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3 Ω

3 Ω. 17 V. A 3. V 3. A 1. 17 V. 5 Ω. 8 Ω. 9 Ω. 7 Ω. A 2. V 3. 3 Ω. 6 Ω. A 4. 10 Ω. V 1. V 2. 12 Ω. A 3. 11 Ω. 4 Ω. 17 V. 5 Ω. 8 Ω. 9 Ω. 7 Ω. A 2. V 3. 3 Ω. 6 Ω. A 4. 10 Ω. V 1. A 1. V 2. 12 Ω. A 3. 11 Ω. 4 Ω. 17 V. 5 Ω. 7 Ω. 17 Ω. A 2. 3 Ω. 6 Ω. A 4.

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3 Ω

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  1. 3Ω 17 V A3 V3

  2. A1 17 V 5Ω 8Ω 9Ω 7Ω A2 V3 3Ω 6Ω A4 10Ω V1 V2 12Ω A3 11Ω 4Ω

  3. 17 V 5Ω 8Ω 9Ω 7Ω A2 V3 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

  4. 17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

  5. 17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 12Ω A3 11Ω 4Ω

  6. 17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 A3 23Ω 4Ω

  7. 17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 10Ω V1 A1 V2 A3 23Ω 4Ω

  8. 17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 V1 A1 V2 A3 6.9696Ω 4Ω

  9. 17 V 5Ω 7Ω 17Ω A2 3Ω 6Ω A4 V1 A1 V2 A3 6.9696Ω 4Ω

  10. 17 V 5Ω 7Ω A2 3Ω 6Ω V1 A1 V2 A3 23.9696Ω 4Ω

  11. 17 V 5Ω 7Ω A2 3Ω 6Ω V1 A1 V2 A3 23.9696Ω 4Ω

  12. 17 V 5Ω 5.4178Ω A2 3Ω 6Ω V1 A1 V2 A3 4Ω

  13. 17 V 5Ω 5.4178Ω A2 3Ω 6Ω V1 A1 V2 A3 4Ω

  14. 17 V 5.4178Ω A2 12Ω 6Ω V2 A3

  15. 17 V 5.4178Ω A2 12Ω 6Ω V2 A3

  16. 17 V Finally we have this last series circuit: 5.4178Ω 4Ω V2 A3

  17. 17 V Solving: Rtot = 5.4178 + 4 = 9.4178 5.4178Ω 4Ω V2 A3

  18. 17 V Solving: Rtot = 5.4178 + 4 = 9.4178 I = V/R = 17/9.4178 = 1.8051A (This is the reading on A3) 5.4178Ω 4Ω V2 A3

  19. 17 V Solving: Rtot = 5.4178 + 4 = 9.4178 I = V/R = 17/9.4178 = 1.8051A (This is the reading on A3) Picking off voltages: V2 = 1.8051*5.4178 = 9.7796 V V4Ω = 1.8051*4 = 7.2204 V 5.4178Ω 4Ω V2 A3

  20. 7.2204 V Solving the subcircuit on the left: 4Ω

  21. 7.2204 V Which is really A2 12Ω 6Ω

  22. 7.2204 V The current through the 6Ω is just V/R = 7.2204/6 = 1.2034A Which is the reading on A2. A2 12Ω 6Ω

  23. 7.2204 V This leaves: 12Ω

  24. 7.2204 V 5Ω Which is really: 3Ω V1 A1 4Ω

  25. 7.2204 V 5Ω Solving (series) Rtot = 5 + 3 + 4 = 12Ω 3Ω V1 A1 4Ω

  26. 7.2204 V 5Ω Solving (series) Rtot = 5 + 3 + 4 = 12Ω I = 7.2204/12 = .6017A Which is the reading on A1 3Ω V1 A1 4Ω

  27. 7.2204 V 5Ω Solving (series) Rtot = 5 + 3 + 4 = 12Ω I = 7.2204/12 = .6017A Which is the reading on A1 V1 = IR = .6017*3 = 1.8051V 3Ω V1 A1 4Ω

  28. 17 V Now let’s look at the right subcircuit: 5.4178Ω 4Ω V2 A3

  29. 17 V Now let’s look at the right subcircuit: (the 5.4178Ω resistor) V2 = 1.8051*5.4178 = 9.7796 V 5.4178Ω 4Ω V2 A3

  30. 9.7796 V So it looks like this: 5.4178Ω

  31. 9.7796 V Which is really 8Ω 9Ω 7Ω V3 A4 10Ω 12Ω 11Ω

  32. 9.7796 V But let’s go back to 7Ω 17Ω A4 6.9696Ω

  33. 9.7796 V Which is really 8Ω 7Ω 9Ω V3 A4 6.9696Ω

  34. 9.7796 V Solving the right side which is a series circuit (ignore the 7Ω) 8Ω 7Ω 9Ω V3 A4 6.9696Ω

  35. 9.7796 V Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω 8Ω 7Ω 9Ω V3 A4 6.9696Ω

  36. 9.7796 V Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω I = V/R = 9.7796/23.9696 = .408A which is the reading on A4 8Ω 7Ω 9Ω V3 A4 6.9696Ω

  37. 9.7796 V Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω I = V/R = 9.7796/23.9696 = .408A which is the reading on A4 And finally, V3 = IR = .408*9 = 3.6720V 8Ω 7Ω 9Ω V3 A4 6.9696Ω

  38. 9.7796 V Solving the right side which is a series circuit (ignore the 7Ω) Rtot= 8+9+6.9696 = 23.9696Ω I = V/R = 9.7796/23.9696 = .408A which is the reading on A4 And finally, V3 = IR = .408*9 = 3.6720V Ta Daa! 8Ω 7Ω 9Ω V3 A4 6.9696Ω

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