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neutral current

D

David Williams

Jan 1, 1970
0
Something I wrote yesterday made me think further. I mentioned that if
a pair of outlets are wired "Canadian code", with one neutral wire and
two hots, one on each phase, supplying the two outlets, then the
current in the neutral wire could conceivably be large if the devices
plugged into the outlets are strongly inductive or capacitive.

I did some math, and came up with the following equation:

In^2 = I1^2 + I2^2 - 2 x I1 x I2 x COS(P)

In is the RMS current in the neutral line. I1 and I2 are the currents
in the two hot lines. P is the difference between the phase-angles for
the two devices, so if one device's current lags the supply voltage by
10 degrees and the other by 30 degrees, then P is 20 degrees. I'm
assuming that all the currents are sinusoidal.

If I1 >= I2, then the following condition determines if In > I1, i.e.
the neutral current is greater than the larger of the two hot-line
currents:

2 * COS(P) < I2 / I1

If I1 and I2 are equal, then this condition is satisfied if P is
greater than 60 degrees.

Imagine plugging a 15-amp kettle into one outlet (its element is purely
resistive) and a 15-amp microwave into the other, with a transformer in
its power supply that makes it substantially inductive. If the phase
lag for the microwave exceeds 60 degrees, then the current in the
neutral wire will exceed 15 amps. Not good, if the outlet is wired
with 14/3 !

These conditions probably never come up in reality, but it is
interesting...

dow
 
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