Wildsoldier PE
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This problem ask for voltage drop at point A.
On this problem I assumed that the XL is inductive since is a line....now the problem treat Xl as capacitive by adding the (-) sign for the imaginary component of the complex number. If it is capacitive i think that the problem shall specify that. This is the way I did it...the book did it a little differently but the answer I think its wrong anyway.
[SIZE=12pt]Data provided:[/SIZE]
Per 1000 ft
Rl = .062
Xl = .041
Distance from the transformer 34.5KV/480V transformer to point A = 175ft
Current in the line = 200<25
[SIZE=12pt]Solution:[/SIZE]
Rl @175' = (.062 x 175) / 1000 = .01085
Xl @ 175' = (.041 x 175) / 1000 = .007175
V drop line to ground = Iline x Impedance = (200<25) (.01085 + .007175i) = 2.60<58.47 V
Now we need to change that voltage from line to ground to line to line= 2.60 squre(3) = 4.50 V line to line
The percentage voltage drop = (4.5V/480V) x 100 = .937 %
Any one agrees with me?
On this problem I assumed that the XL is inductive since is a line....now the problem treat Xl as capacitive by adding the (-) sign for the imaginary component of the complex number. If it is capacitive i think that the problem shall specify that. This is the way I did it...the book did it a little differently but the answer I think its wrong anyway.
[SIZE=12pt]Data provided:[/SIZE]
Per 1000 ft
Rl = .062
Xl = .041
Distance from the transformer 34.5KV/480V transformer to point A = 175ft
Current in the line = 200<25
[SIZE=12pt]Solution:[/SIZE]
Rl @175' = (.062 x 175) / 1000 = .01085
Xl @ 175' = (.041 x 175) / 1000 = .007175
V drop line to ground = Iline x Impedance = (200<25) (.01085 + .007175i) = 2.60<58.47 V
Now we need to change that voltage from line to ground to line to line= 2.60 squre(3) = 4.50 V line to line
The percentage voltage drop = (4.5V/480V) x 100 = .937 %
Any one agrees with me?
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