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A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of $10\,\Omega$. Its internal resistance is -
A
2.0 ohm
B
zero
C
1.0 ohm
D
0.5 ohm
Detailed Solution
In a potentiometer the potential difference is proportional to the balancing length.
On open circuit the cell balances at the longer length, so emf $E \propto l_1$ with $l_1 = 110$ cm.
When the cell is shunted by $R = 10\,\Omega$, its terminal voltage $V = \dfrac{E}{R + r}\cdot R \propto l_2$ with $l_2 = 100$ cm.
$\dfrac{E}{V} = \dfrac{R + r}{R} = \dfrac{l_1}{l_2}$
$\Rightarrow r = \left(\dfrac{l_1}{l_2} - 1\right)R = \left(\dfrac{l_1 - l_2}{l_2}\right)R$
$r = \dfrac{110 - 100}{100} \times 10$
$r = 1\,\Omega$
(The larger balancing length always belongs to the open-circuit emf, since $E \gt V$.)
On open circuit the cell balances at the longer length, so emf $E \propto l_1$ with $l_1 = 110$ cm.
When the cell is shunted by $R = 10\,\Omega$, its terminal voltage $V = \dfrac{E}{R + r}\cdot R \propto l_2$ with $l_2 = 100$ cm.
$\dfrac{E}{V} = \dfrac{R + r}{R} = \dfrac{l_1}{l_2}$
$\Rightarrow r = \left(\dfrac{l_1}{l_2} - 1\right)R = \left(\dfrac{l_1 - l_2}{l_2}\right)R$
$r = \dfrac{110 - 100}{100} \times 10$
$r = 1\,\Omega$
(The larger balancing length always belongs to the open-circuit emf, since $E \gt V$.)