Nodal voltages

  • Thread starter celect
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43
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Can someone direct me to a tutorial for solving network
nodal voltages.
Also need to learn how to convert voltage source to a current source.
 
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a text book should have everything you need..your in philly they must have a kick a*s Library there.. :smile:
 
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Yea I was there today, will be back tomorrow.
 
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celect said:
Can someone direct me to a tutorial for solving network
nodal voltages.
Also need to learn how to convert voltage source to a current source.
You cannot convert a voltage source into a current source. What you can do is to convert a voltage source in series with a resistor into a current source in parallel with the same resistor. This is Norton's theorem.
 
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Ok I'm reading that chapter now, I attached a sample I'm working on now.
 

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The solution you posted is wrong. The value of the current source is V/Z and what must be in parallel with the source is the series impedance, not the parallel you draw.
 
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Regarding the right most circuit component, the symbol represents an inductor; yet the units is given in ohms.
 
inductor are coils , therefore i believe they can have resistance,
anyway,Checkout OpenCourseWare at MIT
 
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So if I need to find E/Z = 20/9.9 = 2A

Now all I do is redraw circuit with 2A current source with
impedance parallel in circuit?
 
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If you want to do a source transformation on this circuit, you should probably look up phasors.
 
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celect said:
So if I need to find E/Z = 20/9.9 = 2A

Now all I do is redraw circuit with 2A current source with
impedance parallel in circuit?
No, you must make
[tex]I=\frac{20[cos(20)+jsin(20)]}{5.6+j8.2}[/tex]
 
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I'm confused on how to get proper result

I have now I = 20V <20 degree/ 5.6+J8.2
and I found in book the way to draw circuit.
but I don't understand how to use the equation with cos and j 20 sin.
 
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That's just Euler's formula and phasor notation:
"A-angle-theta" is phasor notation for A*ej*theta, which by Euler's formula is A[cos(theta) + j*sin(theta)], where A is a scalar value.
 
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You can also write 5.6 + j8.2 as 9.93 <55.67
So, [tex]I=\frac{20 <20}{9.93 <55.67}[/tex]
That is an easiest form to evaluate the quocient.
 
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Here is a problem I got wrong, with the solution.
I don't understand how they got answer.


I don understand how you get
5.6 + j8.2 as 9.93 <55.67
how does 5.6 become 9.33
and j.2 = 55.67
 

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celect said:
Here is a problem I got wrong, with the solution.
I don't understand how they got answer.


I don understand how you get
5.6 + j8.2 as 9.93 <55.67
how does 5.6 become 9.33
and j.2 = 55.67
I have transformed the complex number from its cartesian form to its polar form.
Z = x + jy = R<φ
Where R is the modulus of the complex:
[tex]R = \sqrt{x^2 + y^2}[/tex]
and
[tex]\phi = tan^{-1}\frac{y}{x}[/tex]
 
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Thanks its coming together now.

so my current = 20/9.93 = 2A

I don't understand how to get degrees yet
[tex]\phi = tan^{-1}\frac{y}{x}[/tex]

do I take 8.3/5,5 tan and then use above formula?

what is the -1 how do i use on calculator?
 
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celect said:
Thanks its coming together now.

so my current = 20/9.93 = 2A

I don't understand how to get degrees yet
[tex]\phi = tan^{-1}\frac{y}{x}[/tex]

do I take 8.3/5,5 tan and then use above formula?

what is the -1 how do i use on calculator?
[tex]tan^{-1}[/tex] means arctan. It is the arc whose tangent is y/x.
And your current is I = 2 <-35.67.
When dividing complex numbers in the polar form you divide the moduli and subtract the phases.
 

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