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Start with a 120/240 multiwire branch circuit.

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

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Start with a 120/240 multiwire branch circuit.

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

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

  3. We’ll need to begin with some given values. E=120V I= R= P= E=240V I= R= P= E=120V I= R= P=

  4. Now lets add a load to each circuit. E=120V I= R= P= E=240V I= R= P= E=120V I= R= P=

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

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

  20. 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

  21. 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

  22. 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

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

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

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

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

  27. Now add the values of the two loads’ resistances. E= I= R=144W P= E=240V I= R= P= E= I= R=72W P=

  28. 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=

  29. 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=

  30. 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=

  31. 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=

  32. 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=

  33. 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=

  34. 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=

  35. 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=

  36. 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=

  37. 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=

  38. 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=

  39. 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=

  40. 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=

  41. 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=

  42. 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=

  43. 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=

  44. 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=

  45. 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=

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

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