A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when:
- R = 2r
- R = 2r
- R = r2
- R = r
Answer
(4) R = r
Solution
I = ER + r
P = I2R = E2R(R + r)2
For maximum power: dPdR = E2 (R + r)2 - 2R(R + r)(R + r)4 = 0
R + r - 2R = 0
R = r (maximum power transfer)
Answer: (4)