Re: neutral current
Originally posted by sparky 134:Why is this? Can anyone shed some light ??
It is because not all currents are the same, so that you can?t always add the number ?1? to the number ?-1? and get a result of zero. If a bowl on the table contains three apples and five oranges, and if I put in another apple, and if you take out one orange, has the contents of the bowl changed? You might say that it had held eight fruits, and still does hold eight fruits, so the net change is zero. Or you might notice the change in how many of which types of fruits are now in the bowl, and say that the net change is not zero.
On a 120/240V system, the currents leaving each bus are 180 degrees out of phase with each other. Suppose you have only two loads on the panel, one from each side, each drawing the same current. What happens is the current leaves the panel via a breaker on (let us say) the left side. It travels down the ?hot wire? (i.e., ?phase? or ?ungrounded conductor?) to the load, through the load, and out the ?cold wire? (i.e., ?neutral,? or ?grounded conductor?) to the point at which the shared neutral was split to serve the two loads. From here, it travels up the neutral to the other load, through the other load, and along the ?hot wire? back to the panel, and into the breaker on (let us say) the right side. The shared neutral, from the panel to the point at which that shared neutral is split, sees no current.
On a 120/208V system, the currents leaving each bus are 120 degrees out of phase with each other. The current paths discussed above still apply, but the addition is now that of apples plus oranges, not fruits plus fruits. So a different thing takes place (or more properly, the same thing takes place, but with a different result) at the point at which the shared neutral splits. It is because of Kirchhoff?s Current Law: ?The sum of all currents entering a point must be zero.? (Note: To be consistent in the use of this law, you must speak of all currents as entering the point, not as one entering and another leaving.)
In the 120/240 case, the current entering the split point (from the left side breaker) comes in as (let us say) 5 amps at an angle of 0 degrees. The current entering the split point (from the right side breaker) comes in as (let us say) 5 amps at an angle of 180 degrees. The current entering the split point from the shared neutral must be whatever is left, when these two are added together. But add 5 amps (angle 0) to 5 amps (angle 180) and you get zero, so the shared neutral gets zero current.
In the 120/208 case, the current entering the split point (from the left side breaker) comes in as (let us say) 5 amps at an angle of 0 degrees. The current entering the split point (from the right side breaker) comes in as (let us say) 5 amps at an angle of 120 degrees. The current entering the split point from the shared neutral must be whatever is left, when these two are added together. If you add 5 amps (angle 0) to 5 amps (angle 120) and you get 5 amps (angle 240). So the shared neutral gets a current of 5 amps (angle 240). This is the same result as you get from the formula that brian john used.
