A) \[20/7V\]
B) \[40/7V\]
C) \[10/7V\]
D) zero
Correct Answer: D
Solution :
From the given figure current through lower branch of the circuit (resistance are connected in series) Current drawn \[=\frac{Total\text{ }e.m.f}{Total\text{ }resistance}\] \[{{i}_{1}}=\frac{10}{4+3}=\frac{10}{7}amp.\] \[{{V}_{A}}-{{V}_{P}}=4\times {{i}_{1}}=4\times \frac{10}{7}=\frac{40}{7}\,volt.\] Now current through upper branch of the circuit (resistance are connected in series) is given by: \[{{i}_{2}}=\frac{Total\text{ }e.m.f}{~total\text{ }resistance}\] \[=\frac{10}{8+6}=\frac{10}{14}amp.\] \[{{V}_{B}}-{{V}_{P}}=8\times {{i}_{2}}\] \[=8\times \frac{10}{14}=\frac{40}{7}\,volt\] Again applying Kirchhoffs voltage law, \[{{V}_{B}}-{{V}_{A}}=({{V}_{B}}-{{V}_{P}})-({{V}_{A}}-{{V}_{P}})\] \[\Rightarrow \] \[{{V}_{B}}-{{V}_{A}}=\frac{40}{7}-\frac{40}{7}=0\]You need to login to perform this action.
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