First slide

potentiometer

Question

Balancing length of a potentiometer wire of a cell of emf 20 volt is 80 cm. What will be the balancing length of a cell of emf 15 volt in the same potentiometer wire?

Easy

A

65 cm
B

55 cm
C

60 cm
D

70 cm

Solution

$\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = 80 \times \frac{15}{20} c m = 60 c m$

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