question_answer

A)  \[\frac{20}{7}V\]                            

B)  \[\frac{40}{7}V\]

C)  \[\frac{10}{7}V\]                            

D)  Zero

Correct Answer: D

Solution :

                From the given figure, current through lower branch of resistance which are joined in series. \[{{i}_{1}}=\frac{10}{4+3}=\frac{10}{7}A\] Again, current through upper branch of resistance which are also connected in series. \[{{i}_{2}}=\frac{10}{8+6}=\frac{10}{14}A\] Now, according to the Kirchhoffs voltage law \[{{V}_{B}}-{{V}_{A}}=8\times {{i}_{2}}-4\times {{i}_{1}}\] \[=8\times \frac{10}{14}-4\times \frac{10}{7}\] \[=0\]