A power supply of electromotive force (e.m.f.) 12 V and internal resistance 2 Ω is connected in series with a load resistor.

Question 36

A power supply of electromotive force (e.m.f.) 12 V and internal resistance 2 Ω is connected in series with a load resistor. The value of the load resistor is varied from 0.5 Ω to 4 Ω.

Which graph shows how the power P dissipated in the load resistor varies with the resistance of the load resistor?

Reference: Past Exam Paper – November 2011 Paper 12 Q35

Solution:

Answer: A.

e.m.f. of supply = 12V
Internal resistance, r = 2Ω

The resistance of the load resistor, R varies from 0.5 Ω to 4 Ω.

Power P dissipated in the load resistor = I²R


The power dissipated is maximum when the load resistance is equal to the internal resistance. In the graph, when the resistance of the load is equal to 2 Ω (equal to the internal resistance), the curve is a maxima.

Note that if we had used the formula P = V² / R, V is the p.d across the load and not the e.m.f. of the supply. But since the resistance is changing, the p.d. V across the load will also change. This changing values of V should be accounted in the formula.