question_answer

The potential difference between point A and B is:

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,             \[{{V}_{B}}-{{V}_{A}}=({{V}_{B}}-{{V}_{P}})-({{V}_{A}}-{{V}_{P}})\] \[\Rightarrow \]            \[{{V}_{B}}-{{V}_{A}}=\frac{40}{7}-\frac{40}{7}=0\]