Question

For what value of load resistance, the power transfer is maximum when two identical cells each of emf E and internal resistance r are connected (a) in series (b) in parallel.

Answer 1

(a) When cells are connecteed in series to the external load resistance R, then effective emf
`= E + E = 2E`.
Total resistance of the circuit `= R + r + r`
`= R + 2r`
Current in load, `I= (2E)/(R + 2r)`
Power delivered to the external load is,
`P = I^(2)R = ((2E)/(R + 2r))^(2) R`
Power deliverd to the external load will be maximum if external resistance = internal resistance of cells, i.e., `R = 2r`. Then maximum power, `P_(max) = ((2E)/(R + 2r))^(2) xx 2r = E^(2)/(2r)`
(b) When cells are connected in parallel to the external load resistance R, then effective emf = E. Total resistance of the circuit
`R + (r xx r)/(r+r) = (R + r/2)`
Current in load, `I = E/((R + r//2))`
Power delivered to the external load,
`P = I^(2)R = (E/(R + r//2))^(2)R`
Power is maximum when `R = r//2`. Then
`P_(max) = ((E)/(r//2 + r//2))^(2) r//2 = E^(2)/(2r)`
Note that in both the cases, maximum power deliverd to the loas is the same.