# Determining current I_o in this AC circuit

#### salhi

Joined Nov 29, 2023
61
Hello everyone!
Im trying to determine the best way to analyse this circuitry to extract the value of $$I_o$$
So what i think from my basic point of view is using KCL/KVL in each mesh which will be tedious but will do the job just fine i guess?, but im looking for a better optimal less-time consumming approach, infinite thanks to everyone!

#### WBahn

Joined Mar 31, 2012
29,865
Hello everyone!
Im trying to determine the best way to analyse this circuitry to extract the value of $$I_o$$View attachment 308646
So what i think from my basic point of view is using KCL/KVL in each mesh which will be tedious but will do the job just fine i guess?, but im looking for a better optimal less-time consumming approach, infinite thanks to everyone!
That depends on what techniques you already know at this point. If you are dealing with complex impedance and AC sources, then you should have things like mesh current analysis, nodal analysis, and superposition well in hand. Are you familiar with those?

#### salhi

Joined Nov 29, 2023
61
That depends on what techniques you already know at this point. If you are dealing with complex impedance and AC sources, then you should have things like mesh current analysis, nodal analysis, and superposition well in hand. Are you familiar with those?
yes im familiar with them, do you think any of them is useful here time-consumption speaking?

#### WBahn

Joined Mar 31, 2012
29,865
yes im familiar with them, do you think any of them is useful here time-consumption speaking?
Depends on what you consider to be useful time-consuming speaking.

It took me about ten minutes to set up the equations, solve them by hand, and verify the results with no calculator at all.

#### salhi

Joined Nov 29, 2023
61
Depends on what you consider to be useful time-consuming speaking.

It took me about ten minutes to set up the equations, solve them by hand, and verify the results with no calculator at all.
ahah damn i guess its just skill-issue for me can i see the details of your method?

#### WBahn

Joined Mar 31, 2012
29,865
ahah damn i guess its just skill-issue for me can i see the details of your method?
We don't solve your homework for you. You need to show your best effort to get as far as you can, we can then help you overcome the stumbling blocks you encounter.

I'll give you this much: I chose to use mesh analysis because that is generally easier when the circuit contains only voltage sources or when current sources are in the circuit's periphery (since they then solve that mesh's equation directly).

#### salhi

Joined Nov 29, 2023
61
We don't solve your homework for you. You need to show your best effort to get as far as you can, we can then help you overcome the stumbling blocks you encounter.

I'll give you this much: I chose to use mesh analysis because that is generally easier when the circuit contains only voltage sources or when current sources are in the circuit's periphery (since they then solve that mesh's equation directly).
okey here is what i did first i wrote the KCL and KVL for mesh 1,2,3 respectively:
$$i_3 = i_1 + I_o, -i_3 = i_2 +I_x, -4I_x = i_1 + i_2 , i_2 = -V_x ; \\ -4(0°) +I_o \cdot 1 \Omega + 2V_x =0; i_3 \cdot j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0 ; -V_x - i_3 \cdot j \Omega + i_1 \cdot 1 \Omega = 0$$

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#### WBahn

Joined Mar 31, 2012
29,865
okey here is what i did first i wrote the KCL and KVL for mesh 1,2,3 respectively:
$$i_3 = i_1 + I_o, -i_3 = i_2 +I_x, -4I_x = i_1 + i_2 , i_2 = -V_x ; \\ -4(0°) +I_o \cdot 1 \Omega + 2V_x =0; i_3 \cdot j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0 ; -V_x - i_3 \cdot j \Omega + i_1 \cdot 1 \Omega = 0$$
You're not doing mesh current analysis, you are doing branch current analysis.

Mesh current analysis is a systematic way of applying KVL such that KCL is automatically satisfied and doesn't have to be considered. Similarly, node voltage analysis is a systematic way of applying KCL such that KVL is automatically satisfied.

But let's set that aside for the moment and consider the equations you have given thus far.

$$i_3 = i_1 + I_o$$

Look carefully. How many branches connect to the center node? How many of those branches are reflected in this equation.

Remember, it is critical that you get the set-up equations right. That is where ALL of the EE is at. Everything after that is just algebra. If the setup equations are wrong, all the perfect math skills in the world can't save you, because you are finding the solution to a different problem. So it really pays off to very carefully establish your setup equations, and then go back through them and carefully confirm that you actually believe that each one is correct. Approach it from the perspective of an employee giving you their setup equations for you to review and if you let any errors slip by, you are the one that gets fired. It's that critical that this stage be done right.

$$-i_3 = i_2 +I_x \\ -4I_x = i_1 + i_2 \\ i_2 = -V_x$$

In the vein of making your setup equations both easier to accurately generate and easier to efficiently and accurately verify, write them in a way that directly matches the definitions on the schematic by doing "current out equals current in." This also eliminates nearly all of the minus signs, which humans get tripped up on very easily. So I would recommend

$$i_2 + I_x + i_3 = 0\\ i_1 + i_2 + 4I_x = 0$$

$$i_2 = -V_x$$

At a glance you know that this cannot be correct. You are setting a current equal to a voltage. This is like claiming that a table's length is equal to its color.

Now, I'm guessing that what you are trying to is apply Ohm's Law to the 1 Ω resistor on the left edge. This is not a KCL equation, which you implied was what your first row contained (and KVL equations on the second).

Assuming this is the case, ask your self what direction would current flow in if Vx were positive, say +10 V. Would it flow upward or downward through that resistor? Is that consistent with the polarity your equation would give i_2?

Note: I'm assuming you are using conventional current and not electron current. So please correct me if I'm wrong.

If using conventional current, this equation should be

$$V_x \; = \; i_2 \cdot 1\Omega$$

You need to properly track your units from beginning to the end, letting the math dictate what happens to them just like it dictates what happens to the numbers. Do not just tack whatever units you want onto the final answer. You are going to make mistakes. We all do. Most of those mistakes will mess up the units, allowing you to catch them immediately and correct them. But that can only happen if the units are there so that they can get messed up. Proper units tracking is probably the single strongest error detection tool available to the engineer.

Now let's shift attention to your KVL equations:

First, let me do nothing other than compare the way you presented them to what I would claim is a much better way:

$$-4(0°) +I_o \cdot 1 \Omega + 2V_x =0; i_3 \cdot j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0 ; -V_x - i_3 \cdot j \Omega + i_1 \cdot 1 \Omega = 0$$

$$-4(0°) +I_o \cdot 1 \Omega + 2V_x =0\\ i_3 \cdot j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0\\ -V_x - i_3 \cdot j \Omega + i_1 \cdot 1 \Omega = 0$$

Note that all I did was put each equation on a separate line. If you were grading or reviewing this work, which format would you rather be looking at.

This is not a trivial point -- you want to do everything you can to bias the reader of your work in your favor. It doesn't matter if that reader is a grader, a supervisor, or a customer. Well presented work will make them want to look as favorably on your work as they can. It's human nature.

These equations all look correct -- and you properly dealt with the units -- but it took a bit of head scratching to see that because you did some simplification as you went. I would recommend, again, making the setup equations directly reflect the schematic as close as possible. So perhaps:

$$-4\angle 0° + I_o \cdot 1 \Omega + 2V_x = 0 \\ -i_3 \cdot -j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0\\ -V_x + i_3 \cdot -j \Omega + i_1 \cdot 1 \Omega = 0$$

As a personal preference -- because it tends to eliminate minus signs -- I generally write loop equations as the sum of the voltage drops across loads equals the sum of the voltage gains across sources. This yields

$$I_o \cdot 1 \Omega + 2V_x = -4\angle 0° \\ -i_3 \cdot -j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0\\ i_3 \cdot -j \Omega + i_1 \cdot 1 \Omega = -V_x$$

I treated V_x as a source since the information we are using is presented the same way as a source voltage would be. That's a judgement call.

Finally, there's the question of whether or not you have enough equations to solve the problem -- and whether or not those equations are linearly independent. This is the real downside of directly applying KVL and KCL to anything other than very simple circuits -- the number of equations goes up dramatically and it becomes tricky to choose a set that is linearly independent.

If we look at the circuit we see that there are eight branch currents and five node voltages. That means we have thirteen unknowns. We get to arbitrarily assign a voltage to one of the nodes, so that leaves us with twelve. We also have Vx as an unknown, so we are back to needing thirteen equations. You've given seven. You need six more.

By applying KVL around the meshes, you've ensured that they are linearly independent. But you have one mesh that you haven't done this for (the lower-left mesh). So that's one missing equation.

Then you need to apply KCL at every "essential" node (i.e., where more than two branches connect) except one (because that one is a linear combination of the others). In your circuit you have five essential nodes, so you need four KCL equations. You provided three, so you are missing one. The four remaining equations come from applying Ohm's Law to the four resistors. This is actually embedded in the equations you provided, so they aren't missing. You need one more KVL and one more KCL equation.

Okay, so that's a critique on the approach you started. I would recommend that you take a step back and look at your text about doing mesh current analysis. You have four meshes, so define four mesh currents. Then write KVL around each mesh in terms of those currents. That gives you four equations that are normally sufficient to solve the problem. With two dependent sources, you need two additional equations giving the control signals in terms of the the mesh currents. Since one of your mesh currents is trivial, you end up with four equations (which immediately reduces to three) to solve.

Last edited:

#### salhi

Joined Nov 29, 2023
61
You're not doing mesh current analysis, you are doing branch current analysis.

Mesh current analysis is a systematic way of applying KVL such that KCL is automatically satisfied and doesn't have to be considered. Similarly, node voltage analysis is a systematic way of applying KCL such that KVL is automatically satisfied.

But let's set that aside for the moment and consider the equations you have given thus far.

$$i_3 = i_1 + I_o$$

Look carefully. How many branches connect to the center node? How many of those branches are reflected in this equation.

Remember, it is critical that you get the set-up equations right. That is where ALL of the EE is at. Everything after that is just algebra. If the setup equations are wrong, all the perfect math skills in the world can't save you, because you are finding the solution to a different problem. So it really pays off to very carefully establish your setup equations, and then go back through them and carefully confirm that you actually believe that each one is correct. Approach it from the perspective of an employee giving you their setup equations for you to review and if you let any errors slip by, you are the one that gets fired. It's that critical that this stage be done right.

$$-i_3 = i_2 +I_x \\ -4I_x = i_1 + i_2 \\ i_2 = -V_x$$

In the vein of making your setup equations both easier to accurately generate and easier to efficiently and accurately verify, write them in a way that directly matches the definitions on the schematic by doing "current out equals current in." This also eliminates nearly all of the minus signs, which humans get tripped up on very easily. So I would recommend

$$i_2 + I_x + i_3 = 0\\ i_1 + i_2 + 4I_x = 0$$

$$i_2 = -V_x$$

At a glance you know that this cannot be correct. You are setting a current equal to a voltage. This is like claiming that a table's length is equal to its color.

Now, I'm guessing that what you are trying to is apply Ohm's Law to the 1 Ω resistor on the left edge. This is not a KCL equation, which you implied was what your first row contained (and KVL equations on the second).

Assuming this is the case, ask your self what direction would current flow in if Vx were positive, say +10 V. Would it flow upward or downward through that resistor? Is that consistent with the polarity your equation would give i_2?

Note: I'm assuming you are using conventional current and not electron current. So please correct me if I'm wrong.

If using conventional current, this equation should be

$$V_x \; = \; i_2 \cdot 1\Omega$$

You need to properly track your units from beginning to the end, letting the math dictate what happens to them just like it dictates what happens to the numbers. Do not just tack whatever units you want onto the final answer. You are going to make mistakes. We all do. Most of those mistakes will mess up the units, allowing you to catch them immediately and correct them. But that can only happen if the units are there so that they can get messed up. Proper units tracking is probably the single strongest error detection tool available to the engineer.

Now let's shift attention to your KVL equations:

First, let me do nothing other than compare the way you presented them to what I would claim is a much better way:

$$-4(0°) +I_o \cdot 1 \Omega + 2V_x =0; i_3 \cdot j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0 ; -V_x - i_3 \cdot j \Omega + i_1 \cdot 1 \Omega = 0$$

$$-4(0°) +I_o \cdot 1 \Omega + 2V_x =0\\ i_3 \cdot j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0\\ -V_x - i_3 \cdot j \Omega + i_1 \cdot 1 \Omega = 0$$

Note that all I did was put each equation on a separate line. If you were grading or reviewing this work, which format would you rather be looking at.

This is not a trivial point -- you want to do everything you can to bias the reader of your work in your favor. It doesn't matter if that reader is a grader, a supervisor, or a customer. Well presented work will make them want to look as favorably on your work as they can. It's human nature.

These equations all look correct -- and you properly dealt with the units -- but it took a bit of head scratching to see that because you did some simplification as you went. I would recommend, again, making the setup equations directly reflect the schematic as close as possible. So perhaps:

$$-4\angle 0° + I_o \cdot 1 \Omega + 2V_x = 0 \\ -i_3 \cdot -j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0\\ -V_x + i_3 \cdot -j \Omega + i_1 \cdot 1 \Omega = 0$$

As a personal preference -- because it tends to eliminate minus signs -- I generally write loop equations as the sum of the voltage drops across loads equals the sum of the voltage gains across sources. This yields

$$I_o \cdot 1 \Omega + 2V_x = -4\angle 0° \\ -i_3 \cdot -j \Omega + I_x \cdot 1 \Omega - I_o \cdot 1 \Omega = 0\\ i_3 \cdot -j \Omega + i_1 \cdot 1 \Omega = -V_x$$

I treated V_x as a source since the information we are using is presented the same way as a source voltage would be. That's a judgement call.

Finally, there's the question of whether or not you have enough equations to solve the problem -- and whether or not those equations are linearly independent. This is the real downside of directly applying KVL and KCL to anything other than very simple circuits -- the number of equations goes up dramatically and it becomes tricky to choose a set that is linearly independent.

If we look at the circuit we see that there are eight branch currents and five node voltages. That means we have thirteen unknowns. We get to arbitrarily assign a voltage to one of the nodes, so that leaves us with twelve. We also have Vx as an unknown, so we are back to needing thirteen equations. You've given seven. You need six more.

By applying KVL around the meshes, you've ensured that they are linearly independent. But you have one mesh that you haven't done this for (the lower-left mesh). So that's one missing equation.

Then you need to apply KCL at every "essential" node (i.e., where more than two branches connect) except one (because that one is a linear combination of the others). In your circuit you have five essential nodes, so you need four KCL equations. You provided three, so you are missing one. The four remaining equations come from applying Ohm's Law to the four resistors. This is actually embedded in the equations you provided, so they aren't missing. You need one more KVL and one more KCL equation.

Okay, so that's a critique on the approach you started. I would recommend that you take a step back and look at your text about doing mesh current analysis. You have four meshes, so define four mesh currents. Then write KVL around each mesh in terms of those currents. That gives you four equations that are normally sufficient to solve the problem. With two dependent sources, you need two additional equations giving the control signals in terms of the the mesh currents. Since one of your mesh currents is trivial, you end up with four equations (which immediately reduces to three) to solve.
omg , infinite thanks sir, im deeply grateful for what you wrote, alot of effort

#### salhi

Joined Nov 29, 2023
61
i would like to know what should be the correct equation instead of this $$i_3 = i_1 + I_o$$ is the current outputed by the independent 4V voltage source considered?

#### salhi

Joined Nov 29, 2023
61
omg , infinite thanks sir, im deeply grateful for what you wrote, alot of effort
also what do you mean by mesh analysis, because im studying in french , do you mean i analyse each mesh independently of the other? isnt that same thing as KVL/KCL in each node/mesh?

#### WBahn

Joined Mar 31, 2012
29,865
i would like to know what should be the correct equation instead of this $$i_3 = i_1 + I_o$$ is the current outputed by the independent 4V voltage source considered?
Think back to what KCL is. It is nothing more than a statement of conservation of charge under a specific condition, which is that if charge can't be stored on a node, then the sum of the charge entering a node must always be zero, which translates into the current entering a node must sum to zero.

If current is flowing in the 4 V source, then that current is either flowing into or out of the center node and must therefore be included in the sum of the current entering that node.

#### WBahn

Joined Mar 31, 2012
29,865

#### salhi

Joined Nov 29, 2023
61
After doing the mesh analysis method, here are the equations i came up with:
In first mesh (bottom left)
$$4I_x = i_1 \\ 4(0°) + (4I_x - i_2) \cdot 1 \Omega = 0 \\$$
In second mesh (upper left)
$$-V_x - j \Omega \cdot (i_2 - I_x) + 1 \Omega \cdot (i_2 - 4I_x) = 0 \\$$
In third mesh (upper right)
$$i_3 = I_x \\ (i_3-i_2)\cdot (-j \Omega) + I_x \cdot 1 \Omega + (I_x - i_4) \cdot 1 \Omega =0$$
In fourth mesh (lower right)
$$-4(0°) +2 V_x + 1 \Omega \cdot (i_4 - I_x) = 0$$
is this valid, is this enough to solve the problem?

#### The Electrician

Joined Oct 9, 2007
2,969
After doing the mesh analysis method, here are the equations i came up with: View attachment 309022
In first mesh (bottom left)
$$4I_x = i_1 \\ 4(0°) + (4I_x - i_2) \cdot 1 \Omega = 0 \\$$
In second mesh (upper left)
$$-V_x - j \Omega \cdot (i_2 - I_x) + 1 \Omega \cdot (i_2 - 4I_x) = 0 \\$$
In third mesh (upper right)
$$i_3 = I_x \\ (i_3-i_2)\cdot (-j \Omega) + I_x \cdot 1 \Omega + (I_x - i_4) \cdot 1 \Omega =0$$
In fourth mesh (lower right)
$$-4(0°) +2 V_x + 1 \Omega \cdot (i_4 - I_x) = 0$$
is this valid, is this enough to solve the problem?
What do you get when you solve your equations? If you substitute your solution values back into the circuit, do they work?

#### WBahn

Joined Mar 31, 2012
29,865
After doing the mesh analysis method, here are the equations i came up with: View attachment 309022
In first mesh (bottom left)
$$4I_x = i_1 \\ 4(0°) + (4I_x - i_2) \cdot 1 \Omega = 0 \\$$
Your first equation is all you need. The goal is to solve for I1 and you did that, it is 4Ix because there is a current source that defines I1 to be that.

The KVL equation is not needed, but if you are going to provide it, it needs to be correct. Remember, KVL says that the sum of the voltage drops across ALL components around a loop must sum to zero. Not just the ones that are convenient to write equations for. So you need to define a voltage for the voltage across the current source. Call it whatever you like. Vy sounds good, and then use that in your equation. You've now exposed one more unknown that you will need an equation for.

In second mesh (upper left)
$$-V_x - j \Omega \cdot (i_2 - I_x) + 1 \Omega \cdot (i_2 - 4I_x) = 0 \\$$
In third mesh (upper right)
$$i_3 = I_x \\ (i_3-i_2)\cdot (-j \Omega) + I_x \cdot 1 \Omega + (I_x - i_4) \cdot 1 \Omega =0$$
In fourth mesh (lower right)
$$-4(0°) +2 V_x + 1 \Omega \cdot (i_4 - I_x) = 0$$
is this valid, is this enough to solve the problem?
You almost have enough equations. You have two dependent sources, and so you need two additional equations -- one for Ix and one for Vx. You've taken care of Ix, but now you need one for Vx.

I would STRONGLY recommend that you do the absolute minimum of processing when writing your set up equations. Make it blindingly obvious that they match the circuit diagram. Remember, mistakes at this point usually cannot be caught until after you have a solution, and then only if you actually back verify it against the problem (something you should always do). If your setup equations are wrong, you are simply solving a different problem. The math doesn't care.

So write your setup equations and then verify them very, very carefully before proceeding.

Here's how I would have written my setup equations for this problem:

Mesh Equations:
$$1: \;\; I_1 \; = \; 4I_x \\ 2: \;\; (-V_x) \; + \; (I_2 - I_3)(-j1 \; \Omega) \; + \; (I_2 - I_1)(1 \; \Omega) \; = \; 0 \\ 3: \;\; (I_3 - I_2)(-j1 \; \Omega) \; + \; (I_3)(1 \; \Omega) \; + \; (I_3 - I_4)(1 \; \Omega) \; = \; 0\\ 4: \;\; (-4 \angle 0^\circ \; V) \; + \; (I_4 - I_3)(1 \; \Omega) \; + \; (2V_x)$$

Auxiliary Equations:
$$I_x \; = \; I_3 \\ V_x \; = \; (-I_2)(1 \; \Omega)$$

After writing these down, I reviewed them and spotted two mistakes. In the equation for mesh 2, I had the final term as (I2 - I4), and in the equation for mesh 3 I left out the impedance from the first term, which messed up the units because now I had a current being added to a couple of voltages. We ALL make mistakes on a pretty routine basis -- it really pays to adopt systematic ways of doing things that maximize our ability to catch them and do so early.

This is not QUITE how I typically write the equations -- I do take a shortcut and write them in a form that lends itself to simultaneous equations via matrix manipulation. With just a bit of practice, it is easy to write down (and verify) equations in that form. But this is appropriate for the level you are at.

Notice that one thing that I did was I wrote my KVL equations around each mesh starting at the bottom-left point in each mesh and then proceeding clockwise around the mesh. This ensures that I include everything, because I'm not done until I get back to my starting point. It also makes it much easier to verify that the equations are correct before I proceed.