The maximum power drawn out of cell is given by : (E is emf, r is internal resistance)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{E^{2}}{4r} \)
When a cell of emf E and internal resistance r drives a current through an external resistance R, the power delivered to R is
\( \displaystyle P = I^{2}R = \frac{E^{2}R}{(R + r)^{2}} \)
The power is maximum when \( dP/dR = 0 \):
\( \displaystyle \frac{dP}{dR} = E^{2}\,\frac{(R + r)^{2} - 2R(R + r)}{(R + r)^{4}} = 0 \Rightarrow R + r = 2R \Rightarrow R = r \)
This is the maximum power transfer condition. Putting \( R = r \):
\( \displaystyle P_{\max} = \frac{E^{2}r}{(2r)^{2}} = \frac{E^{2}}{4r} \)