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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, |
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Answer» The power generated in `R_(1)` is `6.4J//s`  (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)` |
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