# Help me Ac Loop analysis

Discussion in 'Homework Help' started by xr329, Nov 24, 2014.

1. ### xr329 Thread Starter New Member

Nov 24, 2014
4
0
Hello
I have mini project and I have to submit it in few days
I have to solve this problem using Loop analysis

Thanks

Mar 31, 2012
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3. ### xr329 Thread Starter New Member

Nov 24, 2014
4
0
Find I0 using Loop analysis
I have upload a picture for the problem

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4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I'm still not seeing a question.

I'm seeing a copy of YOUR homework problem.

I'm seeing that it is a miniproject that YOU have been assigned.

I'm seeing that YOUR solution is due in a few days.

I'm seeing that YOU are supposed to use Loop analysis.

I'm seeing that YOU are supposed to find I0.

What I am NOT seeing are:

Just what are you expecting, for someone to work the problem and hand you the solution? This is the "Homework Help" forum, not the "Homework Done For You" forum.

5. ### xr329 Thread Starter New Member

Nov 24, 2014
4
0
I am sorry I am new on this this
I have attempt and what I get these equations for the left top
loop 1
-6<0+I1(1-j1)-I2=0

loop 2
12<0+I2(1-j1)+j1I1-I3=0

loop 3
3I3-I2=0

and the three loops on the bottom I don't know how to deal with them because the 2 current source

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
You need to define all of your various currents. I don't know which loop corresponds to I1, I2, and I3. It is also helpful to indicate whether you are summing up voltage drops or voltage gains and whether you are doing so in the direction of the currents or in the opposite direction. Don't make us (or the grader) guess. Because as soon as you make a mistake, then the reader is lost because they don't know if you make a mistake or they just didn't guess correctly as to what you intended in the first place.

You need to learn to track your units. Had you don't that, you would have seen that all three of your equations are fundamentally wrong because they are dimensionally inconsistent.

The first equation, as you have developed it, should have been:

-6<0V + I1(1Ω-j1Ω) - I2 = 0

Remember that Loop analysis is nothing more than applying KVL, which means that you are summing up a bunch of voltages. So each term must resolve to a voltage. Your last term is just a current, which can't be added to a voltage. Now, I know that you just dropped the resistance since it was '1' (but it's not, it is 1Ω), and so maybe you would end up with a numerically correct answer at the end, but what you are giving up in being sloppy with your units is the ability to take advantage of perhaps the single greatest error detection technique in most fields of engineering since most mistakes you make will mess up the units and you can quickly catch and correct those mistakes IF the units are there to be messed up. So, your first equations should have been written along the lines of:

-6<0V + I1(1Ω-j1Ω) - I2(1Ω) = 0

Now, assuming that I1 is a current in the upper-left mesh (simple loop) and that I2 is in the upper-middle loop, and assuming that your currents are all flowing clockwise, and assuming that you are summing up the voltage drops in the direction of the current (see all of the assumptions I'm having to make just to understand your first equation?) do you see the problem? I1 and I2 share the capacitor, not the resistor.

As for the problems in dealing with the current sources, this is a difficulty in apply KVL to circuits containing current sources (and similar problems arise when applying KCL to circuits with voltage sources). But since circuit can and do have both voltage and current sources, we have to be able to deal with them somehow.

I'm assuming that you are just learning generic loop voltage analysis and have not gotten to mesh current analysis. But I don't know if that is a good assumption or not since you didn't define what, exactly, your I1, I2, and I3 currents were (see what happens when you make people guess and try to read your mind?). The basic idea that you want to use, with either loop voltage analysis or mesh current analysis, is to use loops that bypass the troublesome current sources (these are called supermeshes in mesh current analysis) and then determine the constraint equations (also known as auxiliary equations) imposed by the current sources.

So, with those thoughts, go read your text and, in particular, any examples they work involving loop analysis on circuits that have current sources, and make another attempt.

7. ### xr329 Thread Starter New Member

Nov 24, 2014
4
0
I get stuck in the three loops on the bottom
I use supermesh but what about the middle loop I5 that have 2 current source

What I get from supermeshe from the bottom left I choose I4 , I5, I6

I4(1Ω)+6<0-12<0+1Ω(I6-I3)+I6(j1Ω)=0

8. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I don't think you do get the point because you are still requiring us to guess what YOU have defined I4, I5, and I6 to be.

DRAW THE DIAGRAM!

If you draw the diagram, then much of this becomes very obvious. Without it, we are stuck using words to visualize something that we may or may not be in agreement on.

What is so hard about annotating a diagram? Just use Paint or any other simple, free drawing program. It took me less than eight minutes to take a screen shot of your original diagram and produce the following:

Note that I have made no effort to guess what you called things or the directions you used, because you haven't defined them and I don't feel like guessing. So my equations won't necessarily match yours and your answers for your currents may well be off by a minus sign compared to mine. That's because the answers are referred to the definitions of the variables and that is why it is important to unambiguously define all of your variables.

Notice that if I talk about I5, then there is no ambiguity about what current I am talking about, what components it goes through, and what direction it flows in. I have also labeled all of the nodes so that we can refer to them unambiguously. I have also shown a supermesh current in green so that we can focus on that.

So let's focus on that.

When using a supermesh, you don't assign it a current designation like you do a mesh current. Instead, you go around the supermesh path and write the supermesh equation (which, remember, is nothing more than KVL) along that path in terms of the defined mesh currents. Summing the voltage drops in the direction of the currents, the equation we get going around the supermesh is:

I4(1Ω) + I6(j1Ω) + (I6-I3)(1Ω) + (12V∠0) - (6V∠0) = 0

Notice that our supermesh current varies as we go around the supermesh. Sometimes it's I4, sometimes it's I5, and sometimes it's I6. That's why we can't just define a simple current as the supermesh current. But the current sources force conditions between these three currents that have to be met. What relationship between I4 and I5 does the 2A source impose? What relationship between I5 and I6 does the 4A source impose?