- #1
CasterSkux
- 16
- 0
Hello fellow engineers!
I am a student doing a simple course in Electrical Engineering.
I've got an enquiry regarding this question
"
1. In a three phase 4 wire system with phase to neutral voltage of 230V, a balanced set of resistive loads of 8 ohms are connected between each phase and neutral.
If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?
""
We use a program to calculate the symmetric components. I believe I need to construct the resultant phasor of this system and derive the symmetric components
This is my approach:
1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V
2) Calculate the balanced current that flows through each resistor I = V/R, I = 132.7906V/8 = 16.5988.
3) If Phase C is eliminated from the fuse, I think that the line/phase voltages stay the same (as there is a neutral) and hence the current is the same.
4) Therefore Phase A is 16.5988 amps and so is Phase B. Phase B is offsetted by 120 degrees taking A as the relative point.
5) Phase C is 0 and it's on the 240 degree mark on the phasor diagram.I was wondering if this is all right in approach?
I am a student doing a simple course in Electrical Engineering.
I've got an enquiry regarding this question
"
1. In a three phase 4 wire system with phase to neutral voltage of 230V, a balanced set of resistive loads of 8 ohms are connected between each phase and neutral.
If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?
""
We use a program to calculate the symmetric components. I believe I need to construct the resultant phasor of this system and derive the symmetric components
This is my approach:
1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V
2) Calculate the balanced current that flows through each resistor I = V/R, I = 132.7906V/8 = 16.5988.
3) If Phase C is eliminated from the fuse, I think that the line/phase voltages stay the same (as there is a neutral) and hence the current is the same.
4) Therefore Phase A is 16.5988 amps and so is Phase B. Phase B is offsetted by 120 degrees taking A as the relative point.
5) Phase C is 0 and it's on the 240 degree mark on the phasor diagram.I was wondering if this is all right in approach?