When a 100-Ω resistor is connected across the terminals of a battery of emf ε and internal resistance r,the battery delivers 0.794 W of power to the 100-Ω resistor.When the 100-Ω resistor is replaced by a 200-Ω resistor,the battery delivers 0.401 W of power to the 200-Ω resistor.What are the emf and internal resistance of the battery?
A) ε = 10.0 V, r = 5.02 Ω
B) ε = 4.50 V, r = 4.00 Ω
C) ε = 9.00 V, r = 2.04 Ω
D) ε = 9.00 V, r = 1.01 Ω
E) ε = 12.0 V, r = 6.00 Ω

Question

Quizplus · Accepted Answer

Using the formula for power delivered by a battery:

P = (ε^2 / (r+R)) * R

where R is the external resistance and r is the internal resistance of the battery.

When R = 100 Ω, P = 0.794 W, so we have:

0.794 = (ε^2 / (r+100)) * 100

Similarly, when R = 200 Ω, P = 0.401 W, so we have:

0.401 = (ε^2 / (r+200)) * 200

Solving these equations simultaneously using algebraic manipulation, we get:

ε = 9.00 V 
r = 1.01 Ω

Therefore, the answer is D.

When a 100-Ω Resistor Is Connected Across the Terminals of a Battery