In the electrical network, shown in the diagram `R_(1)=5Omega,R_(2)=2Omega,R_(3)=3Omega` and `E_(1)=2E_(2)=10V`, sources have negligible internal resistance. For this network,

Options

A. The power generated in `R_(1)` is `6.4J//s`
B. if the sources of e.m.f. `E_(1)` and `E_(2)` were physically inter changed, without changing the polarities, the power generated in `R_(1)` will be the same.
C. If the polarities of the sources `E_(1)` and `E_(2)` were both reversed,the power generated in `R_(1)` will be the same.
D. The ratio of the power generated in `R_(2)` and `R_(3)` `1:30`

Answer

A, C, D

Explanation

(A). Power generated P is `R_(1)` is `i_(1)^(2)R_(1)` now `i_(1)=(V)/(R_(1))`
where V is the potential at `P`, assuming that the negatives of the batteries are earthed. Simple application of Kirchhoff's junction and loop laws
yields `i_(3)=(45)/(31)A,i_(2)=-(10)/(31)A`
`i_(1)=i_(2)+i_(3)=(35)/(31)A`
`V=R_(1)i_(1)=(175)/(31)=5.7V`
When `R_(1)=5Omega,R_(2)=2Omega,R_(3)=3Omega`
Hence `P=(V^(2))/(R_(1))=((5.7)^(2))/(5)=6.4(J)/(s)`
(B). Current in `R_(1)` will be different if the sources of emf `E_(1)` and `E_(2)` are interchanged.
(C). Current in `R_(1)` will be reversed in direction but numerically be the same hence P is not changed
(D). Ratio of powers in `R_(2)` and `R_(3)` is given by
`(P_(2))/(P_(3))=((V-5)^(2)//R_(2))/((10-V)^(2)//R_(3))=((0.7)^(2)//2)/((4.3)^(2)//3)approx(1)/(30)`