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0Suppose you manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line, as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with a component of current that is 90° out of phase with the voltage, as well as with current in phase with the voltage. The electric company charges you an extra fee for “reactive volt-amps,” in addition to the amount you pay for the energy you use. You can avoid the extra fee by
installing a capacitor between the power line and your factory. The following problem models this solution.In an RL circuit, a 120-V (rms), 60.0-Hz source is in series with a 25.0-mH inductor and a 20.0-( resistor. What are (a) the rms current and (b) the power factor? (c) What capacitor must be added in series to make the power factor 1? (d) To what value can the supply voltage be reduced, if the power supplied is to be the same as before the capacitor was installed?
bunu cevaplayan gercek aynştayndır beyler.
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0i kiss you
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0@4 sıçtı
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0@4... gibtir git kaybol ya! sıçtın batırdın amın evladı!
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0lisede bu kadar ağır fizik var mıydı lan?
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0which chair
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0yes i fuck you i'm not sorry i'll fuck you tomorrow peach
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0am günü yağ
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0@4 gerizekalı mısın lan sen
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0king of copy paste
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0sum up sucker
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0@11 ilginç bir tecübe olacak peach
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0aynştayn alman bin..
not=aynştaynım -
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0güç faktörünü hesaplamak için kapasitörle indüktörün ohmik direncini bulmak lazım sonra normal dirençle yaptığı acıyı bulacan öle basit bi kaç işlem işte
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+1@17 zütünü gibiyim ben yazıcaktım onu
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0power factor=0.9
@1 cevap bu işte amk
edit-aynştaynım destekleyin beni -
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0e=mc2
inci sözlük hatası: entry metni girilmelidir.