In the potentimeter circuit shown in figure, the balance point with `R = 10 Omega` when switch `S_(1)` is closed and `S_(2)` is open is 50 cm, while that when `S_(2)` is closed and `S_(1)` is open is 60 cm. What is the value of x? What will you do if you fail to find a balance point with the given cell E'?

Explanation

Let K be the potential gradient of the potentiometer wire due to current cell E and I be the current from cell E and I be the current through resistance `10 Omega` and `x Omega` due to cell E'.
Potential difference across `10 Omega = I xx 10` volt
potential difference across `x Omega = I xx x`
When switch `S_(1)` is closed and `S_(2)` is open, the potential difference across `10 Omega` is balanced across length 50 cm is `I xx 10 = K xx 50`
When switch `S_(2)` is closed and `S_(1)` is open, the potential difference across `(10 + x)Omega` is balanced across length 60 cm is
`I (10 + x) = K xx 60`
from (i) and (ii),
`10+x/10 = 60/50 = 6/5 or x = 2 Omega`
If we fail to get a balance point on potentiometer wire, it shows that potential drop across `(10 + x) Omega` is greater than potential difference across potentiometer wire due to cell E. In order to find the balance point, we can perform the following operations ,
(i) we can increase the voltage of cell E by adding one or more cells in series with E or
(ii) We can reduce the current in resistance (10 +x) and hence potential difference across it by using a suitable resistance in series with cell E'.