Two cells of emf $ E_{1} $ and $ E_{2} $ are joined in opposition (such that $ E_{1}E_{2} $ ). If $ r_{1} $ and $ r_{2} $ be the internal resistance and $ R $ be the external resistance, then the terminal potential difference is

Question

Two cells of emf $ E_{1} $ and $ E_{2} $ are joined in opposition (such that $ E_{1}>E_{2} $ ). If $ r_{1} $ and $ r_{2} $ be the internal resistance and $ R $ be the external resistance, then the terminal potential difference is, $ \frac{E_{1}+E_{2}}{r_{1}+r_{2}} \times R $, $ \frac{E_{1}+E_{2}}{r_{1}+r_{2}+R} \times R $, $ \frac{E_{1}-E_{2}}{r_{1}+r_{2}} \times R $, $ \frac{E_{1}-E_{2}}{r_{1}+r_{2}+R} \times R $

MARKS App · Accepted Answer

Here, current through the circuit $ =I=\frac{\text { net emf }}{\text { effective resistance }} $ $ \Rightarrow I=\frac{E_{1}-E_{2}}{r_{1}+r_{2}+R} $ Using ohm's law, $ V=I R $, Terminal potential difference $ =\frac{\left(E_{1}-E_{2}\right)}{\left(r_{1}+r_{2}+R\right)} \times R $

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