5% voltage drop equal impedance value?

[sparks1]

Question for an electrical engineer
I'm using Fluke 1654B multi tester to perform loop test from the furthest outlet.
The meter gives me the total impedance on the circuit all the way to the utility
Transformer. The code in general wants us keep everything below 5% vd. So if my math is correct and my way of thinking would be any value of impedance to and including 5 ohms or less would equal % of voltage drop.
Eg. 3 ohms impedance reading would equal 3% vd,
A 5 ohm reading would 5 % vd
A 20 ohm read would be 20% an so on...
THE CIRCUIT READS A STEADY123 VOLTS BEFORE APPLYING BEFORE APPLYING THE VD,
Mathematically when you convert total impedance to equal percent voltage drop it seems to work
Look forward to you comments

 

[luckylerado]

Question for an electrical engineer
I'm using Fluke 1654B multi tester to perform loop test from the furthest outlet.
The meter gives me the total impedance on the circuit all the way to the utility
Transformer. The code in general wants us keep everything below 5% vd. So if my math is correct and my way of thinking would be any value of impedance to and including 5 ohms or less would equal % of voltage drop.
Eg. 3 ohms impedance reading would equal 3% vd,
A 5 ohm reading would 5 % vd
A 20 ohm read would be 20% an so on...
THE CIRCUIT READS A STEADY123 VOLTS BEFORE APPLYING BEFORE APPLYING THE VD,
Mathematically when you convert total impedance to equal percent voltage drop it seems to work
Look forward to you comments

You are missing the current component. You need to multiply by current and find the %VD by dividing that # by the voltage at the source. I would suggest measuring the feeder / drop separately and apply the 2% and create my loop at the panel board for the branch circuits and apply 3%.

BTW 5 ohms is huge in terms of VD for a 120V branch circuit. Plug a 12 amp vacuum into a 15 amp circuit with 5 ohms you are looking at 50% VD

 

[steve66]

Yes, 5 ohms would be really high. If you are reading numbers like that, I think you may be doing something wrong, although I'm not sure what.

Maybe the readings are in milli-ohms, or maybe the meter is also measuring the resistance of the transformer winding?

And since the current has to be included in the calculation, you will have to use a larger current for feeders and service conductors than for branch circuits. So the total voltage drop calculation will have to include 2 currents and 2 resistances.

I think you will have to find a way to measure each branch circuit resistance, and then measure the feeder resistance. Then each would be multiplied by the calculated current on that portion of the circuit.

Three phase circuits and multiwire circuits add another complication to the formula. As does the circuit voltage (feeders may be 208V, while most branch circuits are probably 120V).

I don't think finding total voltage drop is going to be as easy as just plugging a meter into the outlet and taking a reading.

 

[charlie b]

Question for an electrical engineer

Sorry, ain’t nobody here but us chickens.

The meter gives me the total impedance on the circuit all the way to the utility Transformer

In that case, you are including the impedance of the transformer secondary in your measurements, and that renders the measurement meaningless. If you are dealing with voltage drop along a feeder or branch circuit, it will be equal to the current flowing through the wires times the impedance of the wires. So what you need to know is the impedance of the wires that comprise the feeder or the branch circuit. You are not measuring those values.

So if my math is correct and my way of thinking would be any value of impedance to and including 5 ohms or less would equal % of voltage drop.

So then, let’s do some math.

• If a 120 volt, 20 amp circuit is drawing all 20 amps, then the total resistance of the circuit is 120/20, or 6 ohms.
• We generally like to limit 20 amp branch circuits that use #12 wire to 100 feet or less. So let’s assume we have 100 feet of wire.
• According to table 9, a #12 wire has an effective impedance of 1.7 ohms per 1000 feet.
• That means the impedance of our 100 foot branch circuit will be 0.17 ohms.
• Multiplying the 0.17 ohms times the total current of 20 amps gives a VD across the branch circuit wires of 3.4 volts.
• To get the %VD, we take 3.4 divided by 120, and multiply by 100%. The result is 2.8%.
• In this example, the 2.8%VD is about half the total impedance of 6 ohms.
• If the circuit instead had been drawing 10 amps, then the numbers would have worked out as a total impedance of 12 ohms, a VD across the branch circuit wires of 1.7 volts, and a %VD of 1.4%. That 1.4% VD is about one tenth the total impedance of the circuit.

Mathematically when you convert total impedance to equal percent voltage drop it seems to work

Sorry, but my math does not confirm your conclusion.

 

[ron]

If a 120 volt, 20 amp circuit is drawing all 20 amps, then the total resistance of the circuit is 120/20, or 6 ohms.

The 6 ohms includes the "total" as mentioned, which also "sees" the load resistance

BTW, great reference https://www.youtube.com/watch?v=jmgaTPz63Bw

 

[mivey]

Sorry, ain’t nobody here but us chickens.

In that case, you are including the impedance of the transformer secondary in your measurements, and that renders the measurement meaningless. If you are dealing with voltage drop along a feeder or branch circuit, it will be equal to the current flowing through the wires times the impedance of the wires. So what you need to know is the impedance of the wires that comprise the feeder or the branch circuit. You are not measuring those values.

So then, let’s do some math.

• If a 120 volt, 20 amp circuit is drawing all 20 amps, then the total resistance of the circuit is 120/20, or 6 ohms.
• We generally like to limit 20 amp branch circuits that use #12 wire to 100 feet or less. So let’s assume we have 100 feet of wire.
• According to table 9, a #12 wire has an effective impedance of 1.7 ohms per 1000 feet.
• That means the impedance of our 100 foot branch circuit will be 0.17 ohms.
• Multiplying the 0.17 ohms times the total current of 20 amps gives a VD across the branch circuit wires of 3.4 volts.
• To get the %VD, we take 3.4 divided by 120, and multiply by 100%. The result is 2.8%.
• In this example, the 2.8%VD is about half the total impedance of 6 ohms.
• If the circuit instead had been drawing 10 amps, then the numbers would have worked out as a total impedance of 12 ohms, a VD across the branch circuit wires of 1.7 volts, and a %VD of 1.4%. That 1.4% VD is about one tenth the total impedance of the circuit.

Sorry, but my math does not confirm your conclusion.

Sorry but you have only calculated the drop in only one conductor. For a single-phase circuit you have two conductors (same size & parameters) and your impedance & volt drop would be two times that number. For a three-phase circuit with equal conductors you would have a multiplier of sqrt(3) or 1.732.

 

[sparks1]

Sorry but you have only calculated the drop in only one conductor. For a single-phase circuit you have two conductors (same size & parameters) and your impedance & volt drop would be two times that number. For a three-phase circuit with equal conductors yo
u would have a multiplier of sqrt(3) or 1.732.

Ok let me try again for the chicks
The circuit under test displays the total impedance from the outlet all the way to the utility transformer coil and back to the outlet.Voltage drop begins at source being
the transformer secondary.so we we go actual reading 0.77ohms of impedance
L-N The PSC 158amps
Knowing the the total Z of 0.77= 0.0077 % VD True?
The circuit has no load current applied at that time other then what the meter test fires around the loop

 

[Bugman1400]

Sorry but you have only calculated the drop in only one conductor. For a single-phase circuit you have two conductors (same size & parameters) and your impedance & volt drop would be two times that number. For a three-phase circuit with equal conductors you would have a multiplier of sqrt(3) or 1.732.

Would the 100' of wire mean to the last load and back or are you assuming that is just one-way?

 

[david luchini]

Would the 100' of wire mean to the last load and back or are you assuming that is just one-way?

I read "assume we have 100' feet of wire" to mean a 50' circuit length, but not sure what Charlie intended.

 

[charlie b]

I read "assume we have 100' feet of wire" to mean a 50' circuit length, but not sure what Charlie intended.

I intended for the OP to catch the apparent contradiction. It was a learning opportunity.

 

[luckylerado]

Knowing the the total Z of 0.77= 0.0077 % VD True?

No. Not true. You need a current component to change ohms to volts. Z is expressed in ohms. VD in volts. Dividing Z by 100 just leaves you with ohms.

 

[mivey]

Would the 100' of wire mean to the last load and back or are you assuming that is just one-way?

100' one way. A rough rule of thumb is to upsize every 100 ft: #12, #10, #8, #6, etc.

 

[mivey]

Ok let me try again for the chicks
The circuit under test displays the total impedance from the outlet all the way to the utility transformer coil and back to the outlet.Voltage drop begins at source being
the transformer secondary.so we we go actual reading 0.77ohms of impedance
L-N The PSC 158amps
Knowing the the total Z of 0.77= 0.0077 % VD True?
The circuit has no load current applied at that time other then what the meter test fires around the loop

No.

On a further note, we should account for the load power factor as well.

Just use the formula.

 

[sparks1]

From the hot wire terminal at the outlet to the panel through the breaker through the meter socket through the overhead drop to one side of transformer coil out the other side to the neutral and back all the way to the outlet neutral terminal . The total impedance = 0.77ohms
Not A 100' . not any table from the code book. No load amps applied except the multi tester

 

[sparks1]

No. Not true. You need a current component to change ohms to volts. Z is expressed in ohms. VD in volts. Dividing Z by 100 just leaves you with ohms.

Not volts
Percent of volts

 

[GoldDigger]

Right. Percent of ohms matches percent of volts dimensionally.
Just like the value of pi does not depend on your system of units.

 

[kwired]

Ok let me try again for the chicks
The circuit under test displays the total impedance from the outlet all the way to the utility transformer coil and back to the outlet.Voltage drop begins at source being
the transformer secondary.so we we go actual reading 0.77ohms of impedance
L-N The PSC 158amps
Knowing the the total Z of 0.77= 0.0077 % VD True?
The circuit has no load current applied at that time other then what the meter test fires around the loop

This tester will give you a different result if you change the loads on other branch circuits that share same feeders, service conductors and the source that are present while taking the measurement. It is only telling you the effects of conditions at the time the measurement was taken.