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  #1  
Old 10-06-2011, 12:00 PM
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more circuit analysis questions

So I am stuck on a type of problem for my circuit analysis work. I cannot figure out how to solve.



I need to use mesh/loop analysis to solve for Vo. I know I need to use KVL for each loop but I don't understand how to solve. The algebra is confusing me. It's not like normal math.

help please!

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  #2  
Old 10-06-2011, 12:11 PM
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39.6 Volts

I learned my basic electricity over 40 years ago and don't remember hearing the term mesh/loop analysis.

Naming the reistors R1 through R4 left to right, start by solving for the voltage at the junction to the right of R1. This will be R2 in parallel with R3/R4 in series. R3 + R4 = 6K. These are in parallel with R2, so product over the sum = 3K X 6K / 3K + 6K = 2K. 2K in series with 2K splits the voltage in half for .5 X 120 or 60 Volts at the junction above R2. Then solve for the voltage split between R3 and R4 is simple by looking at it. .66 X the 60 Volts applied is 39.6 Volts.

Since I spent the last 25 years or so in software without ever soldering a circuit during that time, I surprised myself.
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  #3  
Old 10-06-2011, 12:29 PM
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I think I figured it out.....

For I1 I have : 2KI1 + 3K(I1-I2) = 12v

for I2 I have : 4KI1 + 2KI2 + 3K(I1-I2) = 0

Then from there I believe I will have to simplify and solve.

I1 : 2KI1 + 3KI1 - 3KI2 = 12 reduces to 5K1 - 3K2 = 12

I2 : 3KI1 - 3KI2 = 0

then

I2: -3k2 = -3K1 reduced to I2 = 1/1000 or 1/k amps


then using Ohms law take I2 multiply the 4K resistor on the right to get the voltage across it and it will be the same as Vo, which is 4V.
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  #4  
Old 10-06-2011, 12:55 PM
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How about this one?



I cannot seem to get I2 to equal something useful.
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  #5  
Old 10-06-2011, 01:00 PM
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Quote:
Originally Posted by TheDon View Post
I think I figured it out.....

For I1 I have : 2KI1 + 3K(I1-I2) = 12v

for I2 I have : 4KI1 + 2KI2 + 3K(I1-I2) = 0

Then from there I believe I will have to simplify and solve.

I1 : 2KI1 + 3KI1 - 3KI2 = 12 reduces to 5K1 - 3K2 = 12

I2 : 3KI1 - 3KI2 = 0

then

I2: -3k2 = -3K1 reduced to I2 = 1/1000 or 1/k amps


then using Ohms law take I2 multiply the 4K resistor on the right to get the voltage across it and it will be the same as Vo, which is 4V.

This likes the long way around if you're solving for voltage. I just saw that I was working on 120V input rather than 12V so divide my answer by 10.

Solving for currents is not necessary when they answer is looking for voltage. All you have to do is solve for voltage drops.

I have to run right now, so I'll look at the other one later.
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  #6  
Old 10-06-2011, 01:04 PM
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That was the way we were instructed to use. Thanks for your help. Cant wait to see how the second problem is done.
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  #7  
Old 10-06-2011, 01:46 PM
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Originally Posted by TheDon View Post
How about this one?



I cannot seem to get I2 to equal something useful.
This one is tougher. Can you provide an example of the "mesh" equation they cite? Without a few hints, I'm way to rusty to knock this one out quickly.
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Old 10-06-2011, 01:46 PM
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you could probably do it that way, but we need to do it via mesh/loop analysis

The answer in the book said 4 volts. Not 3.6v
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Old 10-06-2011, 01:56 PM
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Quote:
Originally Posted by TheDon View Post
you could probably do it that way, but we need to do it via mesh/loop analysis

The answer in the book said 4 volts. Not 3.6v

I'm not sure which one you're talking about now, but when I blew through and calculated using 120V I got 39.6, which with a 12V source ends up being 3.96 or very close to four volts.

It's been close to 40 years since I took EE classes and even longer since basic electricity and basic electronics. For the life of me I don't remember a "mesh" equation. At my age, however, there are some days I feel lucky to remember how to find my way to the refrigerator.
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  #10  
Old 10-06-2011, 01:59 PM
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maybe this will help refresh your memory?

Loop Current Analysis of Electric Circuits
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  #11  
Old 10-06-2011, 02:06 PM
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It's the multiple polar opposite voltage sources that are throwing me. Kirchoffs Law I remember, but I'm pretty unpracticed and the sources are killing me.
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  #12  
Old 10-06-2011, 02:20 PM
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I'm lost at how to solve lol.

I think I have the passive sign convention thing down.

I1 : 2KI1 - 12v + 2K(I1 - I2) = 0 reduces to 4KI1 - 2KI2 = 12v

I2 : -2K(I1-I2) + 2KI2 = -4v reduces to 4KI2 - 2KI1 = -4V

and I'm not sure if I2 is correct. I think I have the passive sign convention correct.
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  #13  
Old 10-06-2011, 11:24 PM
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I did this about 2 years ago. Yes it is a specific technique. There is a specific "recipe" for how to do it. As I recall you identify every circuit "loop" keeping in mind that some components are shared by more than one loop. You write an equation for each loop, keeping track of all your polarities which can be tricky. The you end up with simultaneous equations. I solved those with matrix algebra but you might get away with substitution. (yeah my teacher made us do the matrix by hand). But I double-checked with my TI-83. What book are you using?
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  #14  
Old 10-06-2011, 11:37 PM
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Brief introduction to circuit analysis by J. David Irwin.

I wish we did the matrix version, looks easier but I dont know how to do it(so its not easy?)
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  #15  
Old 10-09-2011, 06:09 PM
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nothing? anyone?

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