Stuck on this homework problem:

So far I've used mesh current analysis to create the three equations (treating \( V_x \) as a voltage source that creates three meshes):
A: \( -v_{x}+(1.2k)i_{A}+v_{x}=0 \)
B: \( -v_{x}+R_{x}i_{B}+10^{3}v_{x}=0 \)
C: \( 10^{3}v_{x}+(1.6k)i_{C}=0 \)

\( i_{C} \) is the same as \( i_{0} \), so I was able to find an expression for \( i_{0} \) in terms of \( v_{x} \) using equation C:
\( i_{0}=\frac{-10^{3}v_{x}}{1.6k}=\frac{-v_{x}}{1.6} \)

I then used equations A and B to create the equation:
\( v_{s}=(1.2k)i_{A}+R_{x}i_{B} + 10^{3}v_{x} \)

It is here where I think that I'm stuck, as I don't know how to find equivalencies for \( i_{A} \) and \( i_{B} \) that will give me a final gain \( \frac{i_{0}}{v_{s}} \) that is only in terms of \( R_{x} \). I think that there is a way to relate \( i_{A} \) and \( i_{B} \) by making some current \( i_{x}=i_{A}-i_{B} \), but I don't see how to actually proceed.

 

Welcome to AAC and thank you for showing your work and explaining where you are getting stuck. That is SO refreshing!

I've never seen the approach of treating Vx as a source and creating an additional mesh. I can definitely see how it might make things cleaner and is clearly a valid approach, but I'll have to play with things both ways to see if I think it actually makes things easier.

Also, you have an error in your equation for loop A. It should be -Vs and not -Vx as the first term. Possibly just a typo.

You need to track your units properly -- especially when you are fortunate to work with a book that appears to be paying attention to them. If an equation is not dimensionally sound, you know something is wrong. Most mistakes we make, whether conceptual or just silly math/algebra goofs, will mess up the units, provided the units are there to be messed up.

You your mesh equation for A, you have a voltage plus a current plus a voltage is equal to zero. You can't add voltages and currents together -- it makes no sense. So something is wrong. Your middle term should yield a voltage because you are multiplying a resistance by a current. But, as you set it up, your are just multiplying a current by 1200, giving a current that is just a lot bigger.

One thing to note is that you know that the current in the two meshes formed this way are the same, since they have to cancel out to zero current between the points where Vx is being measured. So Ia = Ib = Is.

Taking your equations and making just these changes, we get:

A: \( -v_{s}+(1.2 \; k \Omega)i_{s}+v_{x}=0 \)
B: \( -v_{x}+R_{x}i_{s}+10^{3}v_{x}=0 \)
C: \( 10^{3}v_{x}+(1.6 \; k \Omega)i_{0}=0 \)

As a quick sanity check, add all three equations together and you should get a loop equation for the perimeter of the circuit. If not, you know something is wrong and there is no point going forward until it is resolved.

\( -v_{s}+(1.2 \; k \Omega)i_{s}+R_{x}i_{s}+10^{3}v_{x}-(1.6 \; k \Omega)i_{0}=0 \)

With just a bit of practice, you won't need to write this down, you'll be able to walk across your meshes that touch the perimeter by inspection.

Do you see the problem? The currents are all flowing clockwise through the resistors, but the corresponding voltage drops are not all positive (or, more specifically, are not all the same sign). So something is wrong. Anything you do beyond this point is guaranteed to produce an incorrect answer, except by a fluke of nature (which I have seen happen).

Here is a key thing to remember. Your set-up equations capture all of the EE stuff about the problem. Everything beyond that is simply math. If you make a mistake in your set up equations, your math can be perfect and it doesn't matter because you are solving a different problem. Furthermore, it is frequently very difficult to catch that a mistake has been made because the problem you are actually solving is probably not that different from the one you think you are solving. So take the time to make your set-up equations as blatantly self-evident as you can and then take a step back and carefully verify that each one of them is correct before you proceed. Mistakes caught at this step are usually trivial to correct and, if we make the conscious effort to look for them, they are usually easy to catch, as well. But mistakes not caught at this stage tend to be very costly.

I'll let you find and fix this issue.

Getting back to units, your next equation has a current on the left but voltages on the right. It should be:

\( i_{0}=\frac{-10^{3}v_{x}}{1.6 \; k \Omega}=\frac{-v_{x}}{1.6 \; \Omega} \)

Many think that getting hung up on tracking units is being pedantic. It's not. Proper units tracking is one of the most powerful error-detection and diagnostic tools available to an engineer.

Now, to be pedantic, in a value consisting of a numerical coefficient and units, there should be a space between them. So it's 1.2 kΩ and not 1.2kΩ. This is not always convenient, but it is the internationally accepted convention (as in agreed to by legally-binding treaties). So I would recommend getting in the habit of making the effort to do it correctly. The results are more professional and, now and then, it will get noticed to your benefit.

Rework things back up to this point and then we can discuss how to proceed.

 

Another thing you might consider is that you do not need to solve Part A in order to do Part B.

You are given that the gain is -100 mS. Pick any convenient values for Vs and Io that satisfy that relationship and then walk your way through the circuit and figure out what Rx has to be.

That is actually a very quick and easy thing to do.

Now you have an answer that you can use to validate your result to Part A.

 

Stuck on this homework problem:
View attachment 344982
So far I've used mesh current analysis to create the three equations (treating \( V_x \) as a voltage source that creates three meshes):
A: \( -v_{x}+(1.2k)i_{A}+v_{x}=0 \)
B: \( -v_{x}+R_{x}i_{B}+10^{3}v_{x}=0 \)
C: \( 10^{3}v_{x}+(1.6k)i_{C}=0 \)

\( i_{C} \) is the same as \( i_{0} \), so I was able to find an expression for \( i_{0} \) in terms of \( v_{x} \) using equation C:
\( i_{0}=\frac{-10^{3}v_{x}}{1.6k}=\frac{-v_{x}}{1.6} \)

I then used equations A and B to create the equation:
\( v_{s}=(1.2k)i_{A}+R_{x}i_{B} + 10^{3}v_{x} \)

It is here where I think that I'm stuck, as I don't know how to find equivalencies for \( i_{A} \) and \( i_{B} \) that will give me a final gain \( \frac{i_{0}}{v_{s}} \) that is only in terms of \( R_{x} \). I think that there is a way to relate \( i_{A} \) and \( i_{B} \) by making some current \( i_{x}=i_{A}-i_{B} \), but I don't see how to actually proceed.

Hi,

You can check your work using another method too which is fairly quick.

An interesting feature about this problem is that it looks like the value of Rx comes out to a whole number with no fractional part. It may depend on how you round during your calculations though, if you do round like that. Check and see if you can get a whole number for Rx through the calculations.

I took a snapshot of the original post because the "Reply" function here does not render the equations correctly (probably Latex).

 

Welcome to AAC and thank you for showing your work and explaining where you are getting stuck. That is SO refreshing!

I've never seen the approach of treating Vx as a source and creating an additional mesh. I can definitely see how it might make things cleaner and is clearly a valid approach, but I'll have to play with things both ways to see if I think it actually makes things easier.

Also, you have an error in your equation for loop A. It should be -Vs and not -Vx as the first term. Possibly just a typo.

You need to track your units properly -- especially when you are fortunate to work with a book that appears to be paying attention to them. If an equation is not dimensionally sound, you know something is wrong. Most mistakes we make, whether conceptual or just silly math/algebra goofs, will mess up the units, provided the units are there to be messed up.

You your mesh equation for A, you have a voltage plus a current plus a voltage is equal to zero. You can't add voltages and currents together -- it makes no sense. So something is wrong. Your middle term should yield a voltage because you are multiplying a resistance by a current. But, as you set it up, your are just multiplying a current by 1200, giving a current that is just a lot bigger.

One thing to note is that you know that the current in the two meshes formed this way are the same, since they have to cancel out to zero current between the points where Vx is being measured. So Ia = Ib = Is.

Taking your equations and making just these changes, we get:

A: \( -v_{s}+(1.2 \; k \Omega)i_{s}+v_{x}=0 \)
B: \( -v_{x}+R_{x}i_{s}+10^{3}v_{x}=0 \)
C: \( 10^{3}v_{x}+(1.6 \; k \Omega)i_{0}=0 \)

As a quick sanity check, add all three equations together and you should get a loop equation for the perimeter of the circuit. If not, you know something is wrong and there is no point going forward until it is resolved.

\( -v_{s}+(1.2 \; k \Omega)i_{s}+R_{x}i_{s}+10^{3}v_{x}-(1.6 \; k \Omega)i_{0}=0 \)

With just a bit of practice, you won't need to write this down, you'll be able to walk across your meshes that touch the perimeter by inspection.

Do you see the problem? The currents are all flowing clockwise through the resistors, but the corresponding voltage drops are not all positive (or, more specifically, are not all the same sign). So something is wrong. Anything you do beyond this point is guaranteed to produce an incorrect answer, except by a fluke of nature (which I have seen happen).

Here is a key thing to remember. Your set-up equations capture all of the EE stuff about the problem. Everything beyond that is simply math. If you make a mistake in your set up equations, your math can be perfect and it doesn't matter because you are solving a different problem. Furthermore, it is frequently very difficult to catch that a mistake has been made because the problem you are actually solving is probably not that different from the one you think you are solving. So take the time to make your set-up equations as blatantly self-evident as you can and then take a step back and carefully verify that each one of them is correct before you proceed. Mistakes caught at this step are usually trivial to correct and, if we make the conscious effort to look for them, they are usually easy to catch, as well. But mistakes not caught at this stage tend to be very costly.

I'll let you find and fix this issue.

Getting back to units, your next equation has a current on the left but voltages on the right. It should be:

\( i_{0}=\frac{-10^{3}v_{x}}{1.6 \; k \Omega}=\frac{-v_{x}}{1.6 \; \Omega} \)

Many think that getting hung up on tracking units is being pedantic. It's not. Proper units tracking is one of the most powerful error-detection and diagnostic tools available to an engineer.

Now, to be pedantic, in a value consisting of a numerical coefficient and units, there should be a space between them. So it's 1.2 kΩ and not 1.2kΩ. This is not always convenient, but it is the internationally accepted convention (as in agreed to by legally-binding treaties). So I would recommend getting in the habit of making the effort to do it correctly. The results are more professional and, now and then, it will get noticed to your benefit.

Rework things back up to this point and then we can discuss how to proceed.

Hello and thanks for the welcome; I'm definitely glad to be here and not paying for Chegg or consulting faulty AI to find answers.

Using \( -V_x \) in loop A was indeed a typo that luckily wasn't reflected in my work, just in my post. However, your notes about the voltage signs not matching up were valid and I've since revised my equations.

I appreciate the discussion about using units in my equations and I'll be doing so going forwards in my classes (including the spacing of the values and the units; I've pointed out this error in other people's work before, but I tend to ignore it when doing written work like this), but I don't think that excluding them was going to have a real impact on my final result assuming that I do everything properly. As you point out, including units helps me keep track of where certain numbers come from and provide an extra set of values that I can use to check for correctness (and is therefore worth doing).

The main help you provided was definitely reminding me that \( i_A \) and \( i_B \) would have the same value, since the only reason they are separate is that I conjured a connection where none really existed. Realizing this allowed me to go forwards after revisiting my existing work and find my final expression.

Revising my work, (almost) the same three equations appear:
A: \( -V_s + (1.2 k\Omega) i_A + V_x = 0 \)
B: \( -V_x + R_x i_B - 10^3 V_x = 0 \)
C: \( 10^3 V_x + (1.6 k\Omega) i_C = 0 \)

Using the fact that \( i_C = i_0 \) and \( i_A = i_B = i_s \), these three equations can be rearranged to yield three different ones:
\( i_0 = \frac{-10^3 V_x}{1.6 k\Omega} \)
\( V_x = \frac{R_x i_s}{10^3 + 1} \)
\( V_s = \frac{R_x i_s}{10^3 + 1} + (1.2 k\Omega)i_s \)

I then substituted \( V_x \) into \( i_0 \) before working out \( \frac{i_0}{V_s} \) to eventually yield the expression:
\( \frac{i_0}{V_s} = \frac{-5R_x}{8R_x + 9609600} \)

Part B was much nicer, where I just substituted \( \frac{I_0}{V_s} \) for -100 mS and solved for \( R_x \). I ended up getting \( R_x \) = 2288 \( k\Omega \) (which is a whole number, as MrAI implied it would be).

Thanks a lot for the help WBahn and MrAI, and I'm open to more feedback/advice if you guys see any more errors with my work.

 

Get in the habit of reviewing EVERY line of your work and asking if it is dimensionally consistent. This is how it will become second nature and how you will quickly reach the point where dimensional inconsistencies will jump out at you with big red flags.

Is this equation dimensionally consistent?

On the left hand side you have a current divided by a voltage, which gives units of conductance (or, equivalently, reciprocal resistance).

On the right you have resistance on the top, and on the bottom you have resistance for the first term and just a number for the second term. So there are multiple dimensional inconsistencies that you ignored in getting to this point, which also means that when you solved for Rx and came up with a solution, you just tacked on the units that you believed and wanted the answer to have without making the effort to determine what the units actually are.

Disaster awaits you down that road. This is how airliners full of people run out of fuel in midflight, or space probes get slammed into planets instead of going into orbit. I stood next to a person as they were killed because they couldn't be bothered to track their units through their work (as the accident investigation later revealed).

Part B was much nicer, where I just substituted \( \frac{I_0}{V_s} \) for -100 mS and solved for \( R_x \). I ended up getting \( R_x \) = 2288 \( k\Omega \) (which is a whole number, as MrAI implied it would be).

Have you verified that 2288 kΩ is actually a solution to the problem?

One of the very nice things about most engineering problems, even in the real world, is that the correctness of the answer can be determined from the problem itself.

So let's validate this answer.



We can do this a number of ways. One is to pick a specific value for either Vs or Io and then analyze the circuit and determine the other and then see if they are consistent with the given gain of -0.1 mS. Another is to assume that the circuit has the desired gain and then determine the voltages and see if the required relationships imposed by the dependent source are satisfied.

Let's assume that Io is 1 mA, which the results, by inspection, in

Vc = (1.2 kΩ + 400 Ω)(1 mA) = 1.6 V

Since Vc is also the negative of the voltage across the dependent source:

Vc = (0 V - 1000·Vx) = -1000·Vx

Vx = -Vc / 1000 = -1.6 mV

This is also Vb, by definition.

We now have the voltage across Rx, which allows us to determine Is

Is = (Vb - Vc) / (2288 kΩ) = (-1.6 mV - 1.6 V) / (1.2 kΩ) = (-1.6016 V) / (2288 kΩ) = -700 nA

Knowing Is, we can determine Vs

Vs = Vb + Is·(1.2 kΩ) = -1.6 mV + (-700 nA)(1.2 kΩ) = -1.6 mV - 0.84 mV = -2.44 mV

Buy we can also determine Vs using the specified gain of -100 mS.

Io/Vs = -100 mS

Vs = Io / (-100 mS) = 1 mA / (-100 mS) = -10 mV

There is a major mismatch between these two, so assuming that I didn't make a mistake in doing the validation check (always a possibility), the answer of 2288 kΩ is wrong.

 

With io/Vs = -0.1 S:

-0.8Rx - 960960 = -5Rx
4.2Rx = 960960
Rx = 228.8 k

 

With io/Vs = -0.1 S:

-0.8Rx - 960960 = -5Rx
4.2Rx = 960960
Rx = 228.8 k

View attachment 345078

Please don't just provide answers/solutions in Homework Help. That is not being of help in the long run. Instead, give hints and suggestions aimed at helping the TS identify their mistakes or to get past their immediate stumbling block. The more of the work they struggle through on their own, the more they will learn and the better it will stick.

 

Get in the habit of reviewing EVERY line of your work and asking if it is dimensionally consistent. This is how it will become second nature and how you will quickly reach the point where dimensional inconsistencies will jump out at you with big red flags.



Is this equation dimensionally consistent?

On the left hand side you have a current divided by a voltage, which gives units of conductance (or, equivalently, reciprocal resistance).

On the right you have resistance on the top, and on the bottom you have resistance for the first term and just a number for the second term. So there are multiple dimensional inconsistencies that you ignored in getting to this point, which also means that when you solved for Rx and came up with a solution, you just tacked on the units that you believed and wanted the answer to have without making the effort to determine what the units actually are.

Disaster awaits you down that road. This is how airliners full of people run out of fuel in midflight, or space probes get slammed into planets instead of going into orbit. I stood next to a person as they were killed because they couldn't be bothered to track their units through their work (as the accident investigation later revealed).



Have you verified that 2288 kΩ is actually a solution to the problem?

One of the very nice things about most engineering problems, even in the real world, is that the correctness of the answer can be determined from the problem itself.

So let's validate this answer.

View attachment 345067

We can do this a number of ways. One is to pick a specific value for either Vs or Io and then analyze the circuit and determine the other and then see if they are consistent with the given gain of -0.1 mS. Another is to assume that the circuit has the desired gain and then determine the voltages and see if the required relationships imposed by the dependent source are satisfied.

Let's assume that Io is 1 mA, which the results, by inspection, in

Vc = (1.2 kΩ + 400 Ω)(1 mA) = 1.6 V

Since Vc is also the negative of the voltage across the dependent source:

Vc = (0 V - 1000·Vx) = -1000·Vx

Vx = -Vc / 1000 = -1.6 mV

This is also Vb, by definition.

We now have the voltage across Rx, which allows us to determine Is

Is = (Vb - Vc) / (2288 kΩ) = (-1.6 mV - 1.6 V) / (1.2 kΩ) = (-1.6016 V) / (2288 kΩ) = -700 nA

Knowing Is, we can determine Vs

Vs = Vb + Is·(1.2 kΩ) = -1.6 mV + (-700 nA)(1.2 kΩ) = -1.6 mV - 0.84 mV = -2.44 mV

Buy we can also determine Vs using the specified gain of -100 mS.

Io/Vs = -100 mS

Vs = Io / (-100 mS) = 1 mA / (-100 mS) = -10 mV

There is a major mismatch between these two, so assuming that I didn't make a mistake in doing the validation check (always a possibility), the answer of 2288 kΩ is wrong.

I've tried the math again and got nearly the same result as before while making sure to be mindful of units through every step. I got a bit lost when I started evaluating \( \frac{i_0}{V_s} \), but after I removed the current from the expression all that was left were Ω or S. In the end I got the expression \( \frac{i_0}{V_s} = \frac{-5}{8} \frac{R_x}{R_x+1201200 \Omega} \), which makes sense.

Using the new expression for part B showed that R_x = 228.8 kΩ (as 0ri0n noted), and by using the same method you did for validating the answer it works out.

Full picture of what I'm pretty sure is the correct work:


Thank yall for all of the help with this; it's definitely cleared things up.

 

I've tried the math again and got nearly the same result as before while making sure to be mindful of units through every step. I got a bit lost when I started evaluating \( \frac{i_0}{V_s} \), but after I removed the current from the expression all that was left were Ω or S. In the end I got the expression \( \frac{i_0}{V_s} = \frac{-5}{8} \frac{R_x}{R_x+1201200 \Omega} \), which makes sense.

Using the new expression for part B showed that R_x = 228.8 kΩ (as 0ri0n noted), and by using the same method you did for validating the answer it works out.

Rx = 228.8 kΩ is the correct answer.

But your expression is still not dimensionally consistent.



The left side has units of current over voltage (i.e., amps/volt, which is siemens). This is the same as the reciprocal of resistance, or amp/volt, which is ohms. In fact, the old unit was the 'mho', which is ohm spelled backwards and the symbol was an upside down capital Omega. It was renamed in 1935 by the IEC and internationally recognized in 1971 as a derived unit.

Many people (myself included) don't work with siemens enough to really get comfortable with it, so we replace it with either 1/Ω or even the upside down omega when doing work.

Now, look at your right-hand side. You have a multiplier out front that is -5/8. No units, it's a pure number. The numerator of the next factor is resistance and so is the denominator, meaning that your result is dimensionless. This does NOT make sense, because you need the right-hand side to yield dimensions consistent with current divided by voltage.



Do you see where you dropped the Ω off the 1.6 kΩ in going from the top line to the bottom line? If you hadn't done that, you would have had resistance over resistance-squared, which is consistent with siemens.

Full picture of what I'm pretty sure is the correct work:

If I were a grader I would not be willing to spend very much time trying to follow work that is so disorganized. It's reasonably legible, but all of these arrows sending the reader to the four winds will make them dizzy. Keep in mind that most graders (and the same will apply to supervisors and customers later on) have a very limited amount of time to review your work. It is in your best interest to make their life as easy as possible. If you have an error and your work is hard to follow, they will give up pretty quickly and just give you a poor score. But if your work is easy to follow, they will be much more likely to work through your work methodically to find your actual error and only take off points according to the severity of the actual mistake. Even more, consider that the reader (grader, supervisor, customer, etc.) is usually so bombarded with so much poorly presented work that they quickly get to the point where they look for ways to punish it -- and by the same token, they tend to look so favorably on well-presented work, because it is such a welcome and refreshing relief from the norm, that they look for ways to reward it as much as possible.

Back before I was a grader (and therefore had no idea of the indecipherable crap that so many people turn in), I submitted an assignment in a class that had a policy of no partial credit. One four-part problem I made silly mistake early on that resulted in two being wrong. When I got it back, the score was 100% with a note from the grader that said that I should have lost half the points on that problem, but that the work was so beautiful that he just couldn't bring himself to take off points for such a small mistake. Once I started grading, I completely understood where he was coming from.

So make your work flow linearly from top to bottom so that, as much as possible, the reader can read it like they would a newspaper article.

As an example of what I mean, attached is a scan of the work I did on the problem.

A couple of thing to note:

I tend to work symbolically as much as possible. I don't like throwing in actual values until the very end. Normally, I would even have replaced the gain of the dependent source with a variable, such as 'g'. I just didn't happen to do so this time. If I had, it would have made for quite a bit less writing.

If you look at the units on every line, you'll see that they work out.

For instance, the gain has units of resistance in the numerator and resistance-squared in the denominator, yielding a result with units of 1/resistance, which is consistent siemens.

The equation for Rx has a factor multiplying R1. Thus the factor has to be dimensionless. The numerator is just a number, while the first term of the denominator has as it's numerator a number and as its denominator the gain (siemens) multiplied by a resistance, which is then dimensionless. The second term is just 1, so dimensionless, so there is no problem subtracting it. The end result is that the factor is, indeed, dimensionless as we require.

 

Hello and thanks for the welcome; I'm definitely glad to be here and not paying for Chegg or consulting faulty AI to find answers.

Using \( -V_x \) in loop A was indeed a typo that luckily wasn't reflected in my work, just in my post. However, your notes about the voltage signs not matching up were valid and I've since revised my equations.

I appreciate the discussion about using units in my equations and I'll be doing so going forwards in my classes (including the spacing of the values and the units; I've pointed out this error in other people's work before, but I tend to ignore it when doing written work like this), but I don't think that excluding them was going to have a real impact on my final result assuming that I do everything properly. As you point out, including units helps me keep track of where certain numbers come from and provide an extra set of values that I can use to check for correctness (and is therefore worth doing).

The main help you provided was definitely reminding me that \( i_A \) and \( i_B \) would have the same value, since the only reason they are separate is that I conjured a connection where none really existed. Realizing this allowed me to go forwards after revisiting my existing work and find my final expression.

Revising my work, (almost) the same three equations appear:
A: \( -V_s + (1.2 k\Omega) i_A + V_x = 0 \)
B: \( -V_x + R_x i_B - 10^3 V_x = 0 \)
C: \( 10^3 V_x + (1.6 k\Omega) i_C = 0 \)

Using the fact that \( i_C = i_0 \) and \( i_A = i_B = i_s \), these three equations can be rearranged to yield three different ones:
\( i_0 = \frac{-10^3 V_x}{1.6 k\Omega} \)
\( V_x = \frac{R_x i_s}{10^3 + 1} \)
\( V_s = \frac{R_x i_s}{10^3 + 1} + (1.2 k\Omega)i_s \)

I then substituted \( V_x \) into \( i_0 \) before working out \( \frac{i_0}{V_s} \) to eventually yield the expression:
\( \frac{i_0}{V_s} = \frac{-5R_x}{8R_x + 9609600} \)

Part B was much nicer, where I just substituted \( \frac{I_0}{V_s} \) for -100 mS and solved for \( R_x \). I ended up getting \( R_x \) = 2288 \( k\Omega \) (which is a whole number, as MrAI implied it would be).

Thanks a lot for the help WBahn and MrAI, and I'm open to more feedback/advice if you guys see any more errors with my work.

Hi,

I think you know now that the result of 2288k is not the right result for the value for Rx. Try again and see if you can get it.

Here is a more intuitive method which can be used to check your results. I've included a drawing of the schematic for reference.

If we label the resistors from left to right R1, Rx, R2, R3 with R3=400 Ohms it makes it easier to follow
the text. See attachment which includes part labels and also labels for the nodes. You can ignore "(R4)" as that was just used to be able to address Rx as a numbered resistor R4 just like the others.

Notice that the current through R1 is:
iR1=(Vs-Vx)/R1

and the current through Rx is:
iRx=(Vx-1000*Vx)/Rx

Since this is a series circuit, the current though both must be the same value. This gives you a place
to start to check your result using a different analysis method.
Also note that everything to the right of the dependent source "1000*Vx" is not needed to get the main
part of the solution, and then once you get that you can easily calculate "io" the current through R2 and R3.

I'll provide more details once you try this if you feel like checking your result using this idea.
Since your original problem called for Mesh analysis you will still have to provide that solution too though.

 

Rx = 228.8 kΩ is the correct answer.

But your expression is still not dimensionally consistent.

View attachment 345091

The left side has units of current over voltage (i.e., amps/volt, which is siemens). This is the same as the reciprocal of resistance, or amp/volt, which is ohms. In fact, the old unit was the 'mho', which is ohm spelled backwards and the symbol was an upside down capital Omega. It was renamed in 1935 by the IEC and internationally recognized in 1971 as a derived unit.

Many people (myself included) don't work with siemens enough to really get comfortable with it, so we replace it with either 1/Ω or even the upside down omega when doing work.

Now, look at your right-hand side. You have a multiplier out front that is -5/8. No units, it's a pure number. The numerator of the next factor is resistance and so is the denominator, meaning that your result is dimensionless. This does NOT make sense, because you need the right-hand side to yield dimensions consistent with current divided by voltage.

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Do you see where you dropped the Ω off the 1.6 kΩ in going from the top line to the bottom line? If you hadn't done that, you would have had resistance over resistance-squared, which is consistent with siemens.



If I were a grader I would not be willing to spend very much time trying to follow work that is so disorganized. It's reasonably legible, but all of these arrows sending the reader to the four winds will make them dizzy. Keep in mind that most graders (and the same will apply to supervisors and customers later on) have a very limited amount of time to review your work. It is in your best interest to make their life as easy as possible. If you have an error and your work is hard to follow, they will give up pretty quickly and just give you a poor score. But if your work is easy to follow, they will be much more likely to work through your work methodically to find your actual error and only take off points according to the severity of the actual mistake. Even more, consider that the reader (grader, supervisor, customer, etc.) is usually so bombarded with so much poorly presented work that they quickly get to the point where they look for ways to punish it -- and by the same token, they tend to look so favorably on well-presented work, because it is such a welcome and refreshing relief from the norm, that they look for ways to reward it as much as possible.

Back before I was a grader (and therefore had no idea of the indecipherable crap that so many people turn in), I submitted an assignment in a class that had a policy of no partial credit. One four-part problem I made silly mistake early on that resulted in two being wrong. When I got it back, the score was 100% with a note from the grader that said that I should have lost half the points on that problem, but that the work was so beautiful that he just couldn't bring himself to take off points for such a small mistake. Once I started grading, I completely understood where he was coming from.

So make your work flow linearly from top to bottom so that, as much as possible, the reader can read it like they would a newspaper article.

As an example of what I mean, attached is a scan of the work I did on the problem.

A couple of thing to note:

I tend to work symbolically as much as possible. I don't like throwing in actual values until the very end. Normally, I would even have replaced the gain of the dependent source with a variable, such as 'g'. I just didn't happen to do so this time. If I had, it would have made for quite a bit less writing.

If you look at the units on every line, you'll see that they work out.

For instance, the gain has units of resistance in the numerator and resistance-squared in the denominator, yielding a result with units of 1/resistance, which is consistent siemens.

The equation for Rx has a factor multiplying R1. Thus the factor has to be dimensionless. The numerator is just a number, while the first term of the denominator has as it's numerator a number and as its denominator the gain (siemens) multiplied by a resistance, which is then dimensionless. The second term is just 1, so dimensionless, so there is no problem subtracting it. The end result is that the factor is, indeed, dimensionless as we require.

I've corrected the expression to account for units I missed, and it now reads as:
\( \frac{i_0}{V_s} = \frac{-5}{8\Omega} \frac{R_x}{R_x + 1201200\Omega} \)

I also recognize the sloppiness of the work I've produced, but frankly for the purposes of this particular assignment the work's legibility is irrelevant as the final answers are all that are looked at for scoring. In fact, once the solutions for this assignment are released it'll be up to me to evaluate my own work for errors, not an instructor/grader. I won't argue that keeping cleaner work isn't something to work on in the future though; at this point I'm just looking to have something submitted in all honesty.

 

I've corrected the expression to account for units I missed, and it now reads as:
\( \frac{i_0}{V_s} = \frac{-5}{8\Omega} \frac{R_x}{R_x + 1201200\Omega} \)

I also recognize the sloppiness of the work I've produced, but frankly for the purposes of this particular assignment the work's legibility is irrelevant as the final answers are all that are looked at for scoring. In fact, once the solutions for this assignment are released it'll be up to me to evaluate my own work for errors, not an instructor/grader. I won't argue that keeping cleaner work isn't something to work on in the future though; at this point I'm just looking to have something submitted in all honesty.

This reflects a common -- and quite natural -- attitude that I would suggest is not a particularly good one. To put it another, more blunt, way, it says that I will worry about doing things well when someone's looking and, if no one's looking, it doesn't matter and so it's okay to not mess with it. As I say, it's a very natural attitude and we all take it to varying degrees regarding lots of things in our lives. But consider the mistakes you made in working this problem that worrying about these things likely would have caught. It's about developing habits that become the norm -- they are simply the way you do things. Consider the example I posted. When I worked the problem, I had no expectation that I would share that work with anyone and I certainly knew that it wasn't going to be graded, so I had no reason to make any effort to be neat and organized and legible. But it is how I have trained myself to approach problems and so it is simply how I approach them -- doing it any other way is what has become unnatural to me. And it pays off -- including catching a mistake I made in working this problem. I didn't distribute a factor correctly when I did the algebra in my head and because the units didn't work out, I caught it immediately and looked at the line above and promptly saw what I had done wrong. The error was found and fixed in a few seconds, instead of ending up with a wrong answer that would only have been caught if I had bothered to actually check my result (and why should I have bothered, since no one was looking?) and would have then probably required that I start over from scratch.

Another way to view things -- a common fallacy that is really applicable here is that we tend to believe that we don't have to practice doing things the correct way because the fact that we know what the correct way is somehow means that we will be able to do it correctly when it counts. This is almost never true -- we tend to do things they way we have always done them. If you don't track units, then you won't track units. This is why Navy fighter pilots land on a three-mile long runway the same way that they would land on a 300 ft long carrier landing zone. It's because they know that they will land on that carrier the way they land on everything else, so if it's important to do it a certain way when it really counts, they go to lengths to make is to that doing it that certain way is simply the way they normally do it, even when it doesn't count.

 

I have to agree because the units are not really there for the next reader anyway, they are there for error prevention.