A)
\[\,\frac{\text{20}}{\text{7}}\text{V}\]
B) \[\,\frac{\text{40}}{\text{7}}\text{V}\]
C) \[\,\frac{\text{10}}{\text{7}}\text{V}\]
D) zero
Correct Answer: D
Solution :
From the given figure, current through lower branch of resistances which are joined in series is \[{{i}_{1}}=\frac{10}{4+3}=\frac{10}{7}\,amp\]
Again current through upper branch of resistances which are also connected in series, is \[{{i}_{2}}=\frac{10}{8+6}=\frac{10}{14}amp\] Now according to the Kirchhoffs voltage law \[{{V}_{B}}-{{V}_{A}}=8\times {{i}_{2}}-4\times {{i}_{1}}\]
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