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

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In summary: Here's a correct set of equations. The third equation is given the same label as in the sample answer but the 21 that appears there is not correct. Instead, the 21 should be 25. Note that you don't really need the third equation as it can be derived from the first two, but you can leave it in if it helps you solve the first two.eq #1) -i1*10 + (i1 - i2)*20 = 25Veq #2) (i1 - i2)*20 + i2*24 + 6*i3 = 0eq #3) i3*14 = 25You can rewrite eq #1 as:-i1*10
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orangeincup
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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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Related to How Do You Apply the Mesh-Current Method Correctly in Circuit Analysis?

1. What is the mesh-current method?

The mesh-current method is a technique used in circuit analysis to find the current flowing through individual loops in a circuit. It is based on Kirchhoff's voltage law and is particularly useful for solving circuits with multiple sources and complex configurations.

2. How does the mesh-current method work?

The mesh-current method involves assigning a current variable to each loop in the circuit and writing equations based on Kirchhoff's voltage law for each loop. These equations are then solved simultaneously to find the values of the mesh currents, which can then be used to find the currents through individual components in the circuit.

3. What are the advantages of using the mesh-current method?

The mesh-current method is particularly useful for solving circuits with multiple sources and complex configurations, as it simplifies the analysis process and allows for the solution of circuits with a large number of components. It also provides a systematic approach to solving circuits, making it easier to track and avoid errors.

4. Are there any limitations to the mesh-current method?

The mesh-current method may not be suitable for circuits with nonlinear components, as it assumes that all components in the circuit are linear. It also requires a good understanding of Kirchhoff's voltage law and can be time-consuming for circuits with a large number of loops.

5. Can the mesh-current method be used for AC circuits?

Yes, the mesh-current method can be used for AC circuits. However, in AC circuits, additional equations are required to account for the effects of reactance and impedance. These equations can be derived using the principles of Ohm's law and the rules of complex numbers.

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