Thanks Steven, it helps a lot. An electron is a negatively charged particle. Given the kW and power factor the kVA can be easily worked out.
Where I have issues is when we have loads connected across two phases. If you understand that definition fully, then the rules are nothing more than footnotes to the definition. thanks.. You don't need to combine the current - just take the worse case line current for sizing cables/breakers. It works the same a for a consumer of power.

I measure the Power (kW) required by the electrical motor. Within the body of the post I have added an example of how to derive the formulae. This brings us to the second principle of series circuits: The total resistance of any series circuit is equal to the sum of the individual resistances. Total VA = 50.46, using 70% require 72 VA transformer (chose next standard size up). I deal with unbalanced systems on a frequent basis and am trying to figure out how to calculate the amps drawn on each leg. You can't divide the current by three (power kW or kVA - yes). Electric motors are one of the most widely used items of electrical equipment. I understand a little more now (perhaps not fully). Read the resistance, according to the value or colour code (see Resources) printed on the resistor. Ohm's Law states that the current is equal to the voltage divided by the resistance.

I just went through and got a few strange results, and realised that earlier you state VLL = √3 x VLN Hence I wondered why you would then multiply VLL again by √3.
As a general guide I would just add up the VA, use a 70% load factor.

However, the method we just used to analyze this simple series circuit can be streamlined for better understanding. In the three-resistor example circuit below, we know that we have 9 volts between points 1 and 4, which is the amount of electromotive force driving the current through the series combination of R1, R2, and R3. By remembering that a three phase power (kW or kVA) is simply three  times the single phase power, any three phase problem can be simplified. This is done all the time (fridge or toilet extract fan in your house for example). thank you very much. But if you don't mind, I would like to ask a few more question about the apparent power and complex power. tian, There is always more than one way to do things. Also, thanks again for posting and all of your help. That definitely clarifies the error in my calcs.

Divide kW by the power factor to get the kVA. When calculating the current use the phase voltage which is related to the line voltage by the square root of three. It's not always possible to measure current through a circuit without disturbing it, so you measure the voltage drop across the resistor and calculate the current.

Luckily in practice voltages tend to be fixed or very by only small amounts. We've updated the image. Other things to consider while carrying out calculations may include the efficiency of equipment. The power taken by a circuit (single or three phase) is measured in watts W (or kW). A-N, B-N and C-N). What are “Series” and “Parallel” Circuits? Phase Voltage = VLL/√3 So our mathematical analysis has been vindicated by the computer. We’ll start with a series circuit consisting of three resistors and a single battery: The first principle to understand about series circuits is as follows: The amount of current in a series circuit is the same through any component in the circuit. Power factor 0.86, and what size of breaker and cable should I use. Write down the resistance value.

My advice would be to look at your actual situation and analysis it to see the effect of changing from star to delta. If you are looking at it on a phase by phase you may need to be a little careful. As a practical matter when literature specifies "line" voltage does this correspond to the country standard; ie 480 VAC 3 phase industrial power in the US?

I have now changed it to show kVA. Wrong or not updated diagram: The revised diagram is supposed have node 0 instead of node 4. If it is star connected than the phase voltage is 240/sqrt(3) = 138 V and the current is 56,000/138 = 400 A.

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