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Mesh Current Analysis of Power in an 80V Circuit with an 8Ω Resistor

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This assessment focuses on calculating the power delivered by an 80V source and the power dissipated in an 8Ω resistor using the mesh-current method. Starting with the identification of mesh currents (I1, I2, I3), the voltage polarities are assigned to each mesh. By formulating and solving the mesh-current equations, we determine the mesh currents and subsequently calculate the power delivered by the source and dissipated in the resistor. The power calculations reveal that the source delivers 400W while the resistor dissipates 50W.

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Mesh Current Analysis of Power in an 80V Circuit with an 8Ω Resistor

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  1. Assessment Problem #4.7 Page 107 of the text by Nilsson and Riedel ECE 201 Circuit Theory I

  2. Determine the power delivered by the 80 V source and the power dissipated in the 8 Ω resistor using the mesh-current method. ECE 201 Circuit Theory I

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

  4. Assign the voltage polarities + - I3 - + - + + - + - - + + I2 I1 - - + ECE 201 Circuit Theory I

  5. Write the mesh-current equations (1) + - I3 - + - + + - + - - + + I2 I1 - - + 80 = 31I1 - 26I2 - 5I3 ECE 201 Circuit Theory I

  6. Write the mesh-current equations (2) + - I3 - + - + + - + - - + + I2 I1 - - + 80 = 31I1 - 26I2 - 5I3 0 = -26I1 +124I2 -90I3 ECE 201 Circuit Theory I

  7. Write the mesh-current equations (3) + - 0 = -5I1 -90I2 +125I3 I3 - + - + + - + - - + + I2 I1 - - + 80 = 31I1 - 26I2 - 5I3 0 = -26I1 +124I2 -90I3 ECE 201 Circuit Theory I

  8. Write the equations in matrix form ECE 201 Circuit Theory I

  9. Check for off-diagonal symmetry ECE 201 Circuit Theory I

  10. Solving with TI-89 • Check CATALOG for simult( --- press ENTER • [31,-26,-5;-26,124,-90;-5,-90,125],[80;0;0]) hit ENTER • Resulting column vector is I1 = 5A I2 = 2.5A I3 = 2A ECE 201 Circuit Theory I

  11. Calculate the power delivered by the source Psource = (Vsource)(I1) = (80V)(5A) = 400 W + I1 = 5 A - ECE 201 Circuit Theory I

  12. Calculate the power dissipated in the 8Ω resistor P8Ω = (I2)2R = (2.5)2(8) = 50 W I2 = 2.5 A ECE 201 Circuit Theory I

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