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Solving a Complex Series Circuit with Multiple Resistors and Voltages

This guide breaks down the solution of a complex series circuit involving multiple resistors and voltage sources. By calculating the total resistance, current, and voltage across different circuit components, we arrive at crucial values. The total resistance is found to be 15.5484Ω, with a current of 1.0934A flowing through the circuit. The voltage drop across the 7.5484Ω resistor is 8.2531V, allowing us to derive further values across additional resistors. This systematic analysis highlights key electrical principles in circuit theory.

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Solving a Complex Series Circuit with Multiple Resistors and Voltages

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  1. 3Ω 17 V A3 V3

  2. V2 A3 V1 V3 A2 A1 3 Ω 7 Ω 17 V 13 Ω 11 Ω 5 Ω

  3. V2 3 Ω A3 7 Ω 17 V A2 13 Ω V1 V3 11 Ω A1 5 Ω

  4. V2 3 Ω A3 17 V 13 Ω V1 18 Ω A1 5 Ω

  5. V2 3 Ω A3 17 V 13 Ω V1 18 Ω A1 5 Ω

  6. V2 3 Ω A3 17 V 7.5484 Ω 5 Ω

  7. V2 Now we can solve this series circuit: Rtot = 3+7.5484+5 = 15.5484 Ω I = V/R = 17/15.5484 = 1.0934A So A3 = 1.09A And V2 = IR = 1.0934*3 = 3.28 V Finally, the voltage present across the subcircuit (the “7.5484 Ω” resistor) is IR 1.0934*7.5484 = 8.2531V 3 Ω A3 17 V 7.5484 Ω 5 Ω

  8. This is the subcircuit now: 8.2531 V 7.5484 Ω

  9. Which is really: 7 Ω 8.2531 V A2 13 Ω V1 V3 11 Ω A1

  10. V1 is connected to the “battery” and reads 8.25 V, and the current in A1 is V/R = 8.2531/13 = .635 A 7 Ω 8.2531 V A2 13 Ω V1 V3 11 Ω A1

  11. Finally we have this last series circuit: 7 Ω 8.2531 V A2 V3 11 Ω

  12. Finally we have this last series circuit: Rtot = 7 + 11 = 18 Ω I = V/R = 8.2531/18 = .459A which is the reading on A2 V3 is IR = .459*11 = 5.04 V 7 Ω 8.2531 V A2 V3 11 Ω

  13. Finally we have this last series circuit: Rtot = 7 + 11 = 18 Ω I = V/R = 8.2531/18 = .459A which is the reading on A2 V3 is IR = .459*11 = 5.04 V Ta Daa! 7 Ω 8.2531 V A2 V3 11 Ω

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