Mesh Analysis for AC RLC Circuits: Finding Vo and Io

[jess_88]

hi guys

I am trying to fined Vo and Io using mesh analysis for the following cct

The formulas I made from the cct using mesh analysis don't seem right

here is what I have-
loop1
2(I1 - I2) = 0
loop2
2(I2 - I1) + I2(j4) - 3Vo = 0
loop3
3Vo + I3(-2j) = 0

and
I1 = 4<-30
vo = 2(I1 - I2)

loop1
2(4<-30 - I2) = 0
I2 = 4<-30

loop 2
2(4<-30 - 4<-30) + I2(j4) - 6(4<-30 - 4<-30)

loop 3
6(4<-30 - 4<-30) + I3(-2j) = 0

I must have done something wrong... I just don't know where

 

[jegues]

here is what I have-
loop1
2(I1 - I2) = 0
loop2
2(I2 - I1) + I2(j4) - 3Vo = 0
loop3
3Vo + I3(-2j) = 0

For loop1 no mesh equation is needed, by inspection,

I_{1} = 4 \angle -30^{o}

Let's just keep that in mind because we will apply it later on.

For loop 2, you are mixing up the sign on the voltage source, it should be +3Vo. (You've mixed this up before in one of your previous problems.)

You have the same type of error in loop 3, -3Vo not +3Vo.

Fix those are you should be ready to solve!

 

[jess_88]

ah great thanks!
just to clarify my mistake with 3Vo.
is the polarity reversed due to the polarity of the vo resister?

 

[jegues]

ah great thanks!
just to clarify my mistake with 3Vo.
is the polarity reversed due to the polarity of the vo resister?

I'm not sure what you mean.

You have to remember all you are doing with mesh analysis is KVL combined with Ohm's Law. (It's a little more complex than that because the mesh currents don't always model the physical currents but that's extra details)

If you apply KVL in a clockwise fashion you should find that you are running a "positive" current through the voltage source as per conventional current.

By conventional current, I mean that a positive current is defined as one that flows form positive to negative.

You can read more about it here, http://www.asmcommunity.net/board/index.php?topic=12847.0;wap2

"It all boils down to always ASSUMING that a external flow of positive charges from the pos to the neg terminal of a voltage source always produces a positive current. This causes electron flow to have a MATHEMATICALLY opposite direction with respect its real direction, but it is consistent and correct with respect to positively charged particles. In cases where knowing the real physical particle direction is important, that can be handled on a case by case basis. Most of the time in circuit analysis, the real direction does not matter."

Hopefully this clears things up!

 

[jess_88]

ah geez.
I get it now!
thank you so much :)

 

[jess_88]

quick question.
I have for loop2
2(I2 - I1) + I2(j4) + 3Vo = 0
= -4I2 + j4(I2) = -I1(4)
can I do this next part??
-I2 + j4(I2) = -I1
-I2(1 - j) = -(4<-30)
I2 = (4<-30)/(sqrt(2)<-45)

... is this ok?