Question

A potentiometer wire of length 1 m has a resistance 10 ohm. It is connected to a 6V battery in series with a resistance of 5 ohm. Determine the emf of the primary cell which gives a balance point at 40 cm.

Answer 1

Hint: A potentiometer is a three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. It has many uses and it can be used to find the unknown resistance or to find out the emf of the unknown cell. Potentiometer is in fact a resistor. A potentiometer can work as a rheostat but the opposite does not hold.

Formula used:
We will make use of Ohm’s law: \[V=IR\]

Complete step by step answer:
Given that, the length of the wire is 1m. The resistance of the wire is $10\Omega $ while the external resistance is $5\Omega $. Thus,
$I=\dfrac{V}{R+R'}$
$\Rightarrow I=\dfrac{6}{10+5}$
$\Rightarrow I=\dfrac{6}{15}=0.4A$
So, the current in the wire is 0.4A. Now we have to find the potential drop from the potentiometer wire, thus
$V=IR$
$\Rightarrow 0.4\times 10$
$\Rightarrow V=4V$
Now we know potential gradient is given by $k=V/l$ where l is the length of the wire.
Now the length of the wire is 1m, so the potential gradient comes out to be $4V/m$
Now, to find out the emf of the unknown cell, we use the formula: $e=kl'$
$\Rightarrow e=4\times 0.4 \\
\therefore e=1.6V $

Note:Potentiometer works by varying or changing the position of the sliding key on the wire. In the potentiometer, the voltage is applied across the length of the wire and the length is 1m. The potentiometer has two input terminals. In this the two batteries which are connected, their negative terminals are joined together and positive terminals are connected across the galvanometer.