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, is \[{{i}_{1}}=\frac{10}{4+3}=\frac{10}{7}A\] Again current through upper branch of resistances which are also connected in sereis, is \[{{i}_{2}}=\frac{10}{8+6}=\frac{10}{14}A\] Now, according to the Kirchhoffs voltage law \[{{V}_{B}}-{{V}_{A}}=8\times {{i}_{2}}-4\times {{i}_{1}}\] \[=8\times \frac{10}{14}-4\times \frac{10}{7}=0\]You need to login to perform this action.
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