A cell of internal resistance r is connected to an external resistor R. The conductor for maximum power transfer is?
A. R < r B. R = r C. R = 2r D. R > r
Correct Answer: B
Explanation
Power = I\(^2\)RÂ = \((\frac{E}{R + r})^2 R\) To get point of maximum power transfer, we take the derivative and equate it to 0. \(P = E^2 (\frac{R}{(R + r)^2})\) \(\frac{\mathrm d P}{\mathrm R} = E^2 (\frac{(R + r)^2 - 2R(R + r)}{(R + r)^4})\) = \(E^2 (\frac{(R + r)[(R + r) - 2R}{(R + r)^4})\) \(\frac{\mathrm d P}{\mathrm d R} = E^2 (\frac{r - R}{(R + r)^3})\) Maximum power is gotten when r - R = 0 i.e. r = R. Â