First slide

potentiometer

Question

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_{0}$ and a resistance $r_{1}$ . An unknown e.m.f. E is balanced at a length $l$ of the potentiometer wire. The e.m.f. E will be given by

Moderate

A

$\frac{E_{0} l}{L}$
B

$\frac{L E_{0} r}{(r + r_{1}) l}$
C

$\frac{L E_{0} r}{l r_{1}}$
D

$\frac{E_{0} r}{(r + r_{1})} \cdot \frac{l}{L}$

Solution

The current through the potentiometer wire is
$I = \frac{E_{0}}{(r + r_{1})}$
and the potential difference across the wire is
$V = I r = \frac{E_{0} r}{(r + r_{1})}$
The potential gradient along the potentiometer wire is
$k = \frac{V}{L} = \frac{E_{0} r}{(r + r_{1}) L}$
As the unknown e.m.f. E is balanced against length $l$ of the potentiometer wire,
$∴ E = k l = \frac{E_{0} r}{(r + r_{1})} \frac{l}{L}$

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