Question

The potentiometer wire of length has 100 cm has a resistance of $10\Omega$.It is connected in series with a resistance of $5\Omega$ and an acceleration of emf 3V having magnitude resistance. A source. A source of 1.2 V is balanced against length I' Of the potential wire. Find the value of L.

Answer 1

Current in potentiometer, $J=\frac{3}{R+10}=\frac{3}{5+10}=\frac{1}{5}$

Potential on potential meter wire, $V=J\times 10=\frac{1}{5}\times 10=2V$

Potential per unit length, $\frac{V}{L}=\frac{2}{100}=\frac{1}{50}Vc{m}^{-1}$

For balancing 2 V, length of potentiometer $L^{\prime }=\frac{1.2}{\left(\frac{1}{50}\right)}=60cm$

Hence, length is $60cm$