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- Thread starter Rajan
- Start date

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:

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.

1) 3Ø - 480:

480

2) 1Ø - 277

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=

480

Hi Sam,3Ø - 480:

Vd-Value * sqrt(3)

480

1Ø - 277/480

Vd-Value

277

1) 3Ø - 480:

480

2) 1Ø - 277

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=

480

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