1.

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

Answer»

`1:1`
`R:P`
`P^(2):Q^(2)`
`P^(2):R^(2)`

Solution :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)`


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