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 \[V=IR=\frac{{{E}_{1}}-{{E}_{2}}}{{{r}_{1}}+{{r}_{2}}+R}\times R\]You need to login to perform this action.
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