JEE Mains · Physics · STD 12 - 3. current electricity
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 \(a =\) ...............
- A \(2\)
- B \(1\)
- C \(4\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
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 \(\frac{ E _{1}}{ E _{2}}=\frac{l_{1}}{l_{2}}\)…
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