Question

A resistance of $R\Omega $draws current from a potentiometer. The potentiometer wire, $AB$, has total resistance of ${R_0}\Omega .$. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across $R$when the sliding contact is in the middle of the potentiometer wire.

Answer 1

Hint: Concept of principle of potentiometer . As length increases, the potential drop across the length also increases and vice versa.

Complete step by step answer:
Potentiometer: It is a device that is used to measure unknown emf or potential differences.
Principle of potentiometer: It is based on the principle that when a constant current flows through a wire of uniform composition and cross-sectional area, then the potential drop across any length of wire will be directly proportional to that length as we know that, from ohm’s law
$V = IR$
And $R = \rho \dfrac{\ell }{A}$
So, $V = \dfrac{{I\rho }}{A}\ell $
When V is the potential difference $\rho $is the resistivity of wire A is the area of cross-section.
l is length of wire

R is the resistance
Now, for uniform cross section and area and constant current
\[\dfrac{{TP}}{A} = cons\tan t = K(let)\]
So, $V = k\ell $
$V\alpha \ell $
Which is the basic principle of potentiometer or we can also say potential gradient that is variation of potential with length his constant
$ \Rightarrow \dfrac{V}{\ell } = constant = K$
Now, in the question, let the length of wire be $\ell $.
And $M$be the mid point of wire.
Now, if $V$is the potential applied between A and B points then, according to principle of potentiometer
$V\,\alpha \,\ell $
$V = k\ell $ …..(i)
Where $k$is constant.
Let voltage across $R\,\,be\,\,V'$and is equal to AM where length is ${\mu _2}$so,
$V'\alpha \dfrac{\ell }{2}$
$ \Rightarrow V' = \dfrac{{K\ell }}{2}$ ……(ii)
From (i) and (ii) we get,
 $V' = \dfrac{V}{2}$

Note:
As the potential difference depends on the end point so the potential drop across resistor R will be the same as V' which is in accordance with the principle of potentiometer.