Current Electricity

JEE Main level, single correct, medium

218 of 380

Question 218 of 380

In the circuit shown, the potential difference between A\mathrm{A} and BB is

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Answer: (B) 2 V2 \mathrm{~V}

Given, E1=1 V,E2=2 V,E3=3 V,r1=1ΩE_{1}=1 \mathrm{~V}, \mathrm{E}_{2}=2 \mathrm{~V}, \mathrm{E}_{3}=3 \mathrm{~V}, \mathrm{r}_{1}=1 \Omega

r2=1Ω\mathrm{r}_{2}=1 \Omega and r3=1Ω\mathrm{r}_{3}=1 \Omega

VAB=VCD=E1r1+E2r2+E3r31r1+1r2+1r3=111+11+11+31=63=2 V\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}

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