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Circuit Analysis: Step-by-Step Guide to Solve Complex Parallel and Series Circuits

This comprehensive guide walks you through the process of solving complex electrical circuits combining parallel and series components. Starting from an initial circuit notation, it emphasizes how to simplify circuits into solvable forms. Step 1 focuses on reducing the circuit for ease of solving, while Steps 2 and 3 involve calculating currents and voltages across different components. Ideal for students new to circuit analysis, this resource elaborates on key formulas and techniques required to master circuit problems.

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Circuit Analysis: Step-by-Step Guide to Solve Complex Parallel and Series Circuits

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  1. Simple Network • How to • Whiteboards

  2. A1 A2 3  7  22.5 V 17  V2 13  A3 11  V3 5  V1 Step 1 – reduce to a circuit you can solve Go to: Expansion 1 |

  3. A1 A2 3  7  22.5 V 17  V2 13  A3 11  V3 5  V1 Step 1 – reduce to a circuit you can solve

  4. 22.5 V V2 V1 Step 1 – reduce to a circuit you can solve A1 A2 3  17  13  18  5 

  5. 22.5 V V2 V1 Step 1 – reduce to a circuit you can solve A1 A2 3  17  13  18  5 

  6. 22.5 V V2 V1 Step 1 – reduce to a circuit you can solve A1 A2 3  17  7.5484  5 

  7. 22.5 V V2 V1 Step 1 – reduce to a circuit you can solve A1 A2 3  17  7.5484  5 

  8. A 22.5 V B VAB = 22.5 V Ignore the left side and… Step 2 – Solve We can solve this parallel circuit I17 = 22.5/17 = 1.3235 A I15.5... = 22.5/ 15.5484 = 1.4471 A = A2 A1 reads the sum of these = 1.3235 A + 1.4471 A = 2.7706 A A1 A2 17  15.5484 

  9. A 22.5 V V2 B V1 VAB = 22.5 V Expand Step 3 –Expand The right side is just a series circuit: R total = 3 + 5 + 7.5484 = 15.5484  I = 22.5/15.5484 = 1.4471 A V1 = IR = (1.4471 A)(5 ) = 7.2355 V V2 = (1.4471 A)(7.5484 ) = 10.9232 V A1 A2 3  22.5 V 7.5484  5 

  10. A 22.5 V V2 B V1 VAB = 10.9232 V Step 3 –Expand The right side is just a series circuit: R total = 3 + 5 + 7.5484 = 15.5484  I = 22.5/15.5484 = 1.4471 A V1 = IR = (1.4471 A)(5 ) = 7.2355 V V2 = (1.4471 A)(7.5484 ) = 10.9232 V A1 A2 3  17  7.5484  5 

  11. A V3 B VAB = 10.9232 V Step 3 –Expand 7  10.9232 V A3 11  This is a series circuit I = 10.9232/(7+11) = 0.6068 A (A3 reads this) V3 = IR = (0.6068 A)(11 ) = 6.6753 V Ta Daaa!

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