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In the following star and delta connections, if each individual current enters via Phases R, Y and B, where does the sum of the three currents (that is Ia+Ib+Ic) flow after meeting at the common point of the three phases?

Moreover, in such types of connections, across which terminals is a load connected to these types of systems to draw current/power for its operation?

enter image description here

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    \$\begingroup\$ The sum of the currents is zero. \$\endgroup\$ – Hearth Aug 17 at 11:58
  • \$\begingroup\$ Okay but if I were to keep an ammeter, I can individually measure the value of the three currents right? So how exactly does it become zero? And if the sum is zero, how can I connect a load to the output and draw power from this type of connection? \$\endgroup\$ – noorav Aug 17 at 12:05
  • \$\begingroup\$ remember this is three-phase AC. If you know complex currents already, it's really trivial, just write down the currents as complex numbers and add them. Also, get better material: the material you use was made by someone who uses the inductor symbol for resistors, and that's plain wrong (assuming that's what they wanted to show and this isn't about electrical motors. If it's about motors, you need to learn a lot now – start with complex currents.). \$\endgroup\$ – Marcus Müller Aug 17 at 12:25
  • \$\begingroup\$ You especially need better learning material if the material you're reading now doesn't explain what happens right after it showed this picture. \$\endgroup\$ – Marcus Müller Aug 17 at 12:27
  • \$\begingroup\$ Oh okay. Could you please provide me with some links that you already know of? Would be much appreciated \$\endgroup\$ – noorav Aug 17 at 12:28
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if each individual current enters via Phases R, Y and B, where does the sum of the three currents (that is Ia+Ib+Ic) flow after meeting at the common point of the three phases?

The "sum" current has to be zero because it can't flow anywhere.

if I were to keep an ammeter, I can individually measure the value of the three currents right? So how exactly does it become zero?

Your ammeter doesn't account for the fact that the three currents have a phase relationship that means they cancel out. Individually they can be measured but I_red = -(I_yellow + I_blue)

Or, I_red + I_yellow + I_blue = 0

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enter image description here

Figure 1. Three-phase currents sum to zero.

  1. Here we can see that the black and blue phases are at 0.5 of max current and positive. Meanwhile, red is at -1 of max current. They sum to zero.
  2. Here we see that black is at +1 while red and blue are at -0.5. They sum to zero.
  3. Here we see that blue is at zero. Red is at \$ +\frac{\sqrt 3}{2} \$ max current while black is at \$ -\frac{\sqrt 3}{2} \$. They sum to zero.

The sum to zero will apply at any point along the time axis if you care to do the calculations.


From the comments:

Ok so if the sum of the currents is zero at the meeting point, how does one draw current from this system to power their load?

There is nothing in your question to indicate whether the circuit represents a generator or a load other than the current direction which suggests that the device is consuming current. In any case, it doesn't matter.

Current flows in a circuit. Current in or out of any branch must find its return through the other two. That's what my Figure 1 is demonstrating.

Moreover, is the sum of currents = 0 only at the neutral/ common point of the three phases?

There is no other point in your circuit where the currents are summed.

If, however, you run the three conductors through a current clamp meter you will have a reading of zero as the currents in one direction must balance those in the opposite direction.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. Typical three-phase inverter setup.

Power flows from the battery, to the inverter where it is converted to three-phase AC and from there to the three-phase load.

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  • \$\begingroup\$ importantly, because of Kirchoff's current law, even if the currents are unbalanced (due to the load being unbalanced) they are still forced to sum to zero. \$\endgroup\$ – Hearth Aug 17 at 12:34
  • \$\begingroup\$ Ok so if the sum of the currents is zero at the meeting point, how does one draw current from this system to power their load? \$\endgroup\$ – noorav Aug 17 at 12:58
  • \$\begingroup\$ Moreover, is the sum of currents = 0 only at the neutral/ common point of the three phases? \$\endgroup\$ – noorav Aug 17 at 12:59
  • \$\begingroup\$ See the updates. \$\endgroup\$ – Transistor Aug 17 at 13:24
  • \$\begingroup\$ Ah okay. And across which terminals will I connect the load because all I see in my figure are the three input terminals. Is there another set of three output terminals that the diagram isn't showing? \$\endgroup\$ – noorav Aug 17 at 13:32

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