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.
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation