Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :

NEETNEET 2016Current Electricity

Solution:

$\frac{E_{1} + E_{2}}{E_{1}-E_{2}} = \frac{50}{10} $
$ \Rightarrow \frac{2 E_{1}}{2E_{2}} = \frac{50+10}{50-10} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{3}{2} $