In a potentiometer, a standard cell of emf `5V` and of negligible resistance maintains a steady current through the galvanometer wire of length `5 m`. Two primary cells of emfs `epsilon_(1) and epsilon_(2)` are joined in series with (i) same polarity and (ii) apposite polarity.The combination is connected through it galvanometer and a joined to the potentiometer. The balancing length is the two cases are found to be `350 cm` and 50 cm` respectively
(i) Draw the necessary circuit diagram
(ii) Find the value of emfs of the two cells

Answer

`2.0 V,1.50 V`

Explanation

(i) The current diagram is

(ii) potential gradent of potentimeter wire
`K = (5V)/(5m) = (5V)/(500 cm ) = (1)/(100) V cm^(-1)`
In first case , `epsilon_(1) + epsilon_(2) = Kl_(1) = (1)/(100) xx 350 = 350`......(i)
IN second case ` epsilon_(1) - epsilon_(2) = Kl_(2) = (1)/(100) xx 50 = 0.50 `.....(ii)
Solving (i) and (ii) we get
`epsilon_(1) = 2.0 V and epsilon_(2) = 1.50 V`