Bill,

I don't think I have ever tried to explain the current flow in a multi-wire circuit without the aid of chalk and a blackboard, but I'll give it a try.

First of all, an electrical current cannot flow in two different directions in a conductor at the same time. You may have opposing voltages, but the higher voltage will win and the resultant current will flow in that direction. For example, an automobile alternator must have a charging voltage higher than battery voltage in order for current to flow back through the battery. This voltage should be somewhere around 14.5 volts for a 12-volt system. The net charging voltage would only be between 2 to 2.5 volts.

In a 120/240 volt multi-wire circuit, we have two parallel circuits sharing a common wire so current can return to whatever half of the transformer it was derived from. If the load is balanced, current will flow in through both loads and through the entire winding of the transformer in a 240 volt series circuit. There will be no current in the common wire of the parallel circuits. In this state, the neutral is not needed and could be removed from the circuit with out any ill effects. Yes, the voltage has doubled in the circuit, but so has the load. Current will be the same throughout the circuit. Voltage drop across each load will still be 120 volts and equipment will not be damaged.

Let's look at a circuit with some resistance and current values. I am assuming that the readers are familiar with Ohm's Law calculations and the behavior of resistances connected in series and parallel.

If we take two 60 watt light bulbs and connect one to each line and neutral, we will have a circuit like the one I described above. The resistance of each load will be 240 ohms and current will be .5 amperes (we will disregard change in resistance because of heating of the filaments, etc.) The voltage drop across each load will be 120 volts.

If we now look at the point in the circuit where the two resistances and the neutral conductor are joined, and apply Kirchhoff's First Law, things may become a little clearer. Kirchhoff's First Law states that algebraic sum of the currents at this junction must equal zero (or that the current entering this point must equal the current leaving, which ever you prefer). We will have .5 amperes entering this point from one resistance and .5 amperes leaving the point to the other resistance. No current is entering or leaving on the neutral conductor and Kirchhoff's First Law is satisfied.

Let's now turn on one more 60 watt light bulb. This would connect, in parallel, another 240 ohms of resistance to one side of our circuit. The result would be a resistance drop to 120 ohms and a current increase to 1 ampere in that branch of our circuit. We have not changed anything in the other branch. Back at our junction we now have 1 ampere entering from the side with the increased load, .5 amperes leaving to the side where the load stayed the same and .5 amperes leaving on the neutral conductor to go back the side of the transformer from which it was derived. Again the equation is balanced and all the gods are happy. The voltage drop is still 120 volts across each load because of the neutral connection to the transformer mid-point. What we actually have is part of the current from the side with the higher load travelling in the 240 volt series circuit (the balanced portion) and the remainder in the 120 volt parallel circuit.

Not to beat this thing to death, but let's look at the circuit in one more way. Let's connect the additional 60 watt light bulb to the other parallel branch and look again at the neutral junction point. We now have .5 amperes of current entering the junction from the smaller load, .5 amperes of current entering on the neutral conductor, and 1 ampere of current leaving the point to the side that now has the larger load. The .5 ampere neutral current is now coming from the side of the transformer where it was derived. The neutral current is now traveling in the opposite direction.

It might as well be mentioned that in the circuit with the unbalanced load, if the neutral connection is removed and we revert back to a 240 volt series circuit and the voltage drop across the loads would not be equal. The current in the circuit will be .66 amperes, and the voltage drop across the loads will now be 160 volts across the single 60 watt bulb and 80 volts across the two other bulbs. The 60 watt bulb just might survive this, but we all know what happens to the VCR.

One more thing for those with insomnia. Reading this might help. It seems to have this effect on some people in my classrooms.

R Dimery