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Simple Network How to Whiteboards

Simple Network How to Whiteboards. A 1. A 2. 3 . 7 . 22.5 V. 17 . V 2. 13 . A 3. 11 . V 3. 5 . V 1. Step 1 – reduce to a circuit you can solve. Go to: Expansion 1 |. A 1. A 2. 3 . 7 . 22.5 V. 17 . V 2. 13 . A 3. 11 . V 3. 5 . V 1.

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Simple Network How to Whiteboards

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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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