The adjoining graph shows the variation of terminal potential difference V, across a combination of three cells in series to a resistor, versus the current, I:
(i) Calculate the emf of each cell. (ii) For what current I, will the power dissipation of the circuit be maximum ?

Explanation

(i) We know that V = `epsi - Ir` . From the graph it is clear that when no current is being drawn from the cells (i.e., I =0), voltage is 6.0 volt. As the battery is a combination of 3 cells and in an open circuit terminal potential difference is equal to emf, hence
emf of each cell `epsi = (6.0)/(3) V = 2.0 V`
(ii) From graph, terminal potential difference Vis zero when current drawn is `I_s=2.0 A` . It represents the short circuit condition, where `0 = epsi - I_s. r ` and r is the internal resistance of the cell combination.
` rArr r = epsi/I_s = (6.0V)/(2.0A) = 3Omega`
Power dissipation of the circuit will be maximum when external resistance is equal to internal resistance (R=r) i.e., current drawn is
`I = (epsi)/(r + R) = (epsi)/(2r) = (6V)/(2 xx 3Omega ) = 1A`