# AC Mesh Analysis

Discussion in 'Homework Help' started by brighton53, Jan 16, 2013.

1. ### brighton53 Thread Starter New Member

Jan 16, 2013
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0
Hi, I'm trying to find what I(t) and V2(t).

- Firstly I'm changing to impedances and then predefining a few new variables
- zc1, zr1, zc2, zr2 and zc3
- then I'm going to do mesh analysis

and then the current I'm trying to work out should be the answer of I2.

and the voltage I'm trying to also workout will do Zc3 * I3.

Is all of this correct or have a made a mistake?

Thanks !!

2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,255
311
It looks ok except that you are postmultiplying the voltage vector by the inverse of the impedance matrix. You must premultiply the voltage vector by the inverse of the impedance matrix.

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3. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
The voltage V2(t) is poorly defined because there is no indication of polarity. Is V2(t) the voltage at the top of the capacitor relative to the bottom of the capacitor, or the other way around. Both are valid and completely consistent with the original diagram. So you need to define what YOU choose to interpret V2(t) to mean. A similar thing applies to the voltage source. Whoever wrote the problem was pretty sloppy.

I'll take a look at your work in a few minutes.

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4. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Your mesh equations are fine, but your final matrix equation has problems. I don't do a lot with linear algebra (have never even taken a course, much to my everlasting regret), so I'm not positive what rule you are breaking. But on the RHS you have a the product of two vectors and the result will be a vector/matrix in which the number of rows is equal to the number of rows in the first operand and the number of columns will be equal to the number of columns in the second operand, thus you will end up with a 3x3 matrix instead of a 3x1 column vector.

I may not be super conversant with the rules, but at least I try to understand enough to be able to sanity check results.

I think your problem is that matrix multiplication is not commutative, so if you multiply the impedance matrix by the inverse of it on the left hand side of the left hand equation, then you have to multiply it on the left hand side of the right hand equation in order to maintain equality.

5. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Oh, good. What I said was consistent with what the Electrician noted and I know he is conversant with matrix manipulation.