Question

How will you compare the emf of two cells using a potentiometer?

Answer 1

Hint: Potentiometer is a device which can be used as voltage divider it can also be used for measuring electric potential. The galvanometer is a device which is used for measuring the current and it can also detect the direction of the current.

Complete step by step solution:
In the problem it is asked to compare the e.m.fs two cells using a potentiometer. In a potentiometer a cell of known e.m.f and a cell of unknown e.m.f are connected in parallel by a double switch, galvanometer and a movable jockey. Now by using the double switch first the cell with known e.m.f is connected into the circuit and the jockey is placed at a point where the current flow from the circuit is zero, let the length of the conducting wire be AB.
$\Rightarrow E_{known}\propto I\left(AB\right)$………eq. (1)
Where current is I.
Now using the double switch the connect the unknown e.m.f with the circuit and place the movable jockey at a position where the current flow from the circuit is zero, let the length of the conducting wire be AB’.
$$\Rightarrow E_{unknown}\propto I\left(A{B}^{\prime }\right)$$………eq. (2)
So here we have two equations relating known and unknown e.m.fs.
Dividing equation (1) by equation (2) we get.
$\Rightarrow {\displaystyle \frac{E_{known}}{E_{unknown}}}={\displaystyle \frac{I\left(AB\right)}{I\left(A{B}^{\prime }\right)}}$
$\Rightarrow {\displaystyle \frac{E_{known}}{E_{unknown}}}={\displaystyle \frac{\left(AB\right)}{\left(A{B}^{\prime }\right)}}$
By using this equation we can get the unknown e.m.f value. So we can get the emf of two cells.

Note: The application of potentiometer is control of volume controls and also audio equipment. The conducting wire has resistance and resistance is proportional length of the wire and therefore the current flow will become less as the resistance will increase.