Some heaters are each labelled 230 V, 1.0 kW. The heaters have constant resistance.

Question 18

(a) A lamp is rated as 12 V, 36 W.

(i) Calculate the resistance of the lamp at its working temperature. [2]

(ii) On the axes of Fig. 6.1, sketch a graph to show the current-voltage (I–V)

characteristic of the lamp. Mark an appropriate scale for current on the y-axis.

Fig. 6.1

[3]

(b) Some heaters are each labelled 230 V, 1.0 kW. The heaters have constant resistance.

Determine the total power dissipation for the heaters connected as shown in each of the

diagrams shown below.

[1][1][2]

Reference: Past Exam Paper – November 2010 Paper 21 Q6

Solution:

(a)

(i)

EITHER Power P = V² / R                 OR P = VI and V = IR

Resistance R = (12² / 36 =) 4.0 Ω

(ii)

The vertical axis of the sketch should be labelled appropriately.

The graph is a (straight) line from the origin then curved in the correct direction.

The line passes through 12V, 3.0A {since the resistance is calculated to be 4.0 Ω}.

(b)

(i)

{The heaters are each 230 V, 1.0 kW. This means that when there is a potential difference of 230 V across a heater, the power consumption is 1.0 kW.}

{NOTE: The alternative methods are more advisable.

When electrical components are connected in parallel to a power supply as shown, the p.d. across each component is equal to the e.m.f. of the supply.

p.d. across each heater = 230 V

So, each heater would have a power consumption of 1.0 kW.}

Total power dissipation (= 1.0 + 1.0) = 2.0 kW

Alternatively,

Let resistance of a heater = R

Power dissipation = V² / R

{For a p.d. of 230V, there is a power dissipation of 1.0 kW}

Total resistance in the circuit = [(1/R) + (1/R)]^-1 = R/2

Since R is now halved,

Total Power dissipation = V² / 0.5R = 2 (V² / R) = 2 × 1.0 kW = 2.0 kW

(ii)

{When components are connected in series to a power supply, the total e.m.f of the source is divided into p.d. across the different components (depending on their resistance) such that

e.m.f. of supply = sum of p.d. across components

The sum of p.d. in the circuit is equal to the e.m.f. of the supply (given there is no loss volts).

Since the heaters have the same resistance, the e.m.f is divided equally between the two.

p.d. across one heater = 230 / 2 = 115 V

Power dissipated across a heater = V² / R

The power dissipated is proportional to the square of the p.d.

A p.d. of 230 V results in a power dissipation of 1.0 kW.

Since the p.d. across a heater is half the e.m.f.,

Power dissipated by a heater = (½)² × 1.0 = ¼ × 1.0

Power dissipated in each heater = 1.0 / 4 = 0.25 kW}

Total power dissipated (= 0.25 + 0.25) = 0.5 kW

Alternatively,

Power dissipated = V² / R = 1.0 kW

Total resistance in circuit = R + R = 2R

Since R is now doubled,

Total power dissipation = V² / 2R = 0.5 × (V² / R) = 0.5 × 1.0 = 0.5 kW

(iii)

{The circuit can be simplified as a series combination of a power supply and a heater connected in series to a parallel combination of 2 heaters.

Total resistance in circuit = R + [(1/R) + (1/R)]^-1 = R + R/2 = 3R / 2

Since total resistance is now 3R /2 (= 1.5R)

Total Power dissipation = V² / 1.5R = (2/3) × (V² / R) = (2/3) × 1.0 = 0.67 kW