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Question

(A) The internal resistance $r$ of a cell using a potentiometer is given by the formula:$r = \left( \frac{l_1}{l_2} - 1 \right) R$where $l_1$ is the b

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) The internal resistance $r$ of a cell using a potentiometer is given by the formula:$r = \left( \frac{l_1}{l_2} - 1 ight) R$where $l_1$ is the balancing length when the cell is in an open circuit (i.e.,$R = \infty$) and $l_2$ is the balancing length when an external resistance $R$ is connected across the cell.Given:$l_1 = 3\,m$$l_2 = 2.85\,m$$R = 9.5\,\Omega$Substituting the values into the formula:$r = \left( \frac{3}{2.85} - 1 ight) 	imes 9.5\,\Omega$$r = \left( \frac{3 - 2.85}{2.85} ight) 	imes 9.5\,\Omega$$r = \left( \frac{0.15}{2.85} ight) 	imes 9.5\,\Omega$$r = \frac{15}{285} 	imes 9.5\,\Omega$$r = \frac{1}{19} 	imes 9.5\,\Omega = 0.5\,\Omega$Thus,the internal resistance of the cell is $0.5\,\Omega$.

Similar Questions

$A$ potentiometer has a uniform potential gradient across it. Two cells connected in series $(i)$ to support each other and $(ii)$ to oppose each other are balanced over $6 \ m$ and $2 \ m$ respectively on the potentiometer wire. The ratio of the $e.m.f.$s of the cells is:

In the potentiometer circuit as shown in the figure,the balance length $l = 60 \ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5 \ \Omega$,the balance length $l' = 50 \ cm$. The internal resistance of the cell $C'$ is : .............. $\Omega$

The arrangement as shown in the figure is called:

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