Question 193 of 380

For the circuit shown, with ${R}_{1}=1.0\Omega$ , ${R}_{2}=2.0\Omega$ , ${E}_{1}=2V$ and ${E}_{2}={E}_{3}=4V,$ the potential difference between the points $‘a’$ and $‘b’$ is approximately ( $inV$ ):

A$3.3$B$2.3$C$2.7$D$3.7$

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Answer: (A) $3.3$

If the electric potential at point $a$ is ${V}_{a}$ and the electric potential at point $b$ is ${V}_{b}$ then
${V}_{b}-{V}_{a}=\dfrac{\dfrac{2}{2}+\dfrac{4}{2}+\dfrac{4}{2}}{\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}}$
${V}_{b}-{V}_{a}=\dfrac{1+2+2}{\dfrac{3}{2}}=\dfrac{10}{3}$
$∴{V}_{b}-{V}_{a}=3.3V$