Finding an unknown resistance

Thread Starter

Alcore

Joined Jan 13, 2020
7
Hello. This may be a simple problem, but I'm struggling to obtain the resistance R found in the figure below.
Untitled.png
First what's throwing my are the two voltage sources. I'm used to seeing these present in op-amps, but the way they intersect is throwing me. I'm not sure how to determine their cummulative affect.
My first impulse was to find the voltage drop across the 6K resistor, but since there is no current value given I can't do this.
I then considered nodal analysis, denoting the red node as Vb.
\[ \frac{12-V_b}{6K} + \frac{7-V_b}{0} = \frac{V_b}{R_{eq}} \]
However, since the 7V has no resistance, can I just eliminate it from the equation? Therefore:
\[ \frac{12-V_b}{6K} = \frac{V_b}{R_{eq}} \]
If this is correct, then I can solve for R by calculating the equivalent resistance.
\[ R_{eq} = \frac{10kOhm R}{10kOhm+R} \]
I am mainly suspicious that I'm overcomplicating this and that the nodal analysis is incorrect/unnecessary.
Thanks in advance.
 

daddio1948

Joined Aug 22, 2019
2
two steps :

1) find the equivalent resistance needed on the bottom to give Vout=7V, with Vin=12V and Rtop=6k.
Hint ...
The current (I=E/R) through this series resistor string is : ((12V - 7V) / 6k) = 5V/6k = 0.000833333A
So, (R=E/I) to get 7V with 0.000833333A flowing : (7V / 0.000833333A) = 8.4k

2) then find the value needed in parallel with the 10k to yield the needed 8.4k
 

MrChips

Joined Oct 2, 2009
20,335
Your red node is Vb.
What are you thinking?
Why is the red node different from the green node?

Besides, you cannot divide by 0.
 

Thread Starter

Alcore

Joined Jan 13, 2020
7
two steps :

1) find the equivalent resistance needed on the bottom to give Vout=7V, with Vin=12V and Rtop=6k.
Hint ...
The current (I=E/R) through this series resistor string is : ((12V - 7V) / 6k) = 5V/6k = 0.000833333A
So, (R=E/I) to get 7V with 0.000833333A flowing : (7V / 0.000833333A) = 8.4k

2) then find the value needed in parallel with the 10k to yield the needed 8.4k
So based off of that: 1/8400 = 1/R + 1/10000
R= 52.5kOhm

How did you know to to treat 12V as an input and 7V as an output? I suppose conceptually, it's common to have high input impedances and low output impedances. Why do we treat the current going from one voltage terminal to the next, rather than going to Ground?
 

MrChips

Joined Oct 2, 2009
20,335
Who said current is going out of the 7V terminal?

If current is flowing out of the 7V terminal then the problem cannot be solved.
 

Thread Starter

Alcore

Joined Jan 13, 2020
7
Your red node is Vb.
What are you thinking?
Why is the red node different from the green node?

Besides, you cannot divide by 0.
The red and green nodes were originally the same color, but I marked them differently when creating the diagram. I figured that if you use the red node, you'll have to consider the current going through both resistors... but I realize now that the incoming current is combined so it makes little difference.

Right. I figured that since you can't divide by zero it can't be included, but I wasn't sure of the logic.
 

MrChips

Joined Oct 2, 2009
20,335
The red and green nodes were originally the same color, but I marked them differently when creating the diagram. I figured that if you use the red node, you'll have to consider the current going through both resistors... but I realize now that the incoming current is combined so it makes little difference.

Right. I figured that since you can't divide by zero it can't be included, but I wasn't sure of the logic.
That logic sucks. Division by zero results in an infinite quantity. You cannot just ignore infinity.
 

Thread Starter

Alcore

Joined Jan 13, 2020
7
Because 7V cannot be input and 12V the output.
Oh, ,right that makes intuitive sense.

As far as current going out of the terminal, I goofed. When thinking about Vin and Vout, I imagined current flowing in the same direction. But that's not the case.
 

MrChips

Joined Oct 2, 2009
20,335
You are making life more complicated than it needs to be.
You only need one formula, and that is Ohm's Law, stated as

I = V / R

Use this formula. That is all you need.

What is the current flowing in the 6KΩ resistor?
What is the current flowing in the 10KΩ resistor?
What is the current flowing in the unknown resistor?
 

Thread Starter

Alcore

Joined Jan 13, 2020
7
You are making life more complicated than it needs to be.
You only need one formula, and that is Ohm's Law, stated as

I = V / R

Use this formula. That is all you need.

What is the current flowing in the 6KΩ resistor?
What is the current flowing in the 10KΩ resistor?
What is the current flowing in the unknown resistor?

Thanks for breaking this down. In my original post, I said that I was probably overcomplicating it. I'm trying to learn. I didn't understand that we can treat the 12V and 7V as input and output. I was viewing them as two inputs, which didn't make sense for the problem, hence the confusion. Dumb, but I see the mistake now.

Current through 6Kohm resistor is (12-7)/6K=0.00083 A.
Current through 10KOhm resistor is 7/10K = 0.0007A
Current through R is 0.00083-0.0007=0.000133.
R= 7/0.00013 = 52.5K

Thanks for explaining
 

WBahn

Joined Mar 31, 2012
25,297
Hello. This may be a simple problem, but I'm struggling to obtain the resistance R found in the figure below.
View attachment 196820
First what's throwing my are the two voltage sources. I'm used to seeing these present in op-amps, but the way they intersect is throwing me. I'm not sure how to determine their cummulative affect.
A lot of what I'm going to say has already been said, but I say it anyway in order to put my overall remarks into the proper context.

There are three possibilities with regard to the two voltages indicated.

1) They are both the result of applied voltages from sources at those nodes (which is your assumption here).
2) The 12 V source is due to a voltage source and the 7 V is just the voltage that happens to appear on that node as a consequence.
3) The 7 V is due to a voltage source and the 12 V is just the voltage that happens to appear on that node as a consequence.

If the first one is the case, then there is no way to solve the problem of finding a value of R because it can be anything at all without having any effect whatsoever on the two voltages or the currents in the two known resistors. So we have to exclude this option.

If the second one is the case, then the applied 12 V to the top node results in some current flowing and that current IS impacted by the specific value of R and since the 7 V is related to the 12 V and the voltage drop across the top resistor due to that current, there is (at most) one value of R that will result in 7 V being at that middle node.

If the third one is the case, then we again have a situation in which the current flowing in the unknown resistor has no influence on the current flowing in the top resistor, and hence no way to solve the problem (unless there is a part of the circuit that we can't see that is adjusting the voltage at the top of the circuit in order to keep the current flowing out of the 7 V node to be zero -- we could certainly design an opamp circuit to do that, but there is not any information in the problem as given to justify making any assumption like that or anything similar).

So the second case is the only one that is reasonable to conclude applies with the information given.

Under the second assumption, since the 7 V is just a note saying what the voltage at that point in the circuit happens to be, we know that no current flows in or out of the 7 V terminal (if it can and does, then all bets are off and there simply isn't enough information to solve the problem). Also, since the 12 V is connected to a source, current can flow into or out of that terminal.

My first impulse was to find the voltage drop across the 6K resistor, but since there is no current value given I can't do this.
Why can't you? If the voltage on one side of a component is 120 V and the voltage on the other side of the component is 90 V, what is the voltage drop across that component?

Think back to the definition of voltage difference.

I then considered nodal analysis, denoting the red node as Vb.
\[ \frac{12-V_b}{6K} + \frac{7-V_b}{0} = \frac{V_b}{R_{eq}} \]
However, since the 7V has no resistance, can I just eliminate it from the equation?
Not based on that reasoning. Something that has no resistance will tend to have the maximum current flowing in it, not zero current flowing in it. The reason you need to based your claim on is directly related to the consequences of the conclusion about what the voltage notations mean discussed at the beginning of this post.

Therefore:
\[ \frac{12-V_b}{6K} = \frac{V_b}{R_{eq}} \]
If this is correct, then I can solve for R by calculating the equivalent resistance.
\[ R_{eq} = \frac{10kOhm R}{10kOhm+R} \]
I am mainly suspicious that I'm overcomplicating this and that the nodal analysis is incorrect/unnecessary.
Thanks in advance.
This is the correct equation but for the wrong reason, which is what really matters. So even if you get the right numerical answer, it doesn't really deserve much credit, if only for the potential for misleading you into thinking that your reasoning was valid.

And what is Vb? What do you know about the voltage anywhere along the length of an ideal wire?
 

Thread Starter

Alcore

Joined Jan 13, 2020
7
1) They are both the result of applied voltages from sources at those nodes (which is your assumption here).
2) The 12 V source is due to a voltage source and the 7 V is just the voltage that happens to appear on that node as a consequence.
3) The 7 V is due to a voltage source and the 12 V is just the voltage that happens to appear on that node as a consequence.
Thank you for the thorough writeup. I won't be able to respond to all of it, but these points were what my main misunderstanding was. I incorrectly interpreted the diagram as having multiple applied inputs, rather than being given the values for the source and at a node. It's absolutely clear now.
 

MrChips

Joined Oct 2, 2009
20,335
12V and 7V can both be voltage sources.
You can still calculate the current flowing in the 6KΩ resistor.
You can still calculate the current flowing in the 10KΩ resistor.

However, if 7V is a voltage source, there is no solution for R because you don't know the current flowing through R.

The problem as shown in post #1 is incompletely stated. In order to solve for R one has to assume that no current flows in or out of the 7V label.
 

MrAl

Joined Jun 17, 2014
7,185
Hi,

The solution seems to hinge on what voltage is applied and what voltage is measured and it's not entirely clear from the original diagram. But we seem to be missing the actual problem statement. That would tell us what the author really wanted. Did they want us to calculate the value of R or realize that when a 7v source is applied in that manner results in a resistor value that can be anything except maybe zero?
With this in mind, we get two results:
1. R=Rx (Rx a finite value calculated from the network with 12v source and 7v measured)
2. R>0

So there must have been some original problem statement to go with the circuit. If we could see that we would know which solution applies.

Moral of the story:
Show the problem statement along with the circuit :)
 

MrChips

Joined Oct 2, 2009
20,335
If one can infer the course level of the exercise, one can assume that it is a simple voltage divider problem, because there can be no other exercise that results in R being solvable.
 

WBahn

Joined Mar 31, 2012
25,297
If it is a simple voltage divider, the 10K and R combined = 8.4K. What R is, I don't know. I haven't done this stuff in decades.
Since the TS has already solved it, we can be as explicit as we need to be. The TS's approach, as hinted by MrChips, is probably the most straightfoward way to solve it. To do it in the most generic way, which would lend itself to playing with the various parameters, we can call the 6 kΩ resistor Rt and the 10 kΩ resistor Rb. Then we have Vin (the 12 V) and Vout (the 7 V). We'll call Rp the parallel equivalent of R and Rb

Rp = (R || Rb) = (R·Rb)/(R+Rb)

Vout = Vin * Rp / (Rt + Rp)

Solve first for Rp

(Rt + Rp) / Rp = (Rt/Rp) + 1 = Vin/Vout

(Rt/Rp) = (Vin/Vout) - 1 = (Vin-Vout)/Vout

Rp = (Rt · Vout) / (Vin - Vout) = (R·Rb)/(R+Rb)

(R+Rb)/R = 1 + (Rb/R) = Rb · (Vin - Vout) / (Rt · Vout) = (Rb/Rt) · (Vin - Vout)/Vout

(Rb/R) = [Rb · (Vin - Vout) / (Rt · Vout)] - 1 = [Rb · (Vin - Vout) - (Rt · Vout)] / (Rt · Vout)

R = Rb · (Rt · Vout) / [Rb · (Vin - Vout) - (Rt · Vout)]

R = (Rb · Rt) · Vout / [Rb·Vin - (Rb+Rt)·Vout]

R = [(Rb · Rt) / (Rb+Rt)] / [Rb/(Rb+Rt)·(Vin/Vout) - 1]

R = (Rb || Rt) / { [Vin·Rb/(Rb+Rt)]/Vout - 1}

We can note that

Vo = Vin·Rb/(Rb+Rt)

is the output voltage if R is removed completely, so we get

R = (Rb || Rt) / [(Vo/Vout) - 1]

This let's us see a couple of points that, with some reflection, make complete sense. First, in order to not have a negative value for R, we have to have the actual Vout be no more than the Vo with R removed. Considering that the impact of having R is to reduce the resistance on the bottom, which will cause Vout to go down, thus the most it can be is Vo.

For the particular values given

Rb || Rt = (6 kΩ · 10 kΩ) / (6 kΩ + 10 kΩ) = (60/16) kΩ = 3.75 kΩ

Vo = 12 V · 10 kΩ / (6 kΩ + 10 kΩ) = (120/16) V = 7.5 V

R = 3.75 kΩ / [ (7.5 V / 7 V) - 1 ] = 52.5 kΩ
 
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