1 / 11

Assessment Problem # 4.12

Assessment Problem # 4.12. Mesh-Current Method Special Case Current Source in a branch. Use the mesh-current method to find the power dissipated in the 1 Ω resistor. Identify the mesh currents. I 3. I 2. I 1. Assign the voltage polarities Identify the voltage @ the 2A source. +. -.

leo-wilson
Download Presentation

Assessment Problem # 4.12

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Assessment Problem # 4.12 Mesh-Current Method Special Case Current Source in a branch ECE 201 Circuit Theory I

  2. Use the mesh-current method to find the power dissipated in the 1 Ω resistor. ECE 201 Circuit Theory I

  3. Identify the mesh currents I3 I2 I1 ECE 201 Circuit Theory I

  4. Assign the voltage polaritiesIdentify the voltage @ the 2A source + - I3 - + + - + - V - + I2 I1 + - - + ECE 201 Circuit Theory I

  5. Write the mesh-current equations (1) + - I3 - + + - + - V - + I2 I1 + - V + 2I1 – 2I2 = 10 - + ECE 201 Circuit Theory I

  6. Write the mesh-current equations (2) + - I3 - + + - + - V - + I2 I1 + - -2I1 +5I2 -2I3 = 6 V + 2I1 – 2I2 = 10 - + ECE 201 Circuit Theory I

  7. Write the mesh-current equations (3) + - I3 -V -2I2 +4I3 = 0 - + + - + - V - + I2 I1 + - -2I1 +5I2 -2I3 = 6 V + 2I1 – 2I2 = 10 - + ECE 201 Circuit Theory I

  8. Eliminate V from the equations for meshes 1 and 3 ECE 201 Circuit Theory I

  9. Write the equations in matrix form Solving with a TI-89, [6,-4;-4,5],[18;2])ENTER ECE 201 Circuit Theory I

  10. Solving • Result looks like I1 = 7A I2 = 6A Calculate I3 I3 =I1 – 2 = 5A ECE 201 Circuit Theory I

  11. Find the power dissipated in the 1Ω resistor • P1Ω = (I2)2R = (6)2(1) = 36 W ECE 201 Circuit Theory I

More Related