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#### guitarboyled

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It's been a while...

Question #1: The current measured on the secondary is at what voltage?

Question #2: Based on the readings what are the total kVA being used?

#### [email protected]

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##### OldGeek
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I'll punt….120v and 3.84 KVA.

#### EB Electric

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Homework questions eh? Who's questions, yours or the teachers? The wording sucks I'll tell you that much for now.

#### guitarboyled

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Homework questions eh? Who's questions, yours or the teachers? The wording sucks I'll tell you that much for now.
I figured it would look like homework but it's not...

I've been a member for 6 years lolll

Just not sure of this one... I'm trying to figure our how many kVA are left for a new application

#### guitarboyled

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I'm a little puzzled, it's been a while and I'm more in the HVAC end of things

Available current per line should be 83.3 A.

30 000 VA / (208 V x 1.732) = 83.3 A

If I measure 10 A on line A does that mean 73.3 A are available?

#### EB Electric

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Your secondary voltage is 120V L-N and 208V L-L. In very basic theory you are going to have 3.84kVA load. The total available current on the secondary side is 83A. What you are saying is correct yes.

#### micromind

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I'm a little puzzled, it's been a while and I'm more in the HVAC end of things

Available current per line should be 83.3 A.

30 000 VA / (208 V x 1.732) = 83.3 A

If I measure 10 A on line A does that mean 73.3 A are available?
Yes, that is correct.

#### guitarboyled

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83.3 A per line x 120 V = 10 000 VA

10 000 VA per line x 3 lines = 30 000 VA or 30 kVA

Some at my work believe that the original clamp meter readings (10A, 12A and 10A) are at 208V.

How can I prove that they're at 120V?

#### guitarboyled

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If I add a 6 kVA charge with 208 V single phase input power, how much would that pull on each line?

Is it :

6000 VA / 208 V = 28.8 A

or

6000 VA / 2 lines = 3000 VA / 120 V = 25 A

If I add that charge to my panel via a single phase two pole breaker to the A and B circuits in addition to the already existing 10 A on phase A and 12 A on phase B what would be the current measured with a clamp meter on those two phases?

#### Barjack

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It's been a while...

Question #1: The current measured on the secondary is at what voltage?

Question #2: Based on the readings what are the total kVA being used?
I'm a little rusty on this too, and I hope one of the experts will chime in.

I don't think the 10A (A), 12A (B), and 10A (C) tells the whole story as far as KVA is concerned. You would need to measure each load (not just the feeders) and find the VA per load, then add that up.

3 phase: 208V x A x 1.732
1 phase: 208V x A
+ 1 phase: 120V x A
Total VA

I think?

#### micromind

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If I add a 6 kVA charge with 208 V single phase input power, how much would that pull on each line?

Is it :

6000 VA / 208 V = 28.8 A

or

6000 VA / 2 lines = 3000 VA / 120 V = 25 A

If I add that charge to my panel via a single phase two pole breaker to the A and B circuits in addition to the already existing 10 A on phase A and 12 A on phase B what would be the current measured with a clamp meter on those two phases?
It doesn't matter if the loads are 120, 208 or 3Ø, the transformer in question is good for 83.3 amps per phase. Neutral current doesn't matter, only phase current.

#### guitarboyled

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It doesn't matter if the loads are 120, 208 or 3Ø, the transformer in question is good for 83.3 amps per phase. Neutral current doesn't matter, only phase current.
But from what I understand, to compare apples with apples the 83.3 A is associated/relative to 120 V.

Therefore I'm guessing loads need to be converted to 120 V if I want to do any sort of calculation in regard to current per line.

For example, if I want to know the theoretical value that would be measured with a clamp meter after adding a load.

#### AK_sparky

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Some at my work believe that the original clamp meter readings (10A, 12A and 10A) are at 208V.

How can I prove that they're at 120V?
Current is current regardless of voltage. You have 83.3A per phase, regardless of if you hook it up LL (208V) or LN (120V).

If I add a 6 kVA charge with 208 V single phase input power, how much would that pull on each line?

Is it :

6000 VA / 208 V = 28.8 A

or

6000 VA / 2 lines = 3000 VA / 120 V = 25 A
Since you state it is 208V, why would you hook it up to 120V?

#### guitarboyled

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Current is current regardless of voltage. You have 83.3A per phase, regardless of if you hook it up LL (208V) or LN (120V).

Since you state it is 208V, why would you hook it up to 120V?
Tell me if I'm right...

Let's say I have 30 kVA and therefore 83.3 A available per line/phase

Option A : If I add 6000 VA @ 208/3 phases (3 line wires)

Current = 6000 VA / (208 x 1.732) = 16.7 A

I therefore have 66.6 A left on each line

Option B : If I add 6000 VA @ 208/1 phase (2 line wires)

Current = 6000 VA / 208 = 28.8 A

I therefore have 54.5 A left on line A and B and 83.3 on line C

Option C : If I add 6000 VA @ 120/1 phase (1 line wire + 1 neutral)

Current = 6000 VA / 120 V = 50 A

I therefore have 33.3 A left on line A and 83.3 A on line B and C

#### AK_sparky

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But you already stated that you are using option B (6kVA 208V 1p). What is the question?

#### guitarboyled

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But you already stated that you are using option B (6kVA 208V 1p). What is the question?
I'm trying to figure out the right way of calculating amps per line being used

#### AK_sparky

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So is the charger not 6kVA 208V 1p like you said? Or are you looking at other options that are 120V or 3p?

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I'm a little rusty on this too, and I hope one of the experts will chime in.

I don't think the 10A (A), 12A (B), and 10A (C) tells the whole story as far as KVA is concerned. You would need to measure each load (not just the feeders) and find the VA per load, then add that up.

3 phase: 208V x A x 1.732
1 phase: 208V x A
+ 1 phase: 120V x A
Total VA

I think?
Read AK Sparky's post CURRENT IS CURRENT, the measured current at the transformer terminals is the current the transformer is seeing

Barjack

#### guitarboyled

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Read AK Sparky's post CURRENT IS CURRENT, the measured current at the transformer terminals is the current the transformer is seeing
I finally got it while writing the following post:

Let's say I have 30 kVA and therefore 83.3 A available per line/phase

Option A : If I add 6000 VA @ 208/3 phases (3 line wires)

Current = 6000 VA / (208 x 1.732) = 16.7 A

I therefore have 66.6 A left on each line

Option B : If I add 6000 VA @ 208/1 phase (2 line wires)

Current = 6000 VA / 208 = 28.8 A

I therefore have 54.5 A left on line A and B and 83.3 on line C

Option C : If I add 6000 VA @ 120/1 phase (1 line wire + 1 neutral)

Current = 6000 VA / 120 V = 50 A

I therefore have 33.3 A left on line A and 83.3 A on line B and C

Thanks guys

#### micromind

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I finally got it while writing the following post:

Let's say I have 30 kVA and therefore 83.3 A available per line/phase

Option A : If I add 6000 VA @ 208/3 phases (3 line wires)

Current = 6000 VA / (208 x 1.732) = 16.7 A

I therefore have 66.6 A left on each line

Option B : If I add 6000 VA @ 208/1 phase (2 line wires)

Current = 6000 VA / 208 = 28.8 A

I therefore have 54.5 A left on line A and B and 83.3 on line C

Option C : If I add 6000 VA @ 120/1 phase (1 line wire + 1 neutral)

Current = 6000 VA / 120 V = 50 A

I therefore have 33.3 A left on line A and 83.3 A on line B and C

Thanks guys
Exactly correct.

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