Question

In an experiment to measure the internal resistance of a cell by a potentiometer. It is found that the balance point is at a length of 2 m, when the cell is shunted by a $5\Omega$ resistance; and is at a length 3 m, when the cell is shunted by a $10\Omega$resistance. The internal resistance of the cell is, then
(A) $1.5\Omega$
(B) $10\Omega$
(C) $15\Omega$
(D) $1\Omega$

Answer 1

Hint It should be known to us that the potentiometer is defined as a three-terminal resistor which has a sliding or rotating contact which forms an adjustable voltage driver. If only 2 terminals are used then in that case one end, and the wiper will act as a variable resistor or a rheostat.

Step-by step answer
We know that:
In case of internal resistance measurement by potentiometer the ratio between the two voltages are given as:
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{\left\{ {\dfrac{{E{R_1}}}{{({R_1} + r)}}} \right\}}}{{\left\{ {\dfrac{{E{R_2}}}{{({R_2} + r)}}} \right\}}} = \dfrac{{{R_1}({R_2} + r)}}{{{R_2}({R_1} + r)}}$
We know that in the above expression the values of the variables are given as: ${l_1} = 2m,\;{l_2} = 3m,\;{R_1} = 5\Omega \;and\;{R_2} = 10\Omega$
Now we have to put the values in the above expression to get the following :
$\Rightarrow \dfrac{2}{3} = \dfrac{{5(10 + r)}}{{10(5 + r)}}$
On the further evaluation we get that the value of r is:
$\Rightarrow 20 + 4r = 30 + 3r$
So we get the value of r as :
$\Rightarrow r = 10\Omega$

Hence, the correct answer is Option B.

Note We should know that the main principle behind the working of the potentiometer is that the potential dropped across a segment of a wire which is of uniform cross- section carrying a constant current is directly proportional to its length.
In simple words we can say that the potentiometer is defined to be a simple device which is used to measure the electrical potentials.