In a potentiometer wire experiment, the e.m.f of a battery in the primary circuit is 20 V and its internal resistance is 5Ω . There is a resistance box in series with the battery and the potentiometer wire. The resistance of resistance box can be varied from 120Ω to 170Ω . Resistance of the potentiometer wire is 75Ω. The maximum value of e.m.f/potential differences that can be measured using this potentiometer is xV . Then what is 10 X x value

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answer is 75.

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Detailed Solution

$E_{measure} \leq V_{wire} = \frac{E_{0} R_{0}}{r_{0} + R_{5} + R_{0}} = \frac{20 \times 75}{5 + 120 + 75} = 7 .5 V$

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