The following 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) Let `in` be emf and r the internal resistance of each cell. The equation of terminal potential difference `V=e_(ff)-ir_(nt)` becomes
`V=3e-ir_(nt)`…………(i)
where `r_(nt)` is effective (total) internal resistance
From fig.when i=0,V=6.0V
`therefore From (i) `6=3e-2 implies e=6/3=3V` i.e, emf of each cell =2V
(ii) For maximum power dissipation, the effective internal resistance of cells must be equal to external resistance. From fig., when `V=0,i=2.0A`
`therefore` Equation (i) gives `0=3e-2.0(r_(nt) implies r_(nt)==(3e)/2.0=(3 times 2)/2.0=3 Omega`
`therefore` For maximum power, external resistance `R=r_(nt)=3Omega`
Current in circuit `i=(3e)/(R+r_(nt))=(3 times 2)/(3 +3)=1.0A`
(i) Thus, emf of each cell, e=2v and (ii) For maximum power dissipation, current in circuit = 1.0 A