[" (1) "(1)" ,(ii) "],[" (i),"(" iii) "" ,(iv) "],[" fripter in external resistance "R" is maximum,then "],[qquad " un "],[qquad [" (1) "R" ."],[" (i) Power in "R" is "(E^(2))/(4R)],[" (ith) Input power is "(E^(2))/(2R)]]

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

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 maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

[" Two cells of equal e.m.f.and of internal "],[" resistances "r_(1)" and "r_(2)(r_(1)>r_(2))" are "],[" connected in series.On connecting this "],[" combination to an external resistance "R" ,"],[" it is observed that the potential "],[" difference across the first cell becomes "],[" zero.The value of "R" will be "],[[" (A) "r_(2)(r_(1)>r_(2)),(r_(1)+r_(2))/(2)],[" (B) "r_(1)-r_(2),(r_(1)-r_(2))/(2)]]

Define internal resistance of a cell. Prove that r ~= ((E)/(V) - 1)R , where R is the external resistance used.