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In a potentiometer wire experiment the emf of a battery in the primary circuit is 20volt and its internal resistance is 5 $Ω$ . There is a resistance box (in series with the battery and the potentiometer wire) whose resistance can be varied from 120 $Ω$ to 170 $Ω$ . Resistance of the potentiometer wire is 75 $Ω$ . The following potential differences can be measured using this potentiometer

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Detailed Solution

$\begin{array}{l} i_{\max} = \frac{20}{5 + 75 + 120} = \frac{1}{10} a m p \\ V_{\max} = i_{\max} R_{75} = \frac{1}{10} \times 75 = 7.5 V \end{array}$

Range of potentiometer 0 to 7.5 V

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