2. E= I= R= P= E= I= R= P= E= I= R= P= We’ll need to begin with some given values.
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=
4. Now lets add a load to each circuit. E=120V I= R= P= E=120V I= R= P= E=240V I= R= P=
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=
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=
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=
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=
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=
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=
11. Then calculate amps and resistance for each load. E=120V I=0.8333A R=144 W P=100W E=120V I= R= P=200W E=240V I= R= P=
12. Then calculate amps and resistance for each load. E=120V I=0.8333A R=144 W P=100W E=120V I=P/E R= P=200W E=240V I= R= P=
13. Then calculate amps and resistance for each load. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R= P=200W E=240V I= R= P=
14. Then calculate amps and resistance for each load. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R=E²/P P=200W E=240V I= R= P=
15. Then calculate amps and resistance for each load. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
16. With these values known, we can begin to understand what happens when the neutral is opened. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
17. First, we’ll open the neutral conductor. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
18. First, we’ll open the neutral conductor. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
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=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
20. So we will erase the figures we have calculated so far. E=120V I=0.8333A R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
21. So we will erase the figures we have calculated so far. E= I=0.8333A R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
22. So we will erase the figures we have calculated so far. E= I= R=144 W P=100W E=120V I=1.6667A R=72 W P=200W E=240V I= R= P=
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=144 W P= E=120V I=1.6667A R=72 W P=200W E=240W I= R= P=
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=144 W P= E= I=1.6667A R=72 W P=200W E=240V I= R= P=
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=144 W P= E= I= R=72 W P=200W E=240V I= R= P=
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=144 W P= E= I= R=72 W P= E=240V I= R= P=
27. Now add the values of the two loads’ resistances. E= I= R=144 W P= E= I= R=72 W P= E=240V I= R= P=
28. Now add the values of the two loads’ resistances. E= I= R=144 W P= E= I= R=72 W P= E=240V I= R=144+72 P=
29. Now add the values of the two loads’ resistances. E= I= R=144 W P= E= I= R=72 W P= E=240V I= R=216 W P=
30. Now calculate I and P for the entire circuit. E= I= R=144 W P= E= I= R=72 W P= E=240V I= R=216 W P=
31. Now calculate I and P for the entire circuit. E= I= R=144 W P= E= I= R=72 W P= E=240V I=E/R R=216 W P=
32. Now calculate I and P for the entire circuit. E= I= R=144 W P= E= I= R=72 W P= E=240V I=1.1111A R=216 W P=
33. Now calculate I and P for the entire circuit. E= I= R=144 W P= E= I= R=72 W P= E=240V I=1.1111A R=216 W P=E²/R
34. Now calculate I and P for the entire circuit. E= I= R=144 W P= E= I= R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
35. Since this is a series circuit, I is always the same. E= I= R=144 W P= E= I= R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
36. Since this is a series circuit, I is always the same. E= I=1.1111A R=144 W P= E= I= R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
37. Since this is a series circuit, I is always the same. E= I=1.1111A R=144 W P= E= I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
38. Now calculate the voltage across each of the two loads. E= I=1.1111A R=144 W P= E= I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
39. Now calculate the voltage across each of the two loads. E=R*I I=1.1111A R=144 W P= E= I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
40. Now calculate the voltage across each of the two loads. E=160V I=1.1111A R=144 W P= E= I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
41. Now calculate the voltage across each of the two loads. E=160V I=1.1111A R=144 W P= E=R*I I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
42. Now calculate the voltage across each of the two loads. E=160V I=1.1111A R=144 W P= E=80V I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
43. Now you can see what happens when you lose the neutral on a multiwire branch circuit. E=160V I=1.1111A R=144 W P= E=80V I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
44. The more resistance, the higher the voltage.... E= 160V I=1.1111A R= 144 W P= E=80V I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W
45. The less resistance, the lower the voltage. E=160V I=1.1111A R=144 W P= E= 80V I=1.1111A R= 72 W P= E=240V I=1.1111A R=216 W P=266.7W
46. This is Ohm’s Law in a most practical use, and the reason Article 300.13(B) exists. E=160V I=1.1111A R=144 W P= E=80V I=1.1111A R=72 W P= E=240V I=1.1111A R=216 W P=266.7W