If power in external resistance `R` is maximum, then

(i) `R = r`
(ii) Power in `R` is `(E^2)/(4R)`
(iii) Input power is `(E^2)/(2R)`
(iv) Efficiency is `50%`

Electric power =I^(2)R

Electric power =V^(2)//R .

(a) A cell of emf E and internal resistance r is connected to a resistance R . (i) Relate r and R , so that power in R is maximum. (ii) Maximum power consumed by R . (iii) Efficiency. (b) Find value r in terms of R so that power in external circuit is maximum. (c) A battery of internal resistance 4Omega is connected to the network of resistance as shown in the figure. what must be the value of R so that maximum power is delivered to the network? what is tha maximum power?

Show that the power supplied by a battery to an external reistance is maximum when it is equal to the internal resistance of the batter. Further show that this maximum power is (e^(2))/(4r) where is the emf of the battery and r is internal reisistance.

The maximum power dissipated by external resistance R by a cell of an external omf E and internal resistance r is E^(2)//4R which is obtained for

The maximum power dissipated by external resistance R by a cell of an external omf E and internal resistance r is E^(2)//4R which is obtained for

The power dissipated in the resistance R, in the circuit as shown in the figure, is maximum if R is equal to