Voltage drop

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xformer

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Would I be right in assuming that 310.10 could be used to enforce voltage drop?
Let me explain... If Circuits with low voltage cause current rises in conductors and that current exceeds the ampacity of the conductor, is not the operating temperature also exceeded?
 

charlie b

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I would not accept that basis for making voltage drop a code violation. For starters, many circuits would see a reduced current, if you have a reduced voltage. So your reasoning falls apart. Secondly, even with motor loads that do have increased current when voltage is low, there is no way to make a broad statement that any specific amount of voltage drop will take the temperature of a conductor beyond its limitations. There are too many variables to create a basis for requiring any specific minimum for percentage of rated voltage. Finally, I don't see anything in 310.10 that says we can't overheat a conductor.
 

realolman

Senior Member
Would I be right in assuming that 310.10 could be used to enforce voltage drop?
Let me explain... If Circuits with low voltage cause current rises in conductors and that current exceeds the ampacity of the conductor, is not the operating temperature also exceeded?

Circuits with enough resistance or impedance to cause appreciable voltage drop do not have higher current.... they have less current.... because of that resistance. Making the conductors larger reduces the resistance and there is increased current... but less voltage drop.
 

infinity

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Circuits with enough resistance or impedance to cause appreciable voltage drop do not have higher current.... they have less current.... because of that resistance. Making the conductors larger reduces the resistance and there is increased current... but less voltage drop.

You need to clarify what type of load you're referring to. Resistive and inductive loads have different affects on current when voltage is reduced.
 

realolman

Senior Member
I am referring to the resistance or impedance in the circuit conductors that cause the voltage drop.

How about some clarification from you.:)
 

infinity

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I am referring to the resistance or impedance in the circuit conductors that cause the voltage drop.

How about some clarification from you.:)

OK, so if the load is purely resistive, like a heater, how would adding the resistance of the condcutors in the resistive circuit make the current increase?
 

realolman

Senior Member
I don't think it would

Circuits with enough resistance or impedance to cause appreciable voltage drop do not have higher current.... they have less current.... because of that resistance. Making the conductors larger reduces the resistance and there is increased current... but less voltage drop.

I get the feeling you have a follow up post ready.:)
 

realolman

Senior Member
NO NO NO NO NO... i won't have any of that.. You don't need to apologize for anything... I'm just about always wrong.. It was just habit ....you can't be faulted for that...;)
 

squaredan

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Pennsylvania
Circuits with enough resistance or impedance to cause appreciable voltage drop do not have higher current.... they have less current.... because of that resistance. Making the conductors larger reduces the resistance and there is increased current... but less voltage drop.

What you say makes Sense but why am i alwaying hearing when you have a fixed load if you lower the voltage you increase the amps.. P=I x E Same with Voltage Drop Evd= IxR .. Not Disagreeing just asking...

Dan
 

infinity

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What you say makes Sense but why am i alwaying hearing when you have a fixed load if you lower the voltage you increase the amps.. P=I x E Same with Voltage Drop Evd= IxR .. Not Disagreeing just asking...

Dan


For a fixed resistive load if you lower the voltage you lower the current.

E=120 volts
R=12 ohms
I=E/R= 120/12=10 amps

Now cut the voltage in half, cut the current in half:

I=E/R= 60/12=5 amps

For an inductive load like a motor the opposite is true.
 

squaredan

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Location
Pennsylvania
OK but what abou tif you have a fixed load like a baseboard heater 1500w at 240v your current is 12.5amps.. Lower the voltage with the same heater 1500w at 120v your current is now 6.25amps..voltage goes up and the current goes down.. Same with low voltage undercounter lighting..We use a 10AWG wire cuz the amps goes up when the voltage is lower..What am i missing?? thanks

Dan
 

iwire

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For a fixed resistive load if you lower the voltage you lower the current.

E=120 volts
R=12 ohms
I=E/R= 120/12=10 amps

Now cut the voltage in half, cut the current in half:

I=E/R= 60/12=5 amps

For an inductive load like a motor the opposite is true.

I would say for inductive load the opposite may be true.

Many times lowing the voltage to lightly loaded motors will also reduce the current and therefore save energy. Each case would have to be looked at individually.
 

david luchini

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OK but what about if you have a fixed load like a baseboard heater 1500w at 240v your current is 12.5amps.. Lower the voltage with the same heater 1500w at 120v your current is now 6.25amps..voltage goes up and the current goes down.. Same with low voltage undercounter lighting..We use a 10AWG wire cuz the amps goes up when the voltage is lower..What am i missing?? thanks

Dan

You have your voltages backwards, but it is clear what you are getting at when you say "voltage goes up..."

What you are missing is that, if you took a 1500W, 120V rated heater, and applied 240V to it, the current wouldn't go down to 6.25A, it would go UP to 25A. The heater is a fixed resistive load. The resistance of the 1500W, 120V heater is 9.6Ohm. The current flowing through a 9.6Ohm resistance with 240V applied to it is I=V/R = 240/9.6 = 25A. You have 6000W at 240V, instead of 1500W.
 
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realolman

Senior Member
OK but what abou tif you have a fixed load like a baseboard heater 1500w at 240v your current is 12.5amps.. Lower the voltage with the same heater 1500w at 120v your current is now 6.25amps..voltage goes up and the current goes down........ thanks

Dan

I don't think your post meant what you intended it to....

I think that you must have meant that there were two separate heaters involved...

Both are 1500 watt... one is intended to be operated at 120v and the other is intended to be operated at 240. Operating them both at their intended voltage, the current is less on the higher voltage , but the wattage is the same.

If you used the same heater and applied different voltages the current would go up as the voltage went up, but the wattage would also go up.
 

K8MHZ

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Michigan. It's a beautiful peninsula, I've looked
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Electrician
For an inductive load like a motor the opposite is true.

The reason an electric motor may draw more current at a lower voltage is due to the fact that the speed of the motor will be reduced and less counter EMF is produced.

In an inductive load with no moving parts, voltage and current remain proportional like a purely resistive load.
 

squaredan

Senior Member
Location
Pennsylvania
I don't think your post meant what you intended it to....

I think that you must have meant that there were two separate heaters involved...

Both are 1500 watt... one is intended to be operated at 120v and the other is intended to be operated at 240. Operating them both at their intended voltage, the current is less on the higher voltage , but the wattage is the same.

If you used the same heater and applied different voltages the current would go up as the voltage went up, but the wattage would also go up.

Yes that was what i was thinking and i total understand what david was saying with the math in his post.. thanks for making me understand better...
 
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