# Solution for a resistive circuit

#### jhmluna

Joined May 15, 2023
7
Hello.

I am studying electronics and I came across a question that I am not able to resolve.
The question asks for the value of resistance R so that the 5 Ω resistor (R5) in the circuit is subjected to a potential difference of 5 V. The circuit mentioned is in the following figure.
I really tried a lot but I couldn't find a solution.

I redraw the circuit that looks like the figure below and I see that it looks like a wheatstone bridge. The wheatstone bridge is easy to solve if there is balance, which is not the case.

I tried it first by applying Kirchhoff's Current Law:
$$(A): I_4 = I_1 - I_g$$
$$(B): I_3 = I_2 + I_g$$

From equations A and B and going through branches 1 and 2 (Kirchhoff's Voltage Law), we have:
$$(1): 20\cdot I_1 - 10\cdot I_2 + 20\cdot I_g = 0$$
$$(2): 5\cdot I_4 - 20\cdot I_g + R\cdot I_3 = 0$$

Knowing that:
$$I_4 = 1 A$$
and that:
$$I_3 = I_2 + I_g$$

Substituting in equations (1) and (2) we have:

$$(1): 20\cdot (1 + I_g) - 10\cdot I_2 + 20\cdot I_g = 0$$
$$(1): 40\cdot I_g - 10\cdot I_2 + 20 = 0$$

$$(2): 5 - 20\cdot I_g + R\cdot (I_2 + I_g) = 0$$

And this is my limit. Try as I might, I can't get past this point.
That's why I came here to ask for help.
Does anyone have an idea how to resolve this question?
How to find the value of R so that the potential difference across resistor R5 is equal to 5 V?

Thanks for any help.

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#### panic mode

Joined Oct 10, 2011
2,504
Why did you chose positive sign for last term in:
(2) 5⋅I4−20⋅Ig+R⋅I3=0

why do you think that voltage drop across R5 does not depend on V1?

how do you know that I4 is 1A?

Last edited:

#### WBahn

Joined Mar 31, 2012
29,164
First off, welcome to AAC and a heart-felt thank you for being one of the few people that has made the effort to show your work up to the point where you ran into problems. Also, thank you for annotating your diagram with the currents you are using in your work, instead of making us guess. Quite refreshing.

You have a couple issues. First, as @panic mode pointed out, you have a sign error in one of your equations.

This is going to happen from time to time -- we are human and we make mistakes on a pretty regular basis. A key point to remember is that your set up equations capture ALL of the physics and electrical engineering of the problem -- everything else is "just" math. So if you have a mistake in your set up equations, your math can be perfect and all it means is that you have arrived at a correct solution to a different problem, not the problem you actually have. The math doesn't care. So it is worth taking the time and effort to always verify that you agree with the equations that you set up before you proceed to start trying to solve them. Walk through EACH term of EACH equation and ask if the term itself is correct -- for instance, have you correctly applied Ohm's Law, using the right voltage, current, and/or resistance for that term -- and ask if you have assigned the polarity of that term correctly. A mistake here usually cannot be caught later, at least not easily.

The next thing is to learn to properly track your units. A resistance is not 20 and a voltage is not 5. Nearly every physical measurement has a unit associated with it and the measurement is simply wrong if those units are omitted. They are not convenient labels, the are part of the value and you do math on them just like you do math on the numbers. Track them accordingly. Most mistakes you make -- and, again, you WILL make plenty of mistakes -- will mess up the units. Tracking them properly, from each step to each step, will almost always let you spot immediately when you make such a mistake, allowing you to quickly identify that a mistake happened, identify what the mistake was, fix the mistake, and move on. The alternative is to do what far too many people do -- throw away the units, do a bunch of math on a bunch of pure numbers, and then tack one the units that you want the answer to have onto the final number. People get killed that way.

Your final equation has four unknowns. This shouldn't be too surprising because there's a lot of information about that circuit that hasn't been taken into account. If someone changes the value of V1, R1, or R6, would you not expect the voltage across R5 to change? Yet none of your equations have taken any of these quantities into account. So you need more equations that do.

I'm going to guess that you haven't been introduced to Mesh Current Analysis or Node Voltage Analysis yet. That seems a bit odd, since in most places, courses are just wrapping up and these are topics that are usually hit about a third of the way into a first course in linear systems. But that might not be the case where you are.

So let's restrict ourselves to KCL, KCL, and Ohm's Law and see if we can come up with the right number of equations. The first step is to see how many unknowns we have.

To help with this, I've fully annotated your schematic, with a couple of tweaks. In order to make the bookkeeping easier, is it useful to have a particular subscript refer to a particular component, so I3 should be the current through R3 and V3 should be the voltage across R3, oriented so that V3 is positive whenever I3 is positive (this is known as the passive sign convention). To accomplish this, I've renamed your R1 to be R0 because the '1' subscript is given in the problem as applying to the supply, and it's best not to have V1 refer to one thing in one drawing and something different in a different drawing. I've also labeled each node with a green node label.

For each component, including the power supply, we have two unknowns, the voltage across it and the current through it. Since we have eight components, we have sixteen voltage and current unknowns. We also have an additional unknown in that we don't know the resistance of R7. So actually have seventeen unknowns and we need seventeen equations. Sounds like a lot, but we can whittle it down pretty severely pretty quickly.

There a bit of low hanging fruit here. For instance, we can immediately write down an equation giving the voltage across the source, namely

V1 = 101 V

Similarly, we are given the voltage across R5

V5 = 5 V

For each resistor, we can relate the voltage across that component to the current through that component using Ohm's Law, so that supplies us with another seven equations, giving us a total of nine so far. We need eight more.

Here is were KCL comes into play. There are some trivial equations because of components that are in series, namely

I1 = I0
I1 = I6

Node that these are just KCL written at nodes A and G respectively, and give us two more equations.

It might be tempting to say that we also have I0 = I6, which is true, but it is also redundant since this comes about directly from combining the two equations above. Mathematically speaking, only two of these three equations are "linearly independent " That's part of the problem with approaching circuit analysis at this level -- we have a hard time identifying which equations actually capture information not already present in some combination of the other equations, i.e., choose a set of equations that are all linearly independent.

Next, we can write KCL at nodes B, C, D, and E, but while it's not very obvious, only three of these capture new information (pick any three, and the fourth can be obtained from all of the others), giving us three more equations.

This brings the total number of equations to fourteen. We just need three more, and we get those from KVL by applying KVL around any set of three independent loops. Here, again, we actually have a lot of loops we could use (I think there are seven, if I counted right) and we have to pick three such that none of them can be written in terms of the others. The easiest way to do this is to use "meshes", which is to consider the circuit as consisting of a bunch of window panes and your equations come from going around each pane of glass. In this circuit, there are three such panes (that this happens to be the exact same as the number of equations we need is NOT a coincidence).

Normally, we don't go into this level of detail, but I thought it would be good for you to see it at least once. Usually, we apply Ohm's Law as we write down our KCL and KVL equations. We also don't give different current labels to currents that are obviously in series, instead we apply KCL at nodes that only have two components by inspection and just assign a single current label for both branches.

This still leaves us with quite a few equations to solve, which is not only inconvenient, it's a source of easy errors as we crank through the math.

This is where techniques like Mesh Current Analysis and Node Voltage Analysis come into play -- mesh analysis is nothing more than the systematic application of KVL in such a way that Ohm's Law and KCL are satisfied automatically, while nodal analysis is the systematic application of KCL in such a way that Ohm's Law and KVL are satisfied automatically. They also allow us to unambiguously identify exactly the right number of linearly independent equations that are needed to solve the problem. So, if you haven't covered these topics yet, pay very close attentions when you do -- they will make your life MUCH easier.

#### jhmluna

Joined May 15, 2023
7
Hi @panic mode, thanks for the remarks.

Why did you chose positive sign for last term in:
(2) 5⋅I4−20⋅Ig+R⋅I3=0
It really is wrong. Thanks for the note.

why do you think that voltage drop across R5 does not depend on V1?
I know it depends on V1, but I got stuck before I even included V1 in the problem.

how do you know that I4 is 1A?
The question says that the potential difference over R5 must be 5 V. As the value of R5 is known and equal to 5 Ω, the current can be found by direct relationship according to Ohm's Law and is equal to 1 A.

#### jhmluna

Joined May 15, 2023
7
First off, welcome to AAC and a heart-felt thank you for being one of the few people that has made the effort to show your work up to the point where you ran into problems. Also, thank you for annotating your diagram with the currents you are using in your work, instead of making us guess. Quite refreshing.

You have a couple issues. First, as @panic mode pointed out, you have a sign error in one of your equations.

This is going to happen from time to time -- we are human and we make mistakes on a pretty regular basis. A key point to remember is that your set up equations capture ALL of the physics and electrical engineering of the problem -- everything else is "just" math. So if you have a mistake in your set up equations, your math can be perfect and all it means is that you have arrived at a correct solution to a different problem, not the problem you actually have. The math doesn't care. So it is worth taking the time and effort to always verify that you agree with the equations that you set up before you proceed to start trying to solve them. Walk through EACH term of EACH equation and ask if the term itself is correct -- for instance, have you correctly applied Ohm's Law, using the right voltage, current, and/or resistance for that term -- and ask if you have assigned the polarity of that term correctly. A mistake here usually cannot be caught later, at least not easily.

The next thing is to learn to properly track your units. A resistance is not 20 and a voltage is not 5. Nearly every physical measurement has a unit associated with it and the measurement is simply wrong if those units are omitted. They are not convenient labels, the are part of the value and you do math on them just like you do math on the numbers. Track them accordingly. Most mistakes you make -- and, again, you WILL make plenty of mistakes -- will mess up the units. Tracking them properly, from each step to each step, will almost always let you spot immediately when you make such a mistake, allowing you to quickly identify that a mistake happened, identify what the mistake was, fix the mistake, and move on. The alternative is to do what far too many people do -- throw away the units, do a bunch of math on a bunch of pure numbers, and then tack one the units that you want the answer to have onto the final number. People get killed that way.

Your final equation has four unknowns. This shouldn't be too surprising because there's a lot of information about that circuit that hasn't been taken into account. If someone changes the value of V1, R1, or R6, would you not expect the voltage across R5 to change? Yet none of your equations have taken any of these quantities into account. So you need more equations that do.

I'm going to guess that you haven't been introduced to Mesh Current Analysis or Node Voltage Analysis yet. That seems a bit odd, since in most places, courses are just wrapping up and these are topics that are usually hit about a third of the way into a first course in linear systems. But that might not be the case where you are.

So let's restrict ourselves to KCL, KCL, and Ohm's Law and see if we can come up with the right number of equations. The first step is to see how many unknowns we have.

To help with this, I've fully annotated your schematic, with a couple of tweaks. In order to make the bookkeeping easier, is it useful to have a particular subscript refer to a particular component, so I3 should be the current through R3 and V3 should be the voltage across R3, oriented so that V3 is positive whenever I3 is positive (this is known as the passive sign convention). To accomplish this, I've renamed your R1 to be R0 because the '1' subscript is given in the problem as applying to the supply, and it's best not to have V1 refer to one thing in one drawing and something different in a different drawing. I've also labeled each node with a green node label.

View attachment 294285
For each component, including the power supply, we have two unknowns, the voltage across it and the current through it. Since we have eight components, we have sixteen voltage and current unknowns. We also have an additional unknown in that we don't know the resistance of R7. So actually have seventeen unknowns and we need seventeen equations. Sounds like a lot, but we can whittle it down pretty severely pretty quickly.

There a bit of low hanging fruit here. For instance, we can immediately write down an equation giving the voltage across the source, namely

V1 = 101 V

Similarly, we are given the voltage across R5

V5 = 5 V

For each resistor, we can relate the voltage across that component to the current through that component using Ohm's Law, so that supplies us with another seven equations, giving us a total of nine so far. We need eight more.

Here is were KCL comes into play. There are some trivial equations because of components that are in series, namely

I1 = I0
I1 = I6

Node that these are just KCL written at nodes A and G respectively, and give us two more equations.

It might be tempting to say that we also have I0 = I6, which is true, but it is also redundant since this comes about directly from combining the two equations above. Mathematically speaking, only two of these three equations are "linearly independent " That's part of the problem with approaching circuit analysis at this level -- we have a hard time identifying which equations actually capture information not already present in some combination of the other equations, i.e., choose a set of equations that are all linearly independent.

Next, we can write KCL at nodes B, C, D, and E, but while it's not very obvious, only three of these capture new information (pick any three, and the fourth can be obtained from all of the others), giving us three more equations.

This brings the total number of equations to fourteen. We just need three more, and we get those from KVL by applying KVL around any set of three independent loops. Here, again, we actually have a lot of loops we could use (I think there are seven, if I counted right) and we have to pick three such that none of them can be written in terms of the others. The easiest way to do this is to use "meshes", which is to consider the circuit as consisting of a bunch of window panes and your equations come from going around each pane of glass. In this circuit, there are three such panes (that this happens to be the exact same as the number of equations we need is NOT a coincidence).

Normally, we don't go into this level of detail, but I thought it would be good for you to see it at least once. Usually, we apply Ohm's Law as we write down our KCL and KVL equations. We also don't give different current labels to currents that are obviously in series, instead we apply KCL at nodes that only have two components by inspection and just assign a single current label for both branches.

This still leaves us with quite a few equations to solve, which is not only inconvenient, it's a source of easy errors as we crank through the math.

This is where techniques like Mesh Current Analysis and Node Voltage Analysis come into play -- mesh analysis is nothing more than the systematic application of KVL in such a way that Ohm's Law and KCL are satisfied automatically, while nodal analysis is the systematic application of KCL in such a way that Ohm's Law and KVL are satisfied automatically. They also allow us to unambiguously identify exactly the right number of linearly independent equations that are needed to solve the problem. So, if you haven't covered these topics yet, pay very close attentions when you do -- they will make your life MUCH easier.
@WBahn
Thank you very much.
I am flattered by an answer that is a real lesson.
I'll do my best to apply what you've given me and try to resolve the quetion.
As soon as I get it, I'll come back to leave the answer.
Thank you very much.

#### jhmluna

Joined May 15, 2023
7
So, after many scrap sheets thrown away, I arrived at a new point after applying the Mesh Current Method mentioned by @WBahn
Once again, I got stuck at one point not understanding my mistake.
I saw in books that when solving a circuit by the method of mesh currents, we end up with a linear system of N equations to solve.
It turns out that my system turned out to be something non-linear as can be seen in the following development.
I apologize to @WBahn for not correctly tracking the units in the equations. I think it would be very confusing if I did.

So, to try to solve the circuit, I selected the three mesh currents indicated in orange, blue and pink in the figure below.

Before writing the mesh equations, it is known that:

$$v_1 = 101\ V\\ v_5 = 5\ V\\ Given\ that\ R_5 = 5\ \Omega,\ i_5 = 1\ A \\ i_0 = i_1 \\ i_6 = i_1 \\ i_0 = i_2+i_3 \\ i_2 = i_4 + i_5 = i_4 + 1 \\ i_6 = i_7 + i_5 = i_7 + 1 \\$$

$$Mesh\ 1:\ v_1 - R_0\cdot i_0 - R_2\cdot i_2 - R_5\cdot i_5 - R_6\cdot i_6 = 0$$

$$Mesh\ 2:\ R_2\cdot i_2 - R_3\cdot i_3 + R_4\cdot i_4 = 0$$

$$Mesh\ 3:\ -R_4\cdot i_4 - R_7\cdot i_7 + R_5\cdot i_5 = 0$$

Rewriting the three mesh equations considering the node currents and known values, we have:

$$Mesh\ 1:\ v_1 - R_0\cdot i_1 - R_2\cdot (i_1 - i_3) - R_5\cdot 1 - R_6\cdot i_1 = 0$$

$$Mesh\ 2:\ R_2\cdot (i_1 - i_3) - R_3\cdot i_3 + R_4\cdot (i_1 - i_3 - 1) = 0$$

$$Mesh\ 3:\ -R_4\cdot (i_1 - i_3 - 1) - R_7\cdot (i_1 - 1) + R_5\cdot 1 = 0$$

Rewriting again after grouping the unknown terms:

$$Mesh\ 1:\ - (R_0 + R_2 + R_6)\cdot i_1 + R_2\cdot i_3 = -v_1 + R_5$$

$$Mesh\ 2:\ (R_2 + R_4)\cdot i_1 - (R_2 + R_3 + R_4)\cdot i_3 = R_4$$

$$Mesh\ 3:\ -(R_4 + R_7)\cdot i_1 + R_4\cdot i_3 + R_7 = -R_5 - R_4$$

There are 3 equations with 3 unknowns:

$$i_1,\ i_3\ e\ R_7$$

After substituting the known values in the equations, we have:

$$Mesh\ 1:\ - (40 + 20 +10)\cdot i_1 + 20\cdot i_3 = -101 + 5$$

$$Mesh\ 2:\ (20 + 20)\cdot i_1 - (20 + 10 + 20)\cdot i_3 = 20$$

$$Mesh\ 3:\ -(20 + R_7)\cdot i_1 + 20\cdot i_3 + R_7 = -5 - 20$$

$$Mesh\ 1:\ -70\cdot i_1 + 20\cdot i_3 = -96$$

$$Mesh\ 2:\ 40\cdot i_1 -50\cdot i_3 = 20$$

$$Mesh\ 3:\ -(20 + R_7)\cdot i_1 + 20\cdot i_3 + R_7 = -25$$

Here was my new stopping point because the system is not linear since I have the multiplication of two unknowns in the mesh 3 equation.
As I reviewed all the calculations and found no error, I came here to ask colleagues if they can tell me what is wrong.

Thank you very much.

#### The Electrician

Joined Oct 9, 2007
2,936
You're making it a lot harder than it needs to be;

Go back to your original circuit and choose 3 mesh currents. Set up your equations using numerical values for all the resistors except the one labeled "R"; keep it as a variable. Solve for the mesh currents. You should get 3 algebraic expressions for I1, I2 and I3 involving the variable "R". The current in R5 will be I1-I3. Subtract the expression for I3 from the expression for I1, and set that result equal to 1 amp. Solve that for the value for "R".

#### panic mode

Joined Oct 10, 2011
2,504
I apologize to @WBahn for not correctly tracking the units in the equations. I think it would be very confusing if I did.
no.... quite the contrary. without tracking units you will wind up with something that makes no sense and cannot be verified.

Before writing the mesh equations, it is known that:
i_2 = i_4 + i_5 = i_4 + 1
i_6 = i_7 + i_5 = i_7 + 1
−(R0+R2+R6)⋅i1+R2⋅i3=−v1+R5

in that equation you are showing that resistance and voltage have same units and can be added/subtracted as comparable terms. if i was one grading your work, i would fail you just for writing nonsense like that even if you did manage to get to result.

you should aim to solve everything symbolically first. substituting values in should come at the very end and units MUST be observed. this is an acceptable example and a world of a difference (all terms in equation are voltage):
−(R0+R2+R6)⋅i1+R2⋅i3=−v1+R5*1A

#### WBahn

Joined Mar 31, 2012
29,164
So, after many scrap sheets thrown away, I arrived at a new point after applying the Mesh Current Method mentioned by @WBahn
Once again, I got stuck at one point not understanding my mistake.
I saw in books that when solving a circuit by the method of mesh currents, we end up with a linear system of N equations to solve.
It turns out that my system turned out to be something non-linear as can be seen in the following development.
I apologize to @WBahn for not correctly tracking the units in the equations. I think it would be very confusing if I did.

So, to try to solve the circuit, I selected the three mesh currents indicated in orange, blue and pink in the figure below.

View attachment 294355

Before writing the mesh equations, it is known that:

$$v_1 = 101\ V\\ v_5 = 5\ V\\ Given\ that\ R_5 = 5\ \Omega,\ i_5 = 1\ A \\ i_0 = i_1 \\ i_6 = i_1 \\ i_0 = i_2+i_3 \\ i_2 = i_4 + i_5 = i_4 + 1 \\ i_6 = i_7 + i_5 = i_7 + 1 \\$$

$$Mesh\ 1:\ v_1 - R_0\cdot i_0 - R_2\cdot i_2 - R_5\cdot i_5 - R_6\cdot i_6 = 0$$

$$Mesh\ 2:\ R_2\cdot i_2 - R_3\cdot i_3 + R_4\cdot i_4 = 0$$

$$Mesh\ 3:\ -R_4\cdot i_4 - R_7\cdot i_7 + R_5\cdot i_5 = 0$$

Rewriting the three mesh equations considering the node currents and known values, we have:

$$Mesh\ 1:\ v_1 - R_0\cdot i_1 - R_2\cdot (i_1 - i_3) - R_5\cdot 1 - R_6\cdot i_1 = 0$$

$$Mesh\ 2:\ R_2\cdot (i_1 - i_3) - R_3\cdot i_3 + R_4\cdot (i_1 - i_3 - 1) = 0$$

$$Mesh\ 3:\ -R_4\cdot (i_1 - i_3 - 1) - R_7\cdot (i_1 - 1) + R_5\cdot 1 = 0$$

Rewriting again after grouping the unknown terms:

$$Mesh\ 1:\ - (R_0 + R_2 + R_6)\cdot i_1 + R_2\cdot i_3 = -v_1 + R_5$$

$$Mesh\ 2:\ (R_2 + R_4)\cdot i_1 - (R_2 + R_3 + R_4)\cdot i_3 = R_4$$

$$Mesh\ 3:\ -(R_4 + R_7)\cdot i_1 + R_4\cdot i_3 + R_7 = -R_5 - R_4$$

There are 3 equations with 3 unknowns:

$$i_1,\ i_3\ e\ R_7$$

After substituting the known values in the equations, we have:

$$Mesh\ 1:\ - (40 + 20 +10)\cdot i_1 + 20\cdot i_3 = -101 + 5$$

$$Mesh\ 2:\ (20 + 20)\cdot i_1 - (20 + 10 + 20)\cdot i_3 = 20$$

$$Mesh\ 3:\ -(20 + R_7)\cdot i_1 + 20\cdot i_3 + R_7 = -5 - 20$$

$$Mesh\ 1:\ -70\cdot i_1 + 20\cdot i_3 = -96$$

$$Mesh\ 2:\ 40\cdot i_1 -50\cdot i_3 = 20$$

$$Mesh\ 3:\ -(20 + R_7)\cdot i_1 + 20\cdot i_3 + R_7 = -25$$

Here was my new stopping point because the system is not linear since I have the multiplication of two unknowns in the mesh 3 equation.
As I reviewed all the calculations and found no error, I came here to ask colleagues if they can tell me what is wrong.

Thank you very much.
You seem to be committing the very sin I warned about. In your work above, is i_3 the current in R_3, or is it the current in Mesh 3?

Be careful not to use the same label for two different things -- it's a recipe for disaster.

The resulting system of equations are linear in terms of voltages and currents. When a resistance is an unknown, it gets trickier.

When using mesh currents, you have one current for each mesh. Normally, they are labeled using the mesh labels, so I_1 would be the current in mesh 1 in the direction indicated. Since we already have a current I_1 defined, I will write the current in mesh 1 as I_11 (and the current in mesh 2 as I_22 and so on).

With a bit of practice, you can write the mesh equations by inspection:

$$\text{Mesh #1:} \; \left( R_0 \; + \; R_2 \; + R_5 \; + \; R_6 \right) I_{11} \; - \; \left( R_2 \right) I_{22} \; - \; \left( R_5 \right) I_{33} \; = \; V_1 \\ \text{Mesh #2:} \; - \left( R_2 \right) I_{11} \; + \; \left( R_2 \; + R_3 \; + \; R_4 \right) I_{22} \; - \; \left( R_4 \right) I_{33} \; = \; 0 \\ \text{Mesh #3:} \; - \left( R_5 \right) I_{11} \; - \; \left( R_4 \right) I_{22} \; + \; \left( R_4 \; + R_5 \; + \; R_7 \right) I_{33} \; = \; 0$$

Notice the symmetry about the diagonal on the left hand side. Also notice that the diagonal terms are all positive and the off-diagonal terms are all negative. These are very valuable checks that you haven't messed things up.

BTW: This is how I caught two mistakes I made -- sign errors on the diagonal terms in the second and third equations because I did a copy/paste/modify of the first equation and got sloppy.

If we put in our known resistances (something that I try not to do until the very end, but I'll go ahead and do it at this point so that we can more easily focus on other things)

$$\text{Mesh #1:} \; \left( 75 \; \Omega \right) I_{11} \; - \; \left( 20 \; \Omega \right) I_{22} \; - \; \left( 5 \; \Omega \right) I_{33} \; = \; 101 \; V \\ \text{Mesh #2:} \; - \left( 20 \; \Omega \right) I_{11} \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 20 \; \Omega \right) I_{33} \; = \; 0 \\ \text{Mesh #3:} \; - \left( 5 \; \Omega \right) I_{11} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 25 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 0$$

At this point, we have three equations and four unknowns. Here is where we can apply our information about V_5

$$I_5 \; = \; \frac{V_5}{R_5} \; = \; I_{11} \; - \; I_{33} \\ V_5 \; = \; 5 \; V$$

Combining these, we have

$$I_{11} \; = \; \left( I_{33} \; + \; 1 \; A \right)$$

Substituting this into our mesh equations, we get

$$\text{Mesh #1:} \; \left( 75 \; \Omega \right) \left( I_{33} \; + \; 1 \; A \right) \; - \; \left( 20 \; \Omega \right) I_{22} \; - \; \left( 5 \; \Omega \right) I_{33} \; = \; 101 \; V \\ \text{Mesh #2:} \; - \left( 20 \; \Omega \right)\left( I_{33} \; + \; 1 \; A \right) \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 20 \; \Omega \right) I_{33} \; = \; 0 \\ \text{Mesh #3:} \; - \left( 5 \; \Omega \right) \left( I_{33} \; + \; 1 \; A \right) \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 25 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 0$$

Collecting common terms:

$$\text{Mesh #1:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \\ \text{Mesh #2:} \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\ \text{Mesh #3:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V$$

At this point, don't try to think of this as three linear equations in three unknowns. That will lead you down a rabbit hole trying to chase some memorized approach.

The first two represent two equations in two unknowns. Can you use just those two equations to solve for I_22 and I_33?

Once you've done that, can you use the third equation to solve for R_7?

#### The Electrician

Joined Oct 9, 2007
2,936
Here was my new stopping point because the system is not linear since I have the multiplication of two unknowns in the mesh 3 equation.
As I reviewed all the calculations and found no error, I came here to ask colleagues if they can tell me what is wrong.

Thank you very much.
The mental breakthrough you need to make is to realize that when you finally obtain your 3 equations, do not treat R7 as an unknown at that point. Treat it as a simple variable which is part of a coefficient. Of course you must fix the errors in deriving your equations that have been pointed out.

Treating R7 as a simple variable you can solve your 3 equations by whatever means you want. You could solve them by hand using Gaussian elimination, or using a computer linear equation solver.

You will then have three algebraic expressions for the 3 mesh currents. The expressions will all involve R7 and now R7 is an unknown. Subtract the expression for I3 from the expression for I1, simplify and set that result equal to 1 amp. This will give you 1 equation in 1 unknown which you can solve for the value of R7

#### jhmluna

Joined May 15, 2023
7
You seem to be committing the very sin I warned about. In your work above, is i_3 the current in R_3, or is it the current in Mesh 3?

Be careful not to use the same label for two different things -- it's a recipe for disaster.

The resulting system of equations are linear in terms of voltages and currents. When a resistance is an unknown, it gets trickier.

When using mesh currents, you have one current for each mesh. Normally, they are labeled using the mesh labels, so I_1 would be the current in mesh 1 in the direction indicated. Since we already have a current I_1 defined, I will write the current in mesh 1 as I_11 (and the current in mesh 2 as I_22 and so on).

This explanation clarified me and now I understand my mistake!
I was mixing the mesh currents with the component currents.
Thank you very much for your willingness to help.
I redesigned the circuit renaming the currents according to the nomenclature you used.

After that, continuing from the system that @WBahn left in the previous post:

$$\text{Mesh #1:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \\ \text{Mesh #2:} \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\ \text{Mesh #3:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V$$

From Mesh #1:

$$\; I_{33}= \frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \;I_{22} \; A$$

Taking the equation from $$I_{33}$$ to the equation of Mesh #2

$$\; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\ \\ \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) \left(\frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \; I_{22} \right ) \; = \; 20 \; V$$

Solving this equation for $$I_{22}$$ we have:

$$I_{22} \; \cong 0.904 \; A$$

Taking the value of $$I_{22}$$ to the equation of $$I_{33}$$, we have:

$$\; I_{33} \; = \; \frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \;I_{22} \; \\ \; I_{33} \; = \; \frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \cdot 0.904A \\ \; I_{33} \; \cong \; 0.629 \; A$$

Now, with the values of $$I_{22}$$ and $$I_{33}$$, we can solve the equation of Mesh 3 for R7:

$$- \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V \\ - \; \left( 20 \; \Omega \right) \cdot 0.904 \; A \; + \; \left( 20 \; \Omega \; + \; R_7 \right) \cdot 0.629 \; A \; = \; 5 \; V \\ R_7 \; = \; 16.69 \; \Omega$$

To confirm the result, the circuit was simulated in Tina TI software. The result of the simulation is in the following figure, which displays the voltage (5 V) and current (1 A) in R5, as requested in the challenge:

I would like to thank all colleagues who helped me.
Thanks @WBahn , @panic mode , @The Electrician

#### WBahn

Joined Mar 31, 2012
29,164
It very good to see that you took steps to verify the correctness of your work. So many people get an answer and then stop there.

Using a simulator to check results is perfectly reasonable -- except when you can't. Like on an exam. So you want to be able to verify the correctness of your results from the problem itself (which, fortunately, is almost always possible with engineering problems).

Once you have I_22 and I_33 (and given that I_5 is a known 1 A). It's trivially easy to find I_0 and I_6 since I_7 = I_33 and I_6 = I_5 + I_33. At this point, it is an extremely simply matter to determine the voltages across R_6, R_7, R_3, and R_0. Sum those up and you should get V_1's 101 V, within acceptable roundoff tolerance.

#### The Electrician

Joined Oct 9, 2007
2,936
Had you followed the procedure I described in post #7 this would be the result:

Last edited:

#### jhmluna

Joined May 15, 2023
7
Had you followed the procedure I described in post #7 this would be the result:

View attachment 294415
I actually tried to follow the procedure you described.
But I didn't get anywhere because I was doing something wrong: I was replacing branch currents with component currents.
That way, I didn't arrive at this system to be solved.
I only realized this after reading @WBahn 's message in post #9.

What software is this that you used to solve the systems and equations?

#### WBahn

Joined Mar 31, 2012
29,164
Had you followed the procedure I described in post #7 this would be the result:

View attachment 294415
You say that the current in R7 is given by I1 - I3. But the current in R7 is just I3.

I1 - I3 is the current in R5.

#### WBahn

Joined Mar 31, 2012
29,164

This explanation clarified me and now I understand my mistake!
I was mixing the mesh currents with the component currents.
Thank you very much for your willingness to help.
I redesigned the circuit renaming the currents according to the nomenclature you used.

View attachment 294411

After that, continuing from the system that @WBahn left in the previous post:

$$\text{Mesh #1:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \\ \text{Mesh #2:} \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\ \text{Mesh #3:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V$$

From Mesh #1:

$$\; I_{33}= \frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \;I_{22} \; A$$

Taking the equation from $$I_{33}$$ to the equation of Mesh #2

$$\; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\ \\ \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) \left(\frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \; I_{22} \right ) \; = \; 20 \; V$$

Solving this equation for $$I_{22}$$ we have:

$$I_{22} \; \cong 0.904 \; A$$

Taking the value of $$I_{22}$$ to the equation of $$I_{33}$$, we have:

$$\; I_{33} \; = \; \frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \;I_{22} \; \\ \; I_{33} \; = \; \frac{26 \; V}{\;70 \; \Omega} + \frac{2}{7} \cdot 0.904A \\ \; I_{33} \; \cong \; 0.629 \; A$$

Now, with the values of $$I_{22}$$ and $$I_{33}$$, we can solve the equation of Mesh 3 for R7:

$$- \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V \\ - \; \left( 20 \; \Omega \right) \cdot 0.904 \; A \; + \; \left( 20 \; \Omega \; + \; R_7 \right) \cdot 0.629 \; A \; = \; 5 \; V \\ R_7 \; = \; 16.69 \; \Omega$$

To confirm the result, the circuit was simulated in Tina TI software. The result of the simulation is in the following figure, which displays the voltage (5 V) and current (1 A) in R5, as requested in the challenge:

View attachment 294414

I would like to thank all colleagues who helped me.
Thanks @WBahn , @panic mode , @The Electrician
While you approach to solving the equations is perfectly valid, it is often easier to create a linear combination of two equations to eliminate an unknown:

$$\text{Mesh #1:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \\ \text{Mesh #2:} \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\ \text{Mesh #3:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V$$

Multiply the first equation by 5 and the second by 2 and then add them to eliminate I_22, letting you solve for I_33 directly.

$$5 \; \left[ \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \; \right] \\ 2 \; \left[ \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \; \right]$$

$$- \; \left( 100 \; \Omega \right) I_{22} \; + \; \left( 350 \; \Omega \right) I_{33} \; = \; 130 \; V \\ + \; \left( 100 \; \Omega \right) I_{22} \; - \; \left( 80 \; \Omega \right) I_{33} \; = \; 40 \; V$$

$$\left( 270 \; \Omega \right) I_{33} \; = \; 170 \; V$$

Which gives

$$I_{33} \; = \; \frac{170 \; V}{270 \; \Omega} \; = \; 629.6 \; mA$$

Then it's a simple matter to substitute I_33 back into any of the equations that just have I_22 and I_33 to get I_22. Let's use the very last one, as it looks the "nicest".

$$I_{22} \; = \; \frac{\left( 80 \; \Omega \; \cdot \; 629.6 \; mA \right) \; + \; 40 \; V}{100 \; \Omega} \; = \; 903.7 \; mA$$

Now it's just a matter of plugging these into the final equation to get R_7.

$$R_7 \; = \; \frac{5 \; V}{I_{33}} \; + \; \left( 20 \; \Omega \right) \frac{I_{22}}{I_{33}} \; - \; 20 \; \Omega \\ R_7 \; = \; \frac{5 \; V}{629.6 \; mA} \; + \; \left( 20 \; \Omega \right) \frac{903.7 \; mA}{629.6 \; mA} \; - \; 20 \; \Omega \; = \; 16.65 \; \Omega$$

As you can hopefully begin to see, properly tracking units is not that hard and is far from confusing. If you apply it consistently, it will quickly become second nature and what you will find confusing is any equation in which the units are missing or do not work out properly. This is a good thing, because (if the units were tracked from the start) this means that errors that mess up the units will jump out at you, preventing you from slugging forward with work that is guaranteed to be a waste of time.

#### jhmluna

Joined May 15, 2023
7
While you approach to solving the equations is perfectly valid, it is often easier to create a linear combination of two equations to eliminate an unknown:

\text{Mesh #1:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \\
\text{Mesh #2:} \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \\
\text{Mesh #3:} \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 20 \; \Omega \; + \; R_7 \right) I_{33} \; = \; 5 \; V

Multiply the first equation by 5 and the second by 2 and then add them to eliminate I_22, letting you solve for I_33 directly.

5 \; \left[ \; - \; \left( 20 \; \Omega \right) I_{22} \; + \; \left( 70 \; \Omega \right) I_{33} \; = \; 26 \; V \; \right] \\
2 \; \left[ \; + \; \left( 50 \; \Omega \right) I_{22} \; - \; \left( 40 \; \Omega \right) I_{33} \; = \; 20 \; V \; \right]

- \; \left( 100 \; \Omega \right) I_{22} \; + \; \left( 350 \; \Omega \right) I_{33} \; = \; 130 \; V \\
+ \; \left( 100 \; \Omega \right) I_{22} \; - \; \left( 80 \; \Omega \right) I_{33} \; = \; 40 \; V
Thanks for this tip.
It really makes the resolution of the system a lot easier.

#### The Electrician

Joined Oct 9, 2007
2,936
You say that the current in R7 is given by I1 - I3. But the current in R7 is just I3.

I1 - I3 is the current in R5.
In post #7 I said " The current in R5 will be I1-I3. "
Typos happen; fixed.

Last edited:

#### The Electrician

Joined Oct 9, 2007
2,936
I actually tried to follow the procedure you described.
But I didn't get anywhere because I was doing something wrong: I was replacing branch currents with component currents.
That way, I didn't arrive at this system to be solved.
I only realized this after reading @WBahn 's message in post #9.

What software is this that you used to solve the systems and equations?
I used Mathematica. The web accessible version of that is https://www.wolframalpha.com/
There are numerous other math programs that can do this such as Mathcad, Matlab, Macsyma, Maple, etc. Some of them are free. If you are attending college one of more of them will probably be available on the school's network.

One advantage of using modern mathematical software for this sort of problem is that the algebra will be done by the computer and there will be no mistakes (assuming you have set up your equations without error ).

#### WBahn

Joined Mar 31, 2012
29,164
I used Mathematica. The web accessible version of that is https://www.wolframalpha.com/
There are numerous other math programs that can do this such as Mathcad, Matlab, Macsyma, Maple, etc. Some of them are free. If you are attending college one of more of them will probably be available on the school's network.

One advantage of using modern mathematical software for this sort of problem is that the algebra will be done by the computer and there will be no mistakes (assuming you have set up your equations without error ).
@jhmluna While using these tools is fine (and good) in "the real world", and for checking your work in school, I would recommend that you NOT jump and in use them to do your work for you. Make sure that YOU are comfortable with solving these equations using nothing more than a basic calculator. That will ensure that you understand and know how to apply the concepts. Then, when you use the tools, they really are being used to just do the grunt work. If you don't, what you are likely to end up with is the tools doing your thinking for you.