Adding Instantaneous AC Current Flows

Here is another easy fundamental concept ....Adding Instantaneous AC Currents. If cable 'A' is feeding a junction that then splits into cables 'B' and 'C' the electric current flow in A is simply the sum of B and C or (A = B + C) at each instant in time. Again, the unit of electric current flow is the Ampere (or amp for short).

Recall that we showed voltage in an AC system as a sine wave. Current flows in AC systems are also in the form of sine waves. So if cables A, B and C all have currents appearing as sine waves and we want to add B plus C to determine A, then we need to add the currents at the same instant of time. Lets look at a couple of points in time in this example and see how A = B + C.

At 0 degrees, B = 0.5, C = 0 and thus A = 0.5

At 90 degrees, B = 0, C = 0.5 and thus A = 0.5

At 135 degrees, B = -0.35, C = 0.35 and thus A = 0
Notice how at 135 degrees, C has the same magnitude as B but is negative, thus flowing in the reverse direction. So B's current is flowing back to the junction and leaves through C's cable. 'A' sees no current flow at this instant.

In this example, A's current is similar in profile to that of a motor. A motor is made up of 2 components of current that add together: "Load Current" and "Magnetizing Current." I'll explain that later (but its very important that you understand this).

Now look at this example and see how the currents in B & C are exactly opposite at every instant, therefore A = 0 all the time. Also notice that the currents of B & C are taking turns feeding each other while A stays at 0.

Power factor theory doesn't get much more complicated than that!

This diagram is similar to what happens when adding a power factor correcting capacitor to a motor. For half of the cycle, the motor's magnetizing current (C) is fed from the capacitor (B) so the motor's cable (A) doesn't need to deliver that extra current. During the other half cycle, the magnetic field is collapsing and pushing current back into the capacitor. Ultimately, cable 'A' isn't carrying any of the magnetizing current.

If you don't understand this, don't worry, I'll explain it again in the "Power Factor Capacitor" section.