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Resistances P, Q, S and R are arranged i...

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

A

`1:1`

B

`R:P`

C

`P^(2):Q^(2)`

D

`P^(2):R^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
For balanced Wheatstone,s 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 Rightarrow (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)/(R)
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