Resistances P, Q, S and R are arranged in a cyclic order to form a balanced Wheatstone's bridge network. The ratio of power consumed in the branches `(P+Q)` and `(R+S)` is

Options

A. `1:1`
B. `R:P`
C. `P^(2):Q^(2)`
D. `P^(2):R^(2)`

Answer

B

Explanation

For balanced Wheatstone bridge `(P)/(Q)=(R )/(S)" "…(i)`
Power dissipation in resistance R with voltage V is `V^(2)//R`.
`therefore (P_(P+Q))/(P_(R+S))=(R+S)/(P+Q)" " ...(ii)`
From equation (i) `(P)/(Q)+1=(R)/(S)+1`
`(P+Q)/(Q)=(R+S)/(S)or(R+S)/(P+Q)=(S)/(Q)`
Using (i), we get
`(R+S)/(P+Q)=(R)/(P), therefore (P_(P+Q))/(P_(R+S))=(R)/(P)`