The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20$\Omega$, the null point on it is found to be a 1000 cm. The resistance of whole wire is :
Solution
Let Resistance per unit length of potentiometer wire = $\lambda$
<br><br>5 = $\lambda$ $\times$ 1000 $\times$ 60 $\times$ 10<sup>-3</sup>
<br><br>$\Rightarrow$ $\lambda$ = ${5 \over {60}}$
<br><br>Resistance of potentiometer wire = 1200 $\times$ ${5 \over {60}}$ = 100 $\Omega$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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