1 / 46

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

vinnie
Download Presentation

Start with a 120/240 multiwire branch circuit.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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=

More Related