In the middle of an AC analysis problem, I am stuck -

In summary, the conversation is about a person trying to solve a complex number equation and getting an incorrect answer. They realize their mistake and learn the correct method to solve the equation.
  • #1
VinnyCee
489
0
OK, I have this complex number equation:

[tex]5\,V\,=\,\left[\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)\right]\,i_L[/tex]

Now I try to solve for [itex]i_L[/itex]:

[tex]i_L\,=\,\frac{5\,V}{\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)}[/tex]

[tex]i_L\,=\,\frac{5\,V}{-20000j^2\,-\,6000j\,+\,1200}[/tex]

Since [itex]j^2[/itex] is just -1:

[tex]i_L\,=\,\frac{5\,V}{21200\,-\,6000j}[/tex]

And since [itex]\frac{1}{j}[/itex] = -j, the final complex numbered answer I get is:

[tex]0.0002358\,+\,0.00083333j[/tex]

However, this is incorrect! I have the answer for the problem, step-by-step, given by the prof. and I have double checked the answer using the Symbulator for the TI-89.

I should be getting:

[tex]0.00021836\,+\,0.0000618j[/tex]

What am I doing wrong?:frown:
 
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  • #2
VinnyCee said:
OK, I have this complex number equation:

[tex]5\,V\,=\,\left[\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)\right]\,i_L[/tex]

Now I try to solve for [itex]i_L[/itex]:

[tex]i_L\,=\,\frac{5\,V}{\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)}[/tex]

[tex]i_L\,=\,\frac{5\,V}{-20000j^2\,-\,6000j\,+\,1200} [/tex]

Since [itex]j^2[/itex] is just -1:

[tex]i_L\,=\,\frac{5\,V}{21200\,-\,6000j}\;\;\; (1)[/tex]

And since [itex]\frac{1}{j}[/itex] = -j, the final complex numbered answer I get is:

[tex]0.0002358\,+\,0.00083333j \;\;\; (2)[/tex]

What've you done to get from (1) to (2)? I would first divide top and bottom by 5 to get
[tex]i_L\,=\,\frac{V}{4240-1200j}[/tex]

Now, you need to get this into the form V(x+jy), so I would multiply top and bottom by the complex conjugate of the denominator

[tex]i_L\,=\,\frac{V}{4240-1200j}\cdot\frac{4240+1200j}{4240+1200j}[/tex]

Now expand out the denominator, and you should obtain a purely real number, and the result will follow from dividing the real and imaginary parts of the numerator by the denominator.
I should be getting:

[tex]0.00021836\,+\,0.0000618j[/tex]
That's what I get too!
 
Last edited:
  • #3
Wow, thanks!

Why does the way I was trying not work?

[tex]\frac{1}{j}\,=\,-j[/tex] <----- Right?

[tex]\frac{5}{21200}\,=\,0.0002358[/tex]

and for the complex part:

[tex]\frac{5}{-6000}\m\left(-j\right)\,=\,0.000833j[/tex]

That is where I get the answer I was getting (wrong):

[tex]0.0002358\,+\,0.000833j[/tex]
 
Last edited:
  • #4
VinnyCee said:
Wow, thanks!

Why does the way I was trying not work?

[tex]\frac{1}{j}\,=\,-j[/tex] <----- Right?

[tex]\frac{5}{21200}\,=\,0.0002358[/tex]

and for the complex part:

[tex]\frac{5}{-6000}\m\left(-j\right)\,=\,0.000833j[/tex]

That is where I get the answer I was getting (wrong):

[tex]0.0002358\,+\,0.000833j[/tex]

Because you made the error of assuming that [tex]\frac{a}{b+cj} = \frac{a}{b}+\frac{a}{cj}[/tex]. This is not true (in general) and you cannot split up a fraction like this.

The trick for solving such a question is to always multiply the fraction by the conjugate of the denominator-- this makes the denominator real, and so the fraction becomes one of the form [tex]\frac{a+bj}{c}[/tex] which of course can be simplified to obtain [tex] \frac{a}{c}+\frac{b}{c}j[/tex]
 
  • #5
Sweet, thanks again! Now I understand.
 

What exactly is an AC analysis problem?

An AC analysis problem is a type of problem in electrical engineering or physics that involves analyzing the behavior of a circuit or system in response to an alternating current (AC) signal. This type of analysis is used to understand the frequency response and performance of a circuit or system, and is an important tool for designing and troubleshooting electronic devices.

Why am I stuck in the middle of an AC analysis problem?

There could be several reasons why you may be stuck in the middle of an AC analysis problem. It could be due to a lack of understanding of the underlying concepts, incorrect calculations or assumptions, or a mistake in setting up the problem. It is important to carefully review your work and seek help from a teacher or classmate if needed.

How can I improve my understanding of AC analysis problems?

One of the best ways to improve your understanding of AC analysis problems is to practice and work through different examples. Additionally, reading textbooks and consulting with knowledgeable individuals can also help clarify any confusion or misunderstanding. It is important to have a strong foundation in the fundamental concepts and equations used in AC analysis.

What are some common mistakes to avoid in AC analysis problems?

Some common mistakes to avoid in AC analysis problems include forgetting to take into account the phase shift of components such as capacitors and inductors, not properly converting between units, and failing to check your calculations for accuracy. It is also important to double check your work and ensure that all assumptions and approximations are valid.

What are some tips for successfully solving AC analysis problems?

To successfully solve an AC analysis problem, it is important to have a clear understanding of the problem statement and the given information. It can be helpful to break down the problem into smaller, more manageable steps and to carefully label and organize your calculations. Additionally, reviewing and practicing with different types of AC analysis problems can also improve your skills and problem-solving abilities.

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