First slide

potentiometer

Question

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 $10 Ω$ Its internal resistance is

Moderate

A

$1.0 Ω$
B

$0.5 Ω$
C

$2 .0 Ω$
D

zero

Solution

This problem is based on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then
$\frac{E}{V} = \frac{l_{1}}{l_{2}}$
where l₁ and l₂ are lengths of potentiometer wire with and without short circuited through a resistance.
$\begin{array}{l} Since \frac{E}{V} = \frac{R + r}{R} [∵ E = I (R + r) and V = IR] \\ ∴ \frac{R + r}{R} = \frac{l_{1}}{l_{2}} \\ or 1 + \frac{r}{R} = \frac{110}{100} or \frac{r}{R} = \frac{10}{100} \\ or r = \frac{1}{10} \times 10 = 1 Ω \end{array}$

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