(a) Three cells of emfs, 1.5 V, 2.0V and 2.5 V are connected in series. Their internal resistance are `0.20 Omega, 0.15 Omega` and `0.15 Omega` respectively. The battery is connected to an external resistor of `5.5 Omega` via a very low resistance ammeter, what would be the reading of ammeter?
(b) If the three cells above were joined in parallel, would they be characterised by a definite emf and internal resistance (independent of their individual internal resistance )? If not, how will you obtain currents in different branches of the circuit ?

Explanation

(a) Here `epsilon_(1)= 1.5 V, epsilon_(2)=2.0V`,
`epsilon_(3)= 2.5V, r_(1) = 0.2 Omega , r_(2)=0.15 Omega`,
`r_(3) = 0.15 Oemga , R=5.5 Omega`
Since the cells are joined in series, the effective emf of all the cells is
`epsilon = epsilon_(1) + epsilon_(2) + epsilon_(3) = 1.5 + 2.0 +2.5 = 6.0 V`
Total internal resistance of all the cell is
`r=r_(1) +r_(2) +r_(3) = 0.20 +0.15 +0.15=0.50 Omega`
Total resistance of the circuit `= R+r`
`=5.5 +0.5=6.0 Omega`
Current, `I=epsilon/R+r = 6/6.0=1A`
(b) No, they will not be characterised by a definite emf and internal resistance, since there is no simple formula for total emf and total internal resistance of cells joined in parallel. However, current in different branches of the circuit can be obtained using Kirchoff's laws.