question_answer

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

A)  \[\frac{{{E}_{1}}+{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}+R}\times R\]

B)  \[\frac{{{E}_{1}}-{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}+R}\times R\]

C)  \[\frac{{{E}_{1}}+{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}}\times R\]

D)  \[\frac{{{E}_{1}}-{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}}\times R\]

Correct Answer: B

Solution :

:  As emfs \[{{E}_{1}}\]  and \[{{E}_{2}}\] are in opposition and \[{{E}_{1}}>{{E}_{2}}\], so equivalent emf is The equivalent resistance of the circuit is \[{{R}_{eq}}={{r}_{1}}+{{r}_{2}}+R\] The current in the circuit is \[I=\frac{{{E}_{eq}}}{{{R}_{eq}}}=\frac{{{E}_{1}}-{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}+R}\] The terminal potential difference is \[V=IR=\frac{{{E}_{1}}-{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}+R}\times R\]