the current flowing through it causes a big change in temperature and we see the most curvature in the graph line as the vibration of the lattice increases.

When it gets to maximum temperature we see a steady straight line as the rate of interaction with the lattice is steady.

[4 marks]

(b) Determine the resistance of the resistor.

V = IR

R = V/I

R = 5.8/0.4

R = 14.5 Ω

Answers between 14Ω - 15Ω gained a mark.

[1 mark]

(c) The resistor and the filament lamp are connected in series with a supply of variable emf and negligible internal resistance.

Determine the emf that produces a current of 0.18 A in the circuit.

See graph

PD across filament lamp = 0.8 V

PD across resistor = 2.6 V

EMF = V + Ir

but r = 0Ω

So EMF = 0.8 + 2.6 = 3.4 V

[3 marks]

(d) The resistor and filament lamp are now connected in parallel.

Determine the resistance of the parallel combination when the emf of the supply is adjusted to be 4.0 V.

EMF = V as r = 0

For the filament lamp

V = IRfilament

4.0 = 0.36 x Rfilament

Rfilament = 4.0/0.36

Rfilament = 11.1 Ω

For the resistor

Resistor has resistance of 14.5 Ω

You worked that out in part (b)

For the parallel arrangement

1/RTotal = 1/Rfilament + 1/Rresistor

1/RTotal = 1/11.1 + 1/14.5 = 0.159

RTotal = 6.3 Ω

[3 marks]

(e) The resistance of the filament lamp at its working temperature is 14 Ω.

The filament has a length of 0.36 m and a diameter of 32 µm.

Calculate the resistivity of the metal that is used for the filament when the lamp is at its working temperature.

Give an appropriate unit for your answer.

ρ = RA/L 

A = πr2

A = ¼πd2

A = ¼π x (32 x 10-6)2

A = 8.04 x 10-10 m2

ρ = 14 x 8.04 x 10-10/0.36

ρ = 3.1 x 10-8 Ωm

[3 marks]
(Total 14 marks)