A cell of internal resistance r is connected to an external resistor R. The conductor for maximum power transfer is?
- R < r
-
R = r
- R = 2r
- R > r
The correct answer is: B
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.