Q : In an experiment with a potentiometer, V_{B} = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E1 of a battery (approx. 8V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the last (4th )segment of the potentiometer, find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Solution : The Null point is obtained only when emf of Primary cell is less than Potential difference across the wire of potentiometer .

The Equivalent Resistance of Potentiometer & Variable Resistor is R_{eq} = 50 + R_{1}

Applied Voltage , V = 10 volt

Therefore current through the main circuit is

$\large I = \frac{10}{50 + R_1}$

The null point is obtained only with 50 ohm resistor when

$\large \frac{10 \times R_1}{10 + R_1} > 8 $

R_{1} > 40