JEE Mains · Physics · STD 12 - 3. current electricity
In the given potentiometer circuit arrangement, the balancing length \({AC}\) is measured to be \(250\) \({cm}\). When the galvanometer connection is shifted from point \((1)\) to point \((2)\) in the given diagram, the balancing length becomes \(400\, {cm}\). The ratio of the emf of two cells, \(\frac{\varepsilon_{1}}{\varepsilon_{2}}\) is -
- A \(\frac{4}{3}\)
- B \(\frac{3}{2}\)
- C \(\frac{5}{3}\)
- D \(\frac{8}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{3}\)
Step-by-step Solution
Detailed explanation
\(E_{1}=k l_{1} \ldots .\) (i) \(E_{1}+E_{2}=k l_{2} \ldots\) (ii) \(\frac{E_{1}}{E_{1}+E_{2}}=\frac{l_{1}}{l_{2}}=\frac{250}{400}=\frac{5}{8}\) \(8 E_{1}=5 E_{1}+5 E_{2}\) \(3 E_{1}=5 E_{2}\) \(\frac{E_{1}}{E_{2}}=\frac{5}{3}\)
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