Start with a 120/240 multiwire branch circuit. Well need to begin - - PowerPoint PPT Presentation

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Aug 24, 2023 388 likes •860 views

Start with a 120/240 multiwire branch circuit. Well need to begin with some given values. E= I= R= P= E= I= R= P= E= I= R= P= Well need to begin with some given values. E=120V I= R= P= E=240V I= R= P= E=120V I= R= P=

SLIDE 1

Start with a 120/240 multiwire branch circuit.

SLIDE 2

E= I= R= P= E= I= R= P= E= I= R= P=

We’ll need to begin with some given values.

SLIDE 3

We’ll need to begin with some given values.

E=120V I= R= P= E=120V I= R= P= E=240V I= R= P=

SLIDE 4

Now lets add a load to each circuit.

E=120V I= R= P= E=120V I= R= P= E=240V I= R= P=

SLIDE 5

Now lets add a load to each circuit.

E=120V I= R= P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 6

Then calculate amps and resistance for each load.

E=120V I= R= P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 7

Then calculate amps and resistance for each load.

E=120V I= R= P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 8

Then calculate amps and resistance for each load.

E=120V I=P/E R= P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 9

Then calculate amps and resistance for each load.

E=120V I=0.8333A R= P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 10

Then calculate amps and resistance for each load.

E=120V I=0.8333A R=E²/P P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 11

Then calculate amps and resistance for each load.

E=120V I=0.8333A R=144Ω P=100W E=120V I= R= P=200W E=240V I= R= P=

SLIDE 12

Then calculate amps and resistance for each load.

E=120V I=0.8333A R=144Ω P=100W E=120V I=P/E R= P=200W E=240V I= R= P=

SLIDE 13

Then calculate amps and resistance for each load.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R= P=200W E=240V I= R= P=

SLIDE 14

Then calculate amps and resistance for each load.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=E²/P P=200W E=240V I= R= P=

SLIDE 15

Then calculate amps and resistance for each load.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 16

With these values known, we can begin to understand what happens when the neutral is opened.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 17

First, we’ll open the neutral conductor.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 18

First, we’ll open the neutral conductor.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 19

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=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 20

So we will erase the figures we have calculated so far.

E=120V I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 21

So we will erase the figures we have calculated so far.

E= I=0.8333A R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 22

So we will erase the figures we have calculated so far.

E= I= R=144Ω P=100W E=120V I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 23

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=144Ω P= E=120V I=1.6667A R=72Ω P=200W E=240W I= R= P=

SLIDE 24

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=144Ω P= E= I=1.6667A R=72Ω P=200W E=240V I= R= P=

SLIDE 25

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=144Ω P= E= I= R=72Ω P=200W E=240V I= R= P=

SLIDE 26

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=144Ω P= E= I= R=72Ω P= E=240V I= R= P=

SLIDE 27

Now add the values of the two loads’ resistances.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I= R= P=

SLIDE 28

Now add the values of the two loads’ resistances.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I= R=144+72 P=

SLIDE 29

Now add the values of the two loads’ resistances.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I= R=216Ω P=

SLIDE 30

Now calculate I and P for the entire circuit.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I= R=216Ω P=

SLIDE 31

Now calculate I and P for the entire circuit.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I=E/R R=216Ω P=

SLIDE 32

Now calculate I and P for the entire circuit.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I=1.1111A R=216Ω P=

SLIDE 33

Now calculate I and P for the entire circuit.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I=1.1111A R=216Ω P=E²/R

SLIDE 34

Now calculate I and P for the entire circuit.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 35

Since this is a series circuit, I is always the same.

E= I= R=144Ω P= E= I= R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 36

Since this is a series circuit, I is always the same.

E= I=1.1111A R=144Ω P= E= I= R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 37

Since this is a series circuit, I is always the same.

E= I=1.1111A R=144Ω P= E= I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 38

Now calculate the voltage across each of the two loads.

E= I=1.1111A R=144Ω P= E= I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 39

Now calculate the voltage across each of the two loads.

E=R*I I=1.1111A R=144Ω P= E= I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 40

Now calculate the voltage across each of the two loads.

E=160V I=1.1111A R=144Ω P= E= I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 41

Now calculate the voltage across each of the two loads.

E=160V I=1.1111A R=144Ω P= E=R*I I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 42

Now calculate the voltage across each of the two loads.

E=160V I=1.1111A R=144Ω P= E=80V I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 43

Now you can see what happens when you lose the neutral

n a multiwire branch circuit.

E=160V I=1.1111A R=144Ω P= E=80V I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 44

The more resistance, the higher the voltage....

E=160V I=1.1111A R=144Ω P= E=80V I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 45

The less resistance, the lower the voltage.

E=160V I=1.1111A R=144Ω P= E=80V I=1.1111A R=72Ω P= E=240V I=1.1111A R=216Ω P=266.7W

SLIDE 46