In the given circuit of potentiometer, the potential difference \(E\) across \(AB\) ( \(10\, m\) length) is larger than \(E _{1}\) and \(E _{2}\) as well. For key \(K _{1}\) (closed), the jockey is adjusted to touch the wire at point \(J_{1}\) so that there is no deflection in the galvanometer. Now the first battery \(\left( E _{1}\right)\) is replaced by second battery \(\left( E _{2}\right)\) for working by making \(K _{1}\) open and \(K _{2}\) closed. The galvanometer gives then null deflection at \(J _{2}\). The value of \(\frac{ E _{1}}{ E _{2}}\) is \(\frac{ a }{ b },\) where \(a =\) ...............

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Answer & Solution

Correct Answer

(B) \(1\)

Detailed explanation

Length of \(AB =10 m\)

For battery \(E _{1}\), balancing length is \(l_{1}\) \(l_{1}=380 cm [\) from end \(A ]\)

For battery \(E _{2}\), balancing length is \(l_{2}\) \(l_{2}=760 cm [\) [from end \(A ]\)

Now, we know that \(\frac{ E _{1}}{ E _{2}}=\frac{l_{1}}{l_{2}}\)

\(\Rightarrow \frac{E_{1}}{E_{2}}=\frac{380}{760}=\frac{1}{2}=\frac{a}{b}\)

\(\therefore a=1 \& b=2\)

\(a =1\)

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