JAMB - Physics (2006 - No. 9)

A cell of internal resistance r is connected to an external resistor R. The condition for maximum power transfer is?
R < r
R = r
R = 2r
R > r

Explanation

Power = I\(^2\)R 

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

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.

 

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