Suppose that I have a 3 phase device (say, a resistive load) that draws X amps. What would be the current draw of a single phase device that consumes the same amount of power.
My hunch is that a single phase device would draw 1.5 times the current of the 3 phase device. In other words, a 3 phase device that draws the same amps would produce 1.5 times as much energy as a single phase device drawing the same amount of amps.
Is that correct?
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Ned Simmons
Close. 1.732 = square root(3)
Ned Simmons
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Ignoramus6304
thank you...
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Bob AZ
If I understand your question correctly:
You are asking for an answer in amps and giving conditions in power and/or amps. For a resistive load.
So for a single phase resistive load and a 3 phase resistive load with the same power consumption, the total amount of amps would be the same. The three phase load would simply be 3 loads, each a third of the single phase load.
For example with a single phase load of 1, the three phase load would be three loads of 3. The same power would be dissapated(sp?) but the current through each of the 3 phase loads would be 1/3 of the single phase load.
With a resistive load any current shifts would not be a factor. Things get interesting when the load is inductive and/or capacitive.
Bob AZ
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Ignoramus6304
I do not think that 3 times the single phase load is the correct answer. It would involve double counting of current. I am not claiming to be a big expert in this, but 3 times cannot be the right answer.
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Don Foreman
If line-to-line voltage is the same in both cases, singlephase line current will be 1.732 x threephase line current for same total power
-- like Ned sed!
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Ignoramus6304
thank you...
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Bob AZ
But we are dealing with a resistive load. No inductance or capacitance. A resistive load will not concern itself with phasing. So the 3 phase thing is moot.
If we had a 3 phase device that has an ouput function, motor, transformer or the like then we would be dealing with other things like is the 3 phase Delta or WYE? And is the loading on the 3 phase, phase-to-phase or phase to neutral/ground. Bob AZ
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Robert Swinney
Bob, The premise under discussion here is multiple phase transmission of power. You might be confusing that with the concept of power factor. Correct me if I'm wrong.
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Robert Nichols
In article , Bob AZ wrote: :But we are dealing with a resistive load. No inductance or capacitance. :A resistive load will not concern itself with phasing. So the 3 phase :thing is moot. : : If we had a 3 phase device that has an ouput function, motor, :transformer or the like then we would be dealing with other things like :is the 3 phase Delta or WYE? And is the loading on the 3 phase, :phase-to-phase or phase to neutral/ground. :Bob AZ
Whether the load is connected delta or WYE does not matter. What _does_ matter is how the voltage is measured, and the article you responded to (and neglected to quote!!) explicitly stated, "If the line-to-line voltage is the same ... ." A WYE connected load will see a line-to-neutral voltage that is (1 / sqrt(3)) times the line-to-line voltage, so the total power for 3 such identical resistive loads is V*I*sqrt(3), where V is the line-to-line voltage and I is the current in any one line.
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Rick
capacitance.
phase
OK, for ease of calculations, letrs say we have a 2400 watt 240 volt single phase heater. The line current is 2400/240, or 10 amps.
Now, a 2400 watt 3 phase heater is 800 watts per phase. The three 800 watt elements are wired in delta. The load current for each element is
800/240, or 3.33 amps.
By drawing the vector diagram of the 3 load currrents and and some trig, the line currrent for each leg is 3.33 * square root of 3, or
5.77 amps
And 5.77 * square root of 3=10
Resistance can be used if you prefer. Say we have 3 heater elements that measure 15 ohms. If one was connected across a 240 volt single phase line, the power dissipated would be 240 * 240/15, or 3840 watts, with a current of 16 amps. Three of those in parallel would be 11,520 watts total, with a current draw of 48 amps.
Now, wire those three elements in delta. Each element is still across
240 volts, so each one still dissipates 3840 watts (for a 11520 watt total) and draws 16 amps. But since it's three phase, the line current is 16 * square root of 3, or 27.7 amps.
And 48 /square root of 3 also =27.7...
It's still early-I hope all those numbers are right, lol...
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