How Do You Apply the Mesh-Current Method Correctly in Circuit Analysis?

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SUMMARY

The mesh-current method is essential for analyzing circuits, particularly in determining voltages and power delivered by dependent sources. In this discussion, the correct application involves defining loop currents clearly and understanding their interactions with resistors. The equations derived include eq #1: -i1*10 + i2*20 = 25V, eq #2: i2*20 + i2*24 + 6iΔ = 0, and eq #3: iΔ*14 = 21, leading to a power calculation of 1.839W. Properly identifying the direction of currents and their contributions to voltage across resistors is crucial for accurate analysis.

PREREQUISITES
  • Understanding of mesh-current analysis in circuit theory
  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of power calculations in electrical circuits (P = IV)
  • Ability to set up and solve simultaneous equations
NEXT STEPS
  • Study the principles of mesh analysis in greater depth
  • Learn how to correctly identify and denote loop currents in circuit diagrams
  • Explore advanced circuit analysis techniques, such as nodal analysis
  • Practice solving circuit problems involving dependent sources
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in circuit analysis who seeks to enhance their understanding of the mesh-current method and its applications in real-world scenarios.

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Homework Statement


Use the mesh-current method to find v0.
Find the power delivered by the dependent source.

Homework Equations


v1+v2+..vn=0
i=v/R
p=iv

The Attempt at a Solution


eq #1)-i1*10+i2*20=25V
eq #2) i2*20+i2*24+6iΔ=0
eq #3) iΔ*14=21

eq#3) iΔ=1.5A
eq#3 sub into eq#2) i2=0.204A
i2 sub into eq#1) -1.6A

Power delivered=
p=iv
p=0.204A*(6*1.5)V = 1.839 W

I posted the solution below. I don't understand how they got it. They added up resistances, and also said i1*20 in the first equation, but i2*20 in the second. Also, I'm not sure how to get the polarities when the + and i signs aren't given. For example, the 24Ω resistor.
 

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eq #1)-i1*10+i2*20=25V
You are not applying mesh analysis correctly. The currents denoted as i1 and i2 are loop currents, this means each flows in a closed loop, so it's best that you show them doing so by marking in each loop current as a complete circle even though the sample answer has set a poor example by not doing so.

When done properly, you can see that the current in the 10Ω resistor is i1, and the current in the 20Ω resistor is the sum of two currents: i1 down and i2 upwards, making it (i1 - i2) total. So the voltage across the 20Ω resistor is going to involve that current, the resistor's total current.

Make a fresh start. Ignore the calculations shown in the sample answer, for the time being.
 
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