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The case of two equal loads on the two phases is a very special case, but does illustrate the fact the the neutral does carry the sum of the two currents, one plus and one minus, but not always 180 degrees out of phase (remember power factor).
But the neutral cannot be eliminated, even in this special case, because it guarantees that there is always 120 volts across each circuit, what our equipment is designed for.
And, if one circuit carries 50 amps and the other 10 amps, and you lose the neutral, you cannot force the 10 amp load to carry 50 amps. For simplicity, let's assume resistive loads: 120 V/50 A=2.4 ohms and 120 V/10 A=12 ohms. When you lose the neutral, 240 V/14.4 ohms=16.7 amps through each load. That means the 2.4 ohm load sees 40 V (2.4 ohms x 16.7 amps) and the 12 ohm load sees 200 V (12 ohms x 16.7 amps) at least momentarily.
Real loads consist of resistive, capacitive, and inductive components which give a more complex result. But the math is fundamentally the same
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