Start with a 120/240 multiwire branch circuit.
We’ll need to begin with some given values.
E=
I=
R=
P=
E=
I=
R=
P=
E=
I=
R=
P=
We’ll need to begin with some given values.
E=120V
I=
R=
P=
E=240V
I=
R=
P=
E=120V
I=
R=
P=
Now lets add a load to each circuit.
E=120V
I=
R=
P=
E=240V
I=
R=
P=
E=120V
I=
R=
P=
Now lets add a load to each circuit.
E=120V
I=
R=
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=
R=
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=
R=
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=P/E
R=
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=E²/P
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=P/E
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=E²/P
P=200W
Then calculate amps and resistance for each load.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
With these values known, we can begin to understand
what happens when the neutral is opened.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
First, we’ll open the neutral conductor.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
First, we’ll open the neutral conductor.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
In doing so, we no longer have two circuits, but only one.
The two loads are now in series.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
So we will erase the figures we have calculated so far.
E=120V
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
So we will erase the figures we have calculated so far.
E=
I=0.8333A
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
So we will erase the figures we have calculated so far.
E=
I=
R=144W
P=100W
E=240V
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
So we will erase the figures we have calculated so far.
Leave the values of R in place. Those will not change.
E=
I=
R=144W
P=
E=240W
I=
R=
P=
E=120V
I=1.6667A
R=72W
P=200W
So we will erase the figures we have calculated so far.
Leave the values of R in place. Those will not change.
E=
I=
R=144W
P=
E=240V
I=
R=
P=
E=
I=1.6667A
R=72W
P=200W
So we will erase the figures we have calculated so far.
Leave the values of R in place. Those will not change.
E=
I=
R=144W
P=
E=240V
I=
R=
P=
E=
I=
R=72W
P=200W
So we will erase the figures we have calculated so far.
Leave the values of R in place. Those will not change.
E=
I=
R=144W
P=
E=240V
I=
R=
P=
E=
I=
R=72W
P=
Now add the values of the two loads’ resistances.
E=
I=
R=144W
P=
E=240V
I=
R=
P=
E=
I=
R=72W
P=
Now add the values of the two loads’ resistances.
E=
I=
R=144W
P=
E=240V
I=
R=144+72
P=
E=
I=
R=72W
P=
Now add the values of the two loads’ resistances.
E=
I=
R=144W
P=
E=240V
I=
R=216W
P=
E=
I=
R=72W
P=
Now calculate I and P for the entire circuit.
E=
I=
R=144W
P=
E=240V
I=
R=216W
P=
E=
I=
R=72W
P=
Now calculate I and P for the entire circuit.
E=
I=
R=144W
P=
E=240V
I=E/R
R=216W
P=
E=
I=
R=72W
P=
Now calculate I and P for the entire circuit.
E=
I=
R=144W
P=
E=240V
I=1.1111A
R=216W
P=
E=
I=
R=72W
P=
Now calculate I and P for the entire circuit.
E=
I=
R=144W
P=
E=240V
I=1.1111A
R=216W
P=E²/R
E=
I=
R=72W
P=
Now calculate I and P for the entire circuit.
E=
I=
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=
R=72W
P=
Since this is a series circuit, I is always the same.
E=
I=
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=
R=72W
P=
Since this is a series circuit, I is always the same.
E=
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=
R=72W
P=
Since this is a series circuit, I is always the same.
E=
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=1.1111A
R=72W
P=
Now calculate the voltage across each of the two loads.
E=
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=1.1111A
R=72W
P=
Now calculate the voltage across each of the two loads.
E=R*I
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=1.1111A
R=72W
P=
Now calculate the voltage across each of the two loads.
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=
I=1.1111A
R=72W
P=
Now calculate the voltage across each of the two loads.
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=R*I
I=1.1111A
R=72W
P=
Now calculate the voltage across each of the two loads.
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=80V
I=1.1111A
R=72W
P=
Now you can see what happens when you lose the neutral
on a multiwire branch circuit.
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=80V
I=1.1111A
R=72W
P=
The more resistance, the higher the voltage....
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=80V
I=1.1111A
R=72W
P=
The less resistance, the lower the voltage.
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=80V
I=1.1111A
R=72W
P=
This is Ohm’s Law in a most practical use, and the reason
Article 300.13(B) exists.
E=160V
I=1.1111A
R=144W
P=
E=240V
I=1.1111A
R=216W
P=266.7W
E=80V
I=1.1111A
R=72W
P=