Here is the circuit and the question is what is the value of R3 when there is 2A through it. I put it into LTS and know that it is 30Ω. I can usually work my way back to a solution but this one has me stumped as to how to analyze it.

 

Time for a Thevenin's equivalent circuit where R3 connects to the circuit.

Without R3 the Thevenin voltage at that point is 120 * 20 / (12+20) = 75V.
Since we normally assume the source V1 is ideal with zero ohms impedance, the Thevenin resistance at the output is then the value of R1 and R2 in parallel or (12*20) / (12+20) = 7.5Ω.
So the total value of resistance needed for 2A is 75V / 2A = 37.5Ω so R3 = 37.5Ω - 7.5Ω = 30Ω, as you found.

Note these better be some healthy wattage resistors.

Magic, eh?

 

Didn't think of Thevenizing as it is in a later chapter. Using Boylestad's Introductory Circuit Analysis and this Chapter 7 covers serial and parallel DC with a lot of ratioing using KVL and KCL. Problem was marked as "Advanced" FWIW. But Thevenizing does fit. Thanks!

 

I would have just used KVL and KCL.

If I is the current from the 120V source,

using KVL
120 = 12I + 20(I -2)

Solving for I = 5A
Voltage across R1 is 60V
Voltage across R2 and R3 = 60V
R3 = 60V / 2A = 30Ω

 

There are several ways to do it. It would be good practice for you to solve it using as many of them as you can.

For a problem like this, Thevenin is probably the quickest and simplest way.

Remove R3 (as it is the load).

Vth = 120 V * 20 Ω / (20 Ω + 12 Ω)] = 75 V
Rth = 20 Ω || 12 Ω = 7.5 Ω

To get 2 A you need a total of

Rtot = 75 V / 2 Ω = 37.5 Ω

So R3 must be

Rtot - Rth = 37.5 Ω - 7.5 Ω = 30 Ω

Let's go to the opposite extreme and do it ad hoc.

Call the voltage at the junction of the resistors Vo.

You know that the current in each resistor is as follows (I'm not providing an annotated diagram, so you need to look at the equations to determine the assigned polarities for the three currents):

I1 = (Vs - Vo) / R1 (Vs is the supply voltage of 120 V)
I2 = Vo / R2
I3 = Vo / R3 = 2 A

The last equation means that

Vo = (2 A)(R3)

This means we can eliminate Vo from the other equations:

I1 = (Vs - (I3)(R3)) / R1
I2 = (I3)(R3/R2)
I3 = 2 A

KCL requires

I1 = I2 + I3

This gives us:

(Vs - (I3)(R3)) / R1 = (I3)(R3)/(R2) + I3

Now just solve for R3

Then you have node voltage analysis and mesh current analysis. Use both of them to find the solution.

Yet another approach you could take, which goes back even further, is to combine things using series/parallel resistance combinations until you get the equivalent total resistance. Then find the current. The find Vo using the voltage drop that current produces in R1, the use Vo to find I3 and set that equal to 2 A.

Doing it that way should convince you once and for all that there really is merit in learning these other techniques.

 

using KVL
120 = 12I + 20(I -2)

OK, but I'm not getting the 20(I -2) part. I guess it is the parallel resistors throwing me. I understand the I-2 but not multiplying the 2A by 20Ω part. It does prove out correct but the 20I-40 part is still not getting through my thick skull... I just don't see it.

Calling it a night. I'll work more on it in the morning once I'm fully awake...

 

Sorry, I skipped over some steps hoping that you would figure it out on your own.

Label the currents going through R1, R2, and R3 as I1, I2, and I3, respectively.

Apply KCL at the junction.
I1 = I2 + I3
I2 = I1 - I3
I3 = 2A
I2 = I1 -2

 

Yet another way to do it, which has utility in situations that don't lend themselves to solving things analytically or in which you are at an impasse and can't figure out how to do it analytically, is to do it iteratively.

First, get a feel for the kind of numbers that are needed.

Let's say that R2 was removed. What would R3 need to be in order to get 2 A through it then?

The total resistance needs to be 120 V / 2 A = 60 Ω, so R3 would need to be 48 Ω.

Since the addition of R2 will shunt some of the current away from R3, the value of R3 will need to be reduced in order to get the current back up to the 2 A target. So we know that the most that R3 can be is 48 Ω.

Since R2 is 20 Ω, let's therefore start off with R3 being 20 Ω because that makes the parallel combination of them easy to get -- it's just 10 Ω. That gives a total resistance seen by the supply of 22 Ω, resulting in a total current of 5.45 A, which is split evenly between R2 and R3, so R3 will have 2.73 A flowing through it.

This tells us that R3 must be increased in order to reduce the current in it. So we already know that

20 Ω < R3 < 48 Ω

At this point, we could take a number of approaches. One would be to just go with the midpoint and try 34 Ω and crank the math. Then, if the current is still too high, split the difference between 34 Ω and 48 Ω and try 41 Ω, while if the current is now too low, split the difference between 34 Ω and 20 Ω and try 27 Ω as the next guess. Keep splitting the difference between the largest known value that is too small and the smallest known value that is too large and within very short order you will have an answer to whatever level of accuracy you want. The first guess got you within 14 Ω, another five iterations puts you within half an ohm of the correct value.

Another way, which may or may not be faster, is to calculate successive values as follows:

First, set R3 to infinity and find the voltage at the junction.

Vo = 120 V * 20 Ω / 32 Ω = 75 V

As the first guess, make R3 the value that would put 2 A through it, which would be 37.5 Ω.

Now, with this value of R3, find the new value of Vo. The total resistance goes to 25.04 Ω and Vo becomes 62.5 V.

So use this to calculate a new value for R3, which would be 31.25 Ω.

Repeating this process again, we get 30.24 Ω.

Repeating yet again, we get 30.048 Ω.

Already we've got our three sig figs and could claim that it is 30.0 Ω.

While this isn't too bad to do with a calculator, it is trivially easy to set up a spreadsheet to do the iterations (and it took about much less time than it took to the calculations manually even for the four iterations to get to this point).

 

Note these better be some healthy wattage resistors.

LOL, there was a note in the book's preface about using standard value resistors to make the problems more realistic. However, I don't think that a 120W 30Ω resistor comes even close to "Standard"!

 

@crutschow et. al. I started with the Thevenin first. Note on this diagram the R3 current is 4A (book errata) although the question text says 2A. So, I worked it for 4A. All the way down to the point circled in the photo. The Vth is 75V but that leaves 45V of the 120Vs which is what is actually going across R3. So I'm not sure which is the correct method. Inserting R3 into the Thevenized circuit would mean that 75V/4A is 18.75Ω for R3. But subtracting the Rth from 18.75Ω gives me 11.25Ω for R3 which is correct and the same answer I would get from 45V/4A = 11.25Ω. The other scribbling is a few calculations ensuring everything in the circuit is copacetic.

 

@crutschow et. al. I started with the Thevenin first. Note on this diagram the R3 current is 4A (book errata) although the question text says 2A. So, I worked it for 4A. All the way down to the point circled in the photo. The Vth is 75V but that leaves 45V of the 120Vs which is what is actually going across R3. So I'm not sure which is the correct method. Inserting R3 into the Thevenized circuit would mean that 75V/4A is 18.75Ω for R3. But subtracting the Rth from 18.75Ω gives me 11.25Ω for R3 which is correct and the same answer I would get from 45V/4A = 11.25Ω. The other scribbling is a few calculations ensuring everything in the circuit is copacetic.
View attachment 302810

Remember, Ohm's Law doesn't allow you to grab just any voltage, current, and resistance and throw it at the equation.

You need to use the voltage ACROSS a resistance, the current through THAT resistance, and the resistance of THAT resistance.

When you compute 75 V / 4 A and get 18.25 Ω, that 18.25 Ω is NOT R3, but rather it is the resistance that the 75 V appears across, which is the series combination of R3 and Rth. So, to find R3, you must subtract Rth from the 18.25 Ω.

 

Or this:

Or superposition:

 

I finally got the 20(I-4A) bit through my thick skull and...

I've been introduced to some very basic nodal analysis but mesh and superposition are yet to come.

Yet another approach you could take, which goes back even further, is to combine things using series/parallel resistance combinations until you get the equivalent total resistance.

Yet another way to do it, which has utility in situations that don't lend themselves to solving things analytically or in which you are at an impasse and can't figure out how to do it analytically, is to do it iteratively.

This is pretty much what I do with LTS and did with this problem and knew what the R3 resistance was, just not how to calculate it. I let LTS do the calculating and just varied the R3 value until it fit but didn't know HOW it was calculating it. I think the KCL & KVL solution was what the book intended at this point as Thevenin & Norton are introduced a few chapters ahead. I thought about a Nodal approach but was thrown off by the parallel resistors. Thank you for the good input as usual!

 

My copy of Boylestad's Introductory Circuit Analysis is the 3rd edition from 1977; chapter 7 does include nodal analysis. With nodal analysis there is no thinking involved; just identify the circuit nodes and write the equation for each node. Conceptual errors will be essentially non-existant. If nodal analysis leads to more complicated algebra, I have Maple to crank the algebra.

 

I finally got the 20(I-4A) bit through my thick skull and...
View attachment 302893
I've been introduced to some very basic nodal analysis but mesh and superposition are yet to come.


This is pretty much what I do with LTS and did with this problem and knew what the R3 resistance was, just not how to calculate it. I let LTS do the calculating and just varied the R3 value until it fit but didn't know HOW it was calculating it. I think the KCL & KVL solution was what the book intended at this point as Thevenin & Norton are introduced a few chapters ahead. I thought about a Nodal approach but was thrown off by the parallel resistors. Thank you for the good input as usual!

Don't let the parallel resistors trip you up. Remember, Nodal analysis is nothing more than a systematic way of applying KCL to a circuit in such a way that KVL is intrinsically satisfied.

In this circuit, there is only one undetermined node, that being the node at the junction of the three resistors -- the other two nodes are already known, one is 0 V (because we called it that) and the other is 120 V (because it is 120 V higher than the node we chose to call "ground"). I think we've called the unknown node Vo in the past, but regardless, it's Vo now.

The nodal equation is simply the sum of the currents flowing out of the node through each branch, which is the node voltage minus the voltage on the other side of the branch, divided by the resistance of that branch. So we have:

(Vo - 120 V)/(12 Ω) + (Vo)/(20 Ω) + (Vo)/(R3) = 0

IF we knew R3, then we would have one equation and one unknown (namely Vo) and we could just solve for Vo directly.

But we have two unknowns, Vo and R3, so we need another equation. The piece of information we haven't used yet is that the current in R3 is 2 A (in the original problem in Post #1). So that would let us write this as

(Vo - 120 V)/(12 Ω) + (Vo)/(20 Ω) + (2 A) = 0

This would let us solve for Vo and then use that, plus the given 2 A and Ohm's Law, to find R3.

Or, we could combine these by doing that last step first. Ohm's Law for R3 says that:

Vo = (R3)(2 A)

Substituting this into our prior equation gets us

((R3)(2 A) - 120 V)/(12 Ω) + (R3)(2 A)/(20 Ω) + (2 A) = 0

Now we have one equation and the only unknown is our desired R3.

Multiply everything by 60 Ω and we get

5*((R3)(2 A) - 120 V) + 3*(R3)(2 A) + 120 V = 0
(R3)(10 A) - 600 V + (R3)(6 A) + 120 V = 0
(R3)(16 A) = 480 V
R3 = 480 V / 16 A = 30 Ω


If you wanted to apply Mesh Current Analysis, you could define the two mesh currents to be I1 in the left mesh and I2 in the right mesh, both circulating in the clockwise direction. The mesh current equations, which are nothing more than a systematic way of apply KVL in such a way that KCL is automatically satisfied, are

(12 Ω + 20 Ω)(I1) - (20 Ω)(I2) = 120 V
-(20 Ω)(I1) + (20 Ω + R3)(I2) = 0

Here we have two equations and three unknowns, so again we need to use our additional constraint which, in this case, is that I2 = 2 A

(32 Ω)(I1) - (20 Ω)(2 A) = 120 V
-(20 Ω)(I1) + (20 Ω + R3)(2 A) = 0

Simplifying a bit, this becomes

(32 Ω)(I1) = 160 V
-(20 Ω)(I1) + (R3)(2 A) = -40 V

Solving for I1, since it is just staring at us, we get

I1 = 160 V / 32 Ω = 5 A

Plugging this into the second equation, we get

-(20 Ω)(5 A) + (R3)(2 A) = -40 V

(R3)(2 A) = -40 V + 100 V = 60 V
R3 = 60 V / 2 A = 30 Ω


I think part of what is tripping you up is that these analysis techniques are geared towards solving for voltages and currents with the component values being givens. As soon as we shift things around and turn some of the component values into unknowns by specifying an equal number of voltages and currents, we lose the ability to just blindly apply the techniques in the same way. At that point we need to be able to sprinkle in a solid understanding of the fundamentals to be able to navigate our way through unfamiliar waters.

 

If nodal analysis leads to more complicated algebra, I have Maple to crank the algebra.

I'm using his 13th Ed and Nodal along with Mesh are in Chap 8 now. I have done some very basic nodal analysis in other texts as well as Thevenin and Norton. I also use Microsoft Mathematics for more complex equations as I am prone to simple error which are disastrous to complex equations. If I can state the relationship in an equation MM will solve it and plot the equation. The plotting is something we had no experience with back in the 60s other than simple slopes and such without programming it in Fortran and keypunching the cards to hopefully compile, run, and print out on the mainframe that took up an entire floor of the main engineering building. Which often did not occur on the first attempt so the exercise could take days to complete. The first time I saw Visicalc running on an Apple IIE I thought I had died and gone to heaven. Instantaneous feeback without having to wait a day to see if the program even ran and gave you an answer.

 

I think part of what is tripping you up is that these analysis techniques are geared towards solving for voltages and currents with the component values being givens.

Exactly! Which is one reason I really like Boylestad over several other texts that I've studied. He not only is better organized and explicit in his explanations but makes you think and apply that knowledge by throwing a few curveballs at you in problems like this one. He is definitely making me work at it instead of simply repeating it verbatim.

 

I let LTS do the calculating and just varied the R3 value until it fit but didn't know HOW it was calculating it.

LTS (or any Spice based simulator) uses iterative calculations to solve for the circuit currents and voltages.
Here is an article on the basics.

 

LTS (or any Spice based simulator) uses iterative calculations

Nice article. Thanks!
Newton-Raphson method goes back quite a ways and is still used but then so does Sir Issac's Calculus. I have SPICE on my long to-do list but currently pretty far down it. Back when I took numerical analysis it was limited to matrices equations and programming them on punch cards in PL1 to run on an IBM 360 which we had full access to. Poor little mainframe didn't have enough capacity to run full-blown Fortan and we had to limit our matrix size to keep it from going into endless loops.