Measurements

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
201018-1444 EDT

At a point in a network the sum of all the instantaneous currents from all paths at that point are zero. You can not replace instantaneous with RMS or average, and have that statement necessarily hold true. Also note that RMS is a type of average.

Not understanding this basic fact resulted in one very long thread.

In a different set of words the same concept exists for voltage. Here it is around a closed loop rather than a point.

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
201019-1717 EDT

My next step in this story is:

Center tapped single phase supply. This means on the secondary side we have two phases that are 180 degrees apart.

If phase 1, L1 to neutral, is loaded with a 10 A resistive load, and phase 2, L2 to neutral, is loaded with a high quality capacitor drawing 10 A, then both load currents are sine waves. What is the wave shape of the neutral current, its phase angle relative to the L1 to neutral voltage, and its magnitude? Its magnitude is not zero.

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winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Start with just the resistive load connected. The L1 current is 10A at 0° out of the L1 terminal and 0° into the N terminal. (Or 180° out of the N terminal.

Now consider only the capacitor load connected. The L2 current is 10A at 90° out of the L2 terminal and 90° into the N terminal.

The total N current is sinusoidal, 14.1A, 45° into the N terminal.

Jon
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
201019-2352 EDT

winnie:

Good description. Hope readers can understand it.

I will continue the journey later. My next step my be very confusing to many.

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
201025-2014 EDT

Now the next step.

Change the Phase 1 loading from said resistor to an ideal inductor with the same current as the resistor. Note: in the real world I have no practical ability to make an ideal inductor.

Now what happens to the neutral current? It goes to 20 A.

Experimentally I can simulate this by:
1. Change the transformer secondary to two separate equal voltage secondaries.
2. Connect the two secondaries together at one point so that the secondaries are in phase. This one point we will still call neutral.
3. Phase 2, L2 to neutral, still has its original capacitor connected.
4. Phase 1, L1 to neutral, now has its phase inverted.
5. Connect another 10 A capacitor from L1 to neutral.
Now I have simulated an ideal inductor for the original circuit.

The neutral current is now 2 times the current of either phase.

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