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The potential difference between points $A$ and $B$ is
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The correct answer is:
zero
From the given figure, current through lower branch of resistances which are joined in series is $i_1=\frac{10}{4+3}=\frac{10}{7} \mathrm{amp}$
Again current through upper branch of resistances which are also connected in series, is
$i_2=\frac{10}{8+6}=\frac{10}{14} \mathrm{amp}$
Now according to the Kirchhoff's voltage law
$\begin{aligned} V_B & -V_A=8 \times i_2-4 \times i_1 \\ & =8 \times \frac{10}{14}-4 \times \frac{10}{7}=0\end{aligned}$
Again current through upper branch of resistances which are also connected in series, is
$i_2=\frac{10}{8+6}=\frac{10}{14} \mathrm{amp}$
Now according to the Kirchhoff's voltage law
$\begin{aligned} V_B & -V_A=8 \times i_2-4 \times i_1 \\ & =8 \times \frac{10}{14}-4 \times \frac{10}{7}=0\end{aligned}$
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