5 resistances `R_(1), R_(2),R_(3),R_(4),R_(5)` are joined as shown in the figure.

The values of `R_(1), R_(2), R_(3) and R_(4)` are so adjusted that the current in the circuit does not change for any value of the resistance `R_(5)`. This is possible for the following relation

Options

A. `R_(1)+R_(2)=R_(3)+R_(4)`
B. `(R_(1))/(R_(4))=(R_(2))/(R_(3))`
C. `R_(1)R_(4)=R_(3)R_(2)`
D. `(1)/(R_(1))+(1)/(R_(3))=(1)/(R_(2))+(1)/(R_(4))`

Answer

C

Explanation

This is balanced wheatstone's network.
`therefore` The relation is `(R_(1))/(R_(2))=(R_(3))/(R_(4)) or R_(1)R_(4)=R_(2)R_(3)`