MWBC voltages

[hardworkingstiff]

On another forum the question came up as to how 123V is showing up on load 1 of the below drawing. Is this book correct that the voltage across load 1 is 123V even though it's supplied by 120V of a MWBC?

 

[roger]

Lou, if the neutral breaks at point A you have 240v series circuit

Roger

 

[hardworkingstiff]

Roger, I understand that, thanks.

But this is supposed to be with the neutral intact. The only thing that is in the neutral is the 1 ohm (I assume wire resistance) resistance shown.

With the neutral intact, how did the book come up with 123V across load 1? That's what I'm trying to understand.

 

[gar]

130118-0844 EST

hardworkingstiff:

It is simple, but I do not know what your circuit is. I just get a red X.

Given two ideal voltage sources in series, additive, V1 and V2, each equal to 120 V, and a 1 ohm neutral. I will assume resistive loads, and one load 10 A, and the other 7 A. The unbalanced current in the neutral is 3 A. Thus, the neutral drop is 3 V. This means the 10 A load has a voltage drop across it of (120-3) V, and the 7 A load has a drop of (120+3) V. You can calculate the individual resistances.

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[hardworkingstiff]

deleted

 

[gar]

1430118-1053 EST

hardworkingstiff:

Your message that was deleted showed up in my e-mail.

The answer is to draw the circuit. The voltage drop on the neutral is common to both loads. I made the assumption of resistive loads to make the analysis simple, and because I did not know what the actual circuit was.

In this case you can replace the AC voltage sources with batteries if it helps to understand the circuit. So let's do that.

You have three loops to consider.

V1 + V2 thru R1 and R2 is one loop and for the moment ignoring Rneutral. The sum of the voltages across R1 and R2 have to equal V1 + V2. At this time you do not know the individual voltages across R1 and R2 because you do not know the voltage of the resistor midpoint relative to the midpoint of V1 and V2.

Now suppose the end of V1 connected to R1 is +120 V relative to the midpoint of V1 and V2, and the non-midpoint end of V2 is -120 V relative to the midpoint of V1 and V2.

If the two load resistances are equal, then the midpoint voltage of R1 and R2 is equal to the midpoint voltage of V1 and V2 relative to any reference point you want. Make that reference point the midpoint of V1 and V2, then the voltage drop from the resistors midpoint to the voltage sources midpoint is zero.

Next connect the negative end of a 3 V battery to the voltage midpoint, the battery positive end to the resistor midpoint. What are the voltage drops across R1, and across R2?

Change the battery to a 1 ohm resistor. Adjust the load resistors to get a + 3 V drop from the resistors midpoint to the voltages midpoint. One possibility is R1 = (120-3)/3 = 39 ohms with R2 = infinity. Another is R1 = 19.5 ohms, I_R1 = 6 A, 117 / 19.5 = 6 A. So when R1 = 19.5 ohms you tell me the value of R2 that produces the 3 V drop in R neutral.

I have not proof read this. See if there are any mistakes.

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[jonathansmith]

Thanks for making me understand this. I was little bit confused but now i m clear.