In the given network, the powers generated in the resistor `R_(1)` and `R_(2)` are `P_(1)` and `P_(2)` and the power generated in `R` is P. For this network.

Options

A. if `P_(1)=P_(2)` it necessarily follows that `R_(1)=R_(2)`
B. if `R=R_(1)+R_(2)`, then `PgtP_(1)+P_(2)`
C. if `P=P_(1)+P_(2)` then `Rle(R_(1)+R_(2))/(4)`
D. if `R_(1)=R_(2)` then it necessarily follows that `P=P_(1)+P_(2)`

Answer

A, B, C

Explanation

(A). If `P_(1)=P_(2),(V^(2))/(R_(1))=(V^(2))/(R_(2))` Where V is the difference of potential from P to Q. Hence `R_(1)=R_(2)`
(B). `P=i^(2)R=i^(2)(R_(1)+R_(2))=i^(2)R_(1)+i^(2)R_(2)gti_(1)^(2)R_(1)+i_(2)^(2)R_(2)`
since `igti_(1)` and `igti_(2)` Hence `PgtP_(1)+P_(2)`
(C). let `P=P_(1)+P_(2)` then `i^(2)R=i_(1)^(2)R_(1)+i_(2)^(2)R_(2)`
but `i=i_(1)+i_(2)implies(i_(1)+i_(2))^(2)R=i_(1)^(2)R_(1)+i_(2)^(2)R_(2)`
now `(i_(1)+i_(2))^(2)=(i_(1)-i_(2))^(2)+4i_(1)i_(2)`
Since `i_(1)` and `i_(2)` are both positive, maximum value of `(i_(1)+i_(2))` is when `i_(1)=i_(2)=(i)/(2)`
Then from (1), `4R=R_(1)+R_(2)`
Hence maximum value R can have consistent with
`P=P_(1)+P_(2)` is `(R_(1)+R_(2))/(4)` Hence `Rle(R_(1)+R_(2))/(4)`