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JEE MAIN 2019
08-04-2019 S1
Question
For the circuit shown, with $\mathrm{R} 1=1.0 \Omega, \mathrm{R} 2=2.0 \Omega$, $\mathrm{E} 1=2 \mathrm{~V}$ and $\mathrm{E} 2=\mathrm{E} 3=4 \mathrm{~V}$, the potential difference between the points ' $a$ ' and ' $b$ ' is approximately (in V)
Question Image
Select the correct option:
A
2.7
B
3.7
C
2.3
D
3.3
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & V=\frac{E_1 r_2 r_3+E_2 r_3 r_1+E_3 r_1 r_2}{r_1 r_2+r_2 r_3+r_3 r_1} \\ & =\frac{2(2 \times 2)+4(2 \times 2)+4(2 \times 2)}{4+4+4} \\ & =\frac{40}{12}=\frac{10}{3}=3.3 \text { Volt }\end{aligned}$
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