lost neutral
TRANSCRIPT
Start with a 120/240 multiwire branch circuit.
E=I=R=P=
E=I=R=P=
E=I=R=P=
We’ll need to begin with some given values.
We’ll need to begin with some given values.
E=120VI=R=P=
E=120VI=R=P=
E=240VI=R=P=
Now lets add a load to each circuit.
E=120VI=R=P=
E=120VI=R=P=
E=240VI=R=P=
Now lets add a load to each circuit.
E=120VI=R=P=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=R=P=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=R=P=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=P/ER=P=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=P=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=E²/PP=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=144WP=100W
E=120VI=R=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=144WP=100W
E=120VI=P/ER=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=P=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=E²/PP=200W
E=240VI=R=P=
Then calculate amps and resistance for each load.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
With these values known, we can begin to understandwhat happens when the neutral is opened.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
First, we’ll open the neutral conductor.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
First, we’ll open the neutral conductor.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
In doing so, we no longer have two circuits, but only one.The two loads are now in series.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
So we will erase the figures we have calculated so far.
E=120VI=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
So we will erase the figures we have calculated so far.
E=I=0.8333AR=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
So we will erase the figures we have calculated so far.
E=I=R=144WP=100W
E=120VI=1.6667AR=72WP=200W
E=240VI=R=P=
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=144WP=
E=120VI=1.6667AR=72WP=200W
E=240WI=R=P=
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=144WP=
E=I=1.6667AR=72WP=200W
E=240VI=R=P=
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=144WP=
E=I=R=72WP=200W
E=240VI=R=P=
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=144WP=
E=I=R=72WP=
E=240VI=R=P=
Now add the values of the two loads’ resistances.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=R=P=
Now add the values of the two loads’ resistances.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=R=144+72P=
Now add the values of the two loads’ resistances.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=R=216WP=
Now calculate I and P for the entire circuit.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=R=216WP=
Now calculate I and P for the entire circuit.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=E/RR=216WP=
Now calculate I and P for the entire circuit.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=1.1111AR=216WP=
Now calculate I and P for the entire circuit.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=1.1111AR=216WP=E²/R
Now calculate I and P for the entire circuit.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=1.1111AR=216WP=266.7W
Since this is a series circuit, I is always the same.
E=I=R=144WP=
E=I=R=72WP=
E=240VI=1.1111AR=216WP=266.7W
Since this is a series circuit, I is always the same.
E=I=1.1111AR=144WP=
E=I=R=72WP=
E=240VI=1.1111AR=216WP=266.7W
Since this is a series circuit, I is always the same.
E=I=1.1111AR=144WP=
E=I=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
Now calculate the voltage across each of the two loads.
E=I=1.1111AR=144WP=
E=I=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
Now calculate the voltage across each of the two loads.
E=R*II=1.1111AR=144WP=
E=I=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
Now calculate the voltage across each of the two loads.
E=160VI=1.1111AR=144WP=
E=I=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
Now calculate the voltage across each of the two loads.
E=160VI=1.1111AR=144WP=
E=R*II=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
Now calculate the voltage across each of the two loads.
E=160VI=1.1111AR=144WP=
E=80VI=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
Now you can see what happens when you lose the neutralon a multiwire branch circuit.
E=160VI=1.1111AR=144WP=
E=80VI=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
The more resistance, the higher the voltage....
E=160VI=1.1111AR=144WP=
E=80VI=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
The less resistance, the lower the voltage.
E=160VI=1.1111AR=144WP=
E=80VI=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W
This is Ohm’s Law in a most practical use, and the reasonArticle 300.13(B) exists.
E=160VI=1.1111AR=144WP=
E=80VI=1.1111AR=72WP=
E=240VI=1.1111AR=216WP=266.7W