Understanding Current as a Vector Property

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Amped_86

Joined Jun 28, 2018
26
If I have an AC current with a magnitude whose vector angle is 0 degrees and another current with the same magnitude but whose vector angle is 180 degrees, can it be said that at a given point in time, these currents are literally moving in opposite directions?

What happens when a multi-phase source where the sources are not in phase is applied to a single load (I'm ignoring the impedance angle of the load for now)? Let's say I have a 3 phase source where the source angles are 120 degrees apart. However I do not connect source B. So I have source A at 0 degrees connected to the top of a single load, and I have source C at +120 degrees connected to the bottom of the load. What is happening with the two currents? They're out of phase so literally I picture them moving back and forth but not at the same time. Are the magnitude and angles of the two currents added together to form one current?
 

MrAl

Joined Jun 17, 2014
11,389
If I have an AC current with a magnitude whose vector angle is 0 degrees and another current with the same magnitude but whose vector angle is 180 degrees, can it be said that at a given point in time, these currents are literally moving in opposite directions?

What happens when a multi-phase source where the sources are not in phase is applied to a single load (I'm ignoring the impedance angle of the load for now)? Let's say I have a 3 phase source where the source angles are 120 degrees apart. However I do not connect source B. So I have source A at 0 degrees connected to the top of a single load, and I have source C at +120 degrees connected to the bottom of the load. What is happening with the two currents? They're out of phase so literally I picture them moving back and forth but not at the same time. Are the magnitude and angles of the two currents added together to form one current?
Hello there,

For the two out of phase sources (A and C or whatever) that have equal amplitudes, the solution is:
Vx=sqrt(3)*sin(t*w+Ph)

where Ph is either plus or minus 60 degrees.

What this tells us is that the two combine to form one source, and the current would then be Vx/R where R is the load resistance.
This solution can be found in various ways and one way is to compute the different voltage across the resistor:
Vxx=sin(w*t)-sin(w*t+ph)

where ph is 120 degrees for the two sources of the three phase system, and that reduces to the above solution for Vx.
Note that for two equal sources, the result phase is plus or minus 1/2 of the source phase difference.

If we had two unequal voltage sources it would be a little different:
Vx=A*sin(w*t)-B*sin(w*t+ph)

and that takes a little more work to reduce that's all, but will also reduce to one single source and the result phase will be between zero and plus or minus 120 degrees not necessarily 1/2 of the difference phase.
 
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