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3 Ω

3 Ω. 17 V. A 3. V 3. 7 Ω. A 3. V 1. 29 Ω. 31 Ω. 9 Ω. V 3. A 1. 37 V. A 4. 23 Ω. 7 Ω. V 2. 7 Ω. 8 Ω. A 2. 3 Ω. 5 Ω. 7 Ω. A 3. V 1. 29 Ω. 31 Ω. 9 Ω. V 3. A 1. 37 V. A 4. 23 Ω. 7 Ω. V 2. 7 Ω. 8 Ω. A 2. 3 Ω. 5 Ω. 7 Ω. A 3. V 1. 29 Ω. 31 Ω. A 1. 37 V. 23 Ω.

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3 Ω

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  1. 3Ω 17 V A3 V3

  2. 7Ω A3 V1 29Ω 31Ω 9Ω V3 A1 37 V A4 23Ω 7Ω V2 7Ω 8Ω A2 3Ω 5Ω

  3. 7Ω A3 V1 29Ω 31Ω 9Ω V3 A1 37 V A4 23Ω 7Ω V2 7Ω 8Ω A2 3Ω 5Ω

  4. 7Ω A3 V1 29Ω 31Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

  5. 7Ω A3 V1 29Ω 31Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

  6. 7Ω V1 14.9833Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

  7. 7Ω V1 14.9833Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

  8. 7Ω V1 21.9833Ω A1 37 V 23Ω 19Ω 8Ω A2 5Ω

  9. 7Ω V1 21.9833Ω A1 37 V 23Ω 19Ω 8Ω A2 5Ω

  10. Finally we have this last series circuit: 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

  11. Solving: Rtot =7+7.0622+5+8=27.0622Ω 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

  12. Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A (the reading on A1) 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

  13. Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A (the reading on A1) V1 = IR = 1.3672*7 = 9.5705V 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

  14. Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A (the reading on A1) V1 = IR = 1.3672*7 = 9.5705V And finally, V7.0622Ω = IR = 1.3672*7.0622 = 9.6556 V which is the voltage across the subcircuit. 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

  15. This is what you have left: 14.9833Ω 9.6556 V 23Ω 7Ω V2 19Ω A2

  16. This is what you have left: Solving for A2 is simple: I = V/R So A2 = 9.6556/23 = .4198A 14.9833Ω 9.6556 V 23Ω 7Ω V2 19Ω A2

  17. Which leaves: 14.9833Ω 9.6556 V 7Ω V2 19Ω

  18. Let’s look at the middle circuit 14.9833Ω 9.6556 V 7Ω V2 19Ω

  19. A3 This is what it really is 29Ω 31Ω 9.6556 V 7Ω V2

  20. But let’s go back to: 14.9833Ω 9.6556 V 7Ω V2

  21. But let’s go back to: We can solve this series circuit for V2 : Rtot = 7 + 14.9833 = 21.9833Ω 14.9833Ω 9.6556 V 7Ω V2

  22. But let’s go back to: We can solve this series circuit for V2 : Rtot = 7 + 14.9833 = 21.9833Ω I = V/R = 9.6556/21.9833 = .4392A 14.9833Ω 9.6556 V 7Ω V2

  23. But let’s go back to: We can solve this series circuit for V2 : Rtot = 7 + 14.9833 = 21.9833Ω I = V/R = 9.6556/21.9833 = .4392A V2 = IR = .4392*7 = 3.0746V And the voltage across the 14.9833Ω resistor is IR = .4392*14.9833 = 6.5810V 14.9833Ω 9.6556 V 7Ω V2

  24. Now, just for fun, let’s look at the 14.9833Ω subcircuit 14.9833Ω 9.6556 V 7Ω V2

  25. Now, just for fun, let’s look at the 14.9833Ω subcircuit Remember that it has a voltage of 6.5810V across it 14.9833Ω 9.6556 V 7Ω V2

  26. So it looks like this: A3 29Ω 31Ω 6.5810V

  27. We can find the current through the 31Ω resistor fairly simply: I = V/R = 6.5810/31 = .2123A A3 29Ω 31Ω 6.5810V

  28. We can find the current through the 31 resistor fairly simply: I = V/R = 6.5810/31 = .2123A But this is not the only current running through the ammeter A3. A3 29Ω 31Ω 6.5810V

  29. It also reads the current going through the 19Ω subcircuit A3 29Ω 31Ω 9.6556 V .2123A 7Ω V2 19Ω

  30. It also reads the current going through the 19Ω subcircuit I 19Ω = V/R = 9.6556/19 = .5082A A3 29Ω 31Ω 9.6556 V .5082A .2123A 7Ω V2 19Ω

  31. It also reads the current going through the 19Ω subcircuit I 19Ω = V/R = 9.6556/19 = .5082A So the reading on A3 is the sum of these currents: = .5082+.2123 = .7205A A3 29Ω 31Ω 9.6556 V .5082A .2123A 7Ω V2 19Ω

  32. It also reads the current going through the 19Ω subcircuit I 19Ω = V/R = 9.6556/19 = .5082A So the reading on A3 is the sum of these currents: = .5082+.2123 = .7205A Ta Daa! A3 29Ω 31Ω 9.6556 V .5082A .2123A 7Ω V2 19Ω

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