NEET · Physics · STD 12 - 3. current electricity
A potentiometer wire is \(100\,\, cm\) long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at \(50\,\, cm\) and \(10\,\, cm\) from the positive end of the wire in the two cases. The ratio of emf's is
- A \(5:4\)
- B \(3:4\)
- C \(3:2\)
- D \(5:1\)
Answer & Solution
Correct Answer
(C) \(3:2\)
Step-by-step Solution
Detailed explanation
Suppose two cells have emfs \(\varepsilon_{1}\) and \(\varepsilon_{2}\) \(\left(\text { also } \varepsilon_{1}>\varepsilon_{2}\right)\)
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