In the circuit shown, the potential difference between $\mathrm{A}$ and $B$ is
In the circuit shown, the potential difference between $\mathrm{A}$ and $B$ is
- $1 \mathrm{~V}$
- $2 \mathrm{~V}$
- $3 V$
- $6 \mathrm{~V}$
Solution
Given, $E_{1}=1 \mathrm{~V}, \mathrm{E}_{2}=2 \mathrm{~V}, \mathrm{E}_{3}=3 \mathrm{~V}, \mathrm{r}_{1}=1 \Omega$
$\mathrm{r}_{2}=1 \Omega$ and $\mathrm{r}_{3}=1 \Omega$
$\mathrm{V}_{\mathrm{AB}}=\mathrm{V}_{\mathrm{CD}}=\frac{\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{E}_{3}}{\mathrm{r}_{3}}}{\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}}=\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{3}{1}}=\frac{6}{3}=2 \mathrm{~V}$
Asked in: JEE Main 2019 (11 Jan Shift 2)
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