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Hello, while conducting nodal analysis on an AC circuit, the book gave one answer. The book did not show any intermediate steps in their approach to simplifying the equation for one of the nodes in the circuit. So I tried to simplify the node equation and came out with a drastically different answer. Could anyone please review the attached .pdf file and let me know what mistake(s) I made in simplifying this equation? Thank you.
Phasor simplification problem
- Thread starter Pavlo138
- Start date
This one isn't actually that bad as the 'j' terms all appear in the denominator and can be cancelled out.
\(\frac{2V_{1}}{-j2.5} + \frac{(V_{1} - V_{2})}{j4} = \frac{V_{2}}{j2}\)
\(-\frac{4V_{1}}{j5} + \frac{V_{1}}{j4} - \frac{V_{2}}{j4} = \frac{V_{2}}{j2}\)
Convert to common denominator of 20 and cancel all 'j' terms:
\(-16V_{1} + 5V_{1} - 5V_{2} = 10V_{2}\)
Therefore:
\(11V_{1} + 15V_{2} = 0\)
\(\frac{2V_{1}}{-j2.5} + \frac{(V_{1} - V_{2})}{j4} = \frac{V_{2}}{j2}\)
\(-\frac{4V_{1}}{j5} + \frac{V_{1}}{j4} - \frac{V_{2}}{j4} = \frac{V_{2}}{j2}\)
Convert to common denominator of 20 and cancel all 'j' terms:
\(-16V_{1} + 5V_{1} - 5V_{2} = 10V_{2}\)
Therefore:
\(11V_{1} + 15V_{2} = 0\)
First, you need to get in the habit of checking your own work when and where possible -- not only in the case of homework, but even more importantly for when you are doing this for real and you don't have the "book" answer or a forum handy.
So what happens when you take your answer, which immediately results in V1 = -6V2, and plug that in to the original equation? SO, without resorting to the book's answer, you know that yours is wrong. Developing the skills and habits that will enable you to routinely catch your mistakes (and you will make them throughout your career, have no illusion about that) is one of the most valuable things you can possibly do.
Your mistake is in your very first line.
Imagine you have
\(
\frac{A}{10} + \frac{B}{2} = \frac{C}{3}
\)
would you start out by claiming that this is the same as
\(
10A + 2B = 3C
\)
If not, then why did you do that here?
If so, then your problem is that you need to take a step back and build up your basic algebra knowledge. If you don't, it WILL haunt you.
So what happens when you take your answer, which immediately results in V1 = -6V2, and plug that in to the original equation? SO, without resorting to the book's answer, you know that yours is wrong. Developing the skills and habits that will enable you to routinely catch your mistakes (and you will make them throughout your career, have no illusion about that) is one of the most valuable things you can possibly do.
Your mistake is in your very first line.
Imagine you have
\(
\frac{A}{10} + \frac{B}{2} = \frac{C}{3}
\)
would you start out by claiming that this is the same as
\(
10A + 2B = 3C
\)
If not, then why did you do that here?
If so, then your problem is that you need to take a step back and build up your basic algebra knowledge. If you don't, it WILL haunt you.
So, j/j= 1 then? What has been throwing me off this whole time has been operations on complex numbers. Thanks.
Anything, other than zero, divided by itself is one.
You can always just treat j as though it were any other variable or parameter. Multiply your entire equation by whatever power of j is required to get rid of all the j's in the denominator. When the dust settles you might then have j raised to various powers running around, but you can simplify those knowing that j^2 is -1.
