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

Explanation

(i) The circuit diagram is shown in figure

(ii) Potential gradient,
`k = (5V)/(5m) = 1V m^(-1) = 1/100 V cm^(-1)`
In case(i)
`epsilon_(1) + epsilon_(2) = k l_(2)= 1/100 xx 350 = 3.50V`....(i)
In case (ii) `epsilon_(1)-epsilon_(2) = k l_(2) = 1/100 xx 50 = 0.50V`.....(ii)
Solving (i) and (ii), we get,
`epsilon_(1) = 2.0V and epsilon_(2) = 1.50V`