1. Start with a 120/240 multiwire branch circuit.

2. E= I= R= P= E= I= R= P= E= I= R= P= We’ll need to begin with some given values.

3. 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