Percent Voltage Drop
- Thread starter Rajan
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Help Support Professional Engineer & PE Exam Forum:
To be a bit more clear and accurate about 3-phase voltage drop, you have to be consistent about whether you are using line-to-neutral or line-to-line voltage drop.
If you are doing line-to-neutral voltage drop, you have to calculate as a percentage of the source line-to-neutral voltage (e.g. 277 V).
If you are doing line-to-line voltage drop, you have to calculate as a percentage of the source line-to-line voltage (e.g. 480 V).
So that means you can't mix and match L-N and L-L voltage. Meaning you cannot do a line-to-line voltage divided by 277 V or a line-to-neutral voltage divided by 480 V in this problem.
If you are doing line-to-neutral voltage drop, you have to calculate as a percentage of the source line-to-neutral voltage (e.g. 277 V).
If you are doing line-to-line voltage drop, you have to calculate as a percentage of the source line-to-line voltage (e.g. 480 V).
So that means you can't mix and match L-N and L-L voltage. Meaning you cannot do a line-to-line voltage divided by 277 V or a line-to-neutral voltage divided by 480 V in this problem.
As Sam replied:
1) 3Ø - 480:
Vd-Value * sqrt(3)
480
2) 1Ø - 277
Vd-Value
277
3) What about 1Ø - 480?
Is this formula below correct for 1Ø - 480?
To calculate for Vd-value, do we use 1Ø formula?
Vd Percent= Vd-Value X100
480
Hi Sam,
1) 3Ø - 480:
Vd-Value * sqrt(3)
480
2) 1Ø - 277
Vd-Value
277
3) What about 1Ø - 480?
Is this formula below correct for 1Ø - 480?
To calculate for Vd-value, do we use 1Ø formula?
View attachment 23158
Vd Percent= Vd-Value X100
480
salmanshu322
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3@ - 480:
@ is equal to theta
Vd-Value * sqrt(3)
480
1@ - 277/480
Vd-Value
277
@ is equal to theta
Vd-Value * sqrt(3)
480
1@ - 277/480
Vd-Value
277
