V=100Vf=50Hz
CaseA:
Ia=8APa=120W
Ra=Pa/Ia2=120/82=1.875Ω
Za=V/Ia=100/8=12.5Ω
Xa=Za2−Ra2=12.52−1.8752=12.36Ω
⟹Laω=12.36,La=12.36/2π50=39.34mH
CaseB:
Ib=10APb=500W
Rb=Pb/Ib2=500/102=5Ω
Zb=V/Ib=100/10=10Ω
Xb=Zb2−Rb2=102−52=8.66Ω
⟹Lbω=8.66,Lb=8.66/2π50=27.56mH
When La and Lb are connected in series current Iab=V/(Xa+Xb)=100/(12.36+8.66)=4.76A
Power= 0 , as both are inductors there is no power loss.
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