A potentiometer wire is supplied with a constant voltage of 3 V. A cell of EMF 1.08 V is balanced by a voltage drop across 216 cm of the wire. Find the total length of the wire.

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Solution

$Working principle of potentiometer : V α L \frac{L_{2}}{L_{1}} = \frac{V_{2}}{V_{1}} L_{2} = 216 \times \frac{3}{1.08} = 600 cm$

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A potentiometer wire is supplied with a constant voltage of 3 V. A cell of EMF 1.08 V is balanced by a voltage drop across 216 cm of the wire. Find the total length of the wire.

Working principle of potentiometer:V α LL2L1=V2V1L2=216×31.08 =600 cm

$Working principle of potentiometer : V α L \frac{L_{2}}{L_{1}} = \frac{V_{2}}{V_{1}} L_{2} = 216 \times \frac{3}{1.08} = 600 cm$