Question

Consider a power supply with fixed emf ε and internal resistance r causing current in a load resistance R. In this problem, R is fixed and r is a variable. The efficiency is defined as the energy delivered to the load divided by the energy delivered by the emf.
(a) When the internal resistance is adjusted for maximum power transfer, what is the efficiency? %
(b) What should be the internal resistance for maximum possible efficiency?

Answer 1

Answer:

Part a)

the efficiency will be 100%

Part b)

for above condition internal resistance of the cell must be ZERO

Explanation:

Electric current supplied by the battery is given as

$i = \frac{E}{R + r}$

now the total power supplied by the cell is given as

$P = E. i$

$P = \frac{E^2}{R + r}$

power supplied to external load is given as

$P_{out} = i^2 R$

$P_{out} = \frac{E^2}{(R + r)^2} R$

Part a)

If output power is adjusted to have maximum value then value of internal resistance must be ZERO

$P_{out} = \frac{E^2}{R}$

for same condition power given by the cell

$P = \frac{E^2}{R}$

so the efficiency will be 100%

Part b)

for above condition internal resistance of the cell must be ZERO