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ric115
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Hi There
I have attempted the following question can anyone confirm what i have done and help with the rest of the question. Thanks
Question
Derive the numerical relationship between the line and phase currents for a balanced three-phase delta-connected load.
Three coils are connected in delta to a three-phase, three-wire, 400V, 50Hz supply and take a line current of 5A 0.8 power factor lagging. Calculate the resistance and inductance of the coils.
If the coils are star-connected to the same supply, calculate the line current and the total power.
Calculate the line currents if one coil becomes open-circuited when the coils are connected in star.
This is what i have;
EL = 400V
IL = 5 A
P.F cos0 = 0.8 lagging
Iph = IL/√3 = 5/√3 = 2.887 A
Impedance of each phase, Zph = Eph/Iph = 400/2.887 = 138.6 Ohms
Resistance of each phase, Rph = Zph cos0 = 138.6 x 0.8 = 110.8 Ohms
Reactance of each phase, Xph = Zph cos0 = 138.6 √1 - (o.8)^2 = 83.2 Ohms
Therefore inductance = Xph/2πf = .265 H
Coils connected in star;
Eph = EL/√3 = 231
Iph = Eph/Zph = 231/138.6 = 1.667 A ( Iph = IL )
Using P = √3 x EL x IL cos o.8
P = √3 x 400 x 1.667 x 0.8
P = 923.9 W
one coil becomes open-circuited; ?
Thanks
I have attempted the following question can anyone confirm what i have done and help with the rest of the question. Thanks
Question
Derive the numerical relationship between the line and phase currents for a balanced three-phase delta-connected load.
Three coils are connected in delta to a three-phase, three-wire, 400V, 50Hz supply and take a line current of 5A 0.8 power factor lagging. Calculate the resistance and inductance of the coils.
If the coils are star-connected to the same supply, calculate the line current and the total power.
Calculate the line currents if one coil becomes open-circuited when the coils are connected in star.
This is what i have;
EL = 400V
IL = 5 A
P.F cos0 = 0.8 lagging
Iph = IL/√3 = 5/√3 = 2.887 A
Impedance of each phase, Zph = Eph/Iph = 400/2.887 = 138.6 Ohms
Resistance of each phase, Rph = Zph cos0 = 138.6 x 0.8 = 110.8 Ohms
Reactance of each phase, Xph = Zph cos0 = 138.6 √1 - (o.8)^2 = 83.2 Ohms
Therefore inductance = Xph/2πf = .265 H
Coils connected in star;
Eph = EL/√3 = 231
Iph = Eph/Zph = 231/138.6 = 1.667 A ( Iph = IL )
Using P = √3 x EL x IL cos o.8
P = √3 x 400 x 1.667 x 0.8
P = 923.9 W
one coil becomes open-circuited; ?
Thanks