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In the circuit shown, the potential difference between A and B is:

 

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P.D. across AB will be equal to battery equivalent across CD 

\mathrm{v}_{\mathrm{AB}}=\mathrm{v}_{\mathrm{cD}}=\frac{\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{E}_{3}}{\mathrm{r}_{3}}}{\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}} 

                     =\frac{\frac{1}{1}+\frac{2}{1}+\frac{3}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=\frac{6}{3}=2 v 

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Satyajeet Kumar

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