In the following circuit, a conducing wire is connected between the terminals B and D, the current in the wire will

Options

A. flow from B to D
B. flow from D to B
C. the zero
D. flow in the direction which will be decided by the voltage V

Answer

B

Explanation

This is an unbalanced Wheatstone's network.
Curent through the arm ABC is given by `I_(1)=(V)/(8) A and " current through " ADC=(V)/(4)A=I_(2)`
`therefore V_(A)-V_(B)=I_(1)R=(V)/(8)xx 4= (V)/(2)`
and `V_(A)-V_(D)=I_(2)R' = (V)/(4)xx1=(V)/(4)`
`therefore (V_(A)-V_(B))-(V_(A)-V_(D))=(V)/(2)-(V)/(4)=(V)/(4)`
`therefore V_(D)-V_(B)=(V)/(4)`
`because V_(D) gt V_(B)`, the current will flow from D to B.