Question:
In the circuit shown, the potential difference between $\mathrm{A}$ and $B$ is:
Correct Option: 2,
Solution:
(2) Given, $E_{1}=1 V, E_{2}=2 V, E_{3}=3 V, r_{1}=1 \Omega$,
$r_{2}=1 \Omega$ and $r_{3}=1 \Omega$
$\mathrm{V}_{\mathrm{AB}}=\mathrm{V}_{\mathrm{CD}}=\frac{\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{E}_{3}}{\mathrm{r}_{3}}}{\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}}=\frac{\frac{1}{1}+\frac{2}{1}+\frac{3}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}$
$=\frac{6}{3}=2 \mathrm{~V}$
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