Mesh analysis with dependent voltage source

In summary: However, if you are asked to find the apparent power, then you can convert it to VA form.Not necessary if apparent power is not asked. You can keep it in the complex form. However, if you are asked to find the apparent power, then you can convert it to VA form.In summary, The complex power delivered by the source is 8-6j. The power factor of the total load is 0.8 leading.
  • #1
eehelp150
237
0

Homework Statement


Find the:
-Complex power delivered by source
-Power factor of total load
-Is the power factor leading, lagging, or in unity?[/B]
upload_2016-10-24_22-39-39.png


Homework Equations


KVL/KCL

The Attempt at a Solution


I want to solve this via mesh analysis to get currents.
Mesh1: -10V + V1 + 0.75V1 = 0
V1 = 40/7V
4I1 = 40/7
I1 = 10/7 A

Mesh2: -0.75V1 = I2(1-j1.5) = 0
I2(1-j1.5) = 0.75 * 40/7
I2 = 30/(7(1-j1.5))

Did I do this right?
 
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  • #2
The controlled source depicted is a current source, not a voltage source. So either employ a supermesh or introduce another variable to represent the potential across it (and you'll need another equation so that you have as many equations as variables in your equation set).
 
  • #3
gneill said:
The controlled source depicted is a current source, not a voltage source. So either employ a supermesh or introduce another variable to represent the potential across it (and you'll need another equation so that you have as many equations as variables in your equation set).
-10V + 4(I1) + I2(1 - j1.5) = 0
I2 = 3/4V1 + I1
V1 = 4(I1)

Are these correct?
 
  • #4
Yes, those look good.
 
  • #5
gneill said:
Yes, those look good.
i1 = 0.8 + 0.6i
i2 = 3.2 + 2.4i
correct?
 
  • #6
That looks good, too.
 
  • #7
gneill said:
That looks good, too.
How would I find complex power supplied by the source?
 
  • #8
eehelp150 said:
How would I find complex power supplied by the source?
Use the current supplied by the source and the source voltage.
 
  • #9
gneill said:
Use the current supplied by the source and the source voltage.
Is current supplied by source I1 - 3/4V1?
How would I find the power factor?
 
  • #10
Presumably the source referred to is the 10 V voltage source. It supplies just I1.

For the power factor you need to look at the power triangle. The complex power will give you the two "legs" of the triangle.
 
  • #11
gneill said:
Presumably the source referred to is the 10 V voltage source. It supplies just I1.

For the power factor you need to look at the power triangle. The complex power will give you the two "legs" of the triangle.
So complex power = 10 * I* = 10 * (0.8 - 0.6i)?
S = 8-6i
Power factor = P/S = 8/10 = 0.8 leading? (Q<0)
Does this triangle look right?
upload_2016-10-26_23-5-16.png
 
  • #12
eehelp150 said:
So complex power = 10 * I* = 10 * (0.8 - 0.6i)?
S = 8-6i
Power factor = P/S = 8/10 = 0.8 leading?
Yes. Also, write 'j' instead of 'i' in the imaginary part of the complex form (e.g
S=8-j6). That's a convention followed in order to avoid confusion with the symbol 'I' for current.
 
  • #13
cnh1995 said:
Yes. Also, write 'j' instead of 'i' in the imaginary part of the complex form (e.g
S=8-j6). That's a convention followed in order to avoid confusion with the symbol 'I' for current.
Can (should) I convert to VA form? (sqrt(P^2+Q^2))
 
  • #14
eehelp150 said:
Can (should) I convert to VA form? (sqrt(P^2+Q^2))
Not necessary if apparent power is not asked. You can keep it in the complex form.
 

1. How do you handle dependent voltage sources in mesh analysis?

In order to handle dependent voltage sources in mesh analysis, you need to first convert them into equivalent current sources. This can be done by using the source transformation technique, where the voltage source is replaced by a current source with the same value and a resistor in series. Once the dependent voltage source has been converted, you can proceed with the usual mesh analysis method.

2. Can dependent voltage sources be ignored in mesh analysis?

No, dependent voltage sources cannot be ignored in mesh analysis. They play a significant role in the circuit and their effects need to be considered in order to accurately analyze the circuit. As mentioned earlier, they can be converted into equivalent current sources and included in the analysis.

3. How do you determine the polarity of a dependent voltage source in mesh analysis?

The polarity of a dependent voltage source is determined by the direction of the current flowing through it. If the current is flowing in the positive direction, the polarity of the dependent voltage source will be positive, and vice versa. This can also be determined by using the passive sign convention, where the positive terminal of the dependent voltage source is placed on the side of the circuit with the higher potential.

4. Can mesh analysis be used for circuits with multiple dependent voltage sources?

Yes, mesh analysis can be used for circuits with multiple dependent voltage sources. Each dependent voltage source can be converted into an equivalent current source, which can then be included in the analysis. The resulting equations may become more complex, but the same principles of mesh analysis can still be applied.

5. What is the advantage of using mesh analysis with dependent voltage sources?

The advantage of using mesh analysis with dependent voltage sources is that it allows for a more accurate analysis of the circuit. By converting the dependent voltage sources into equivalent current sources, their effects can be included in the analysis. This results in more precise calculations and a better understanding of the circuit's behavior.

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