Electrical circuits with resistors and variable voltage sources

[Biker]

Homework Statement

The following picture illustrate how the current in the circuit changes when you change $E_2$. Where slope 1 represents the current flowing through $E_1$ and slope 2 represents the current flowing through $E2$

Homework Equations

Usual circuit laws :P

The Attempt at a Solution

I think there is inconsistency in data.
It didn't give me the direction in current so I assumed the following.
If we look at the point where $E_2$ is equal to 5 ( 5 , 0.1). There is only one possibility of both current being upward so it can flow in R2
So we now know that the positive values must be upward.
and we can easily know R2
$R_2 = \frac{E_2}{0.2} = \frac{5}{0.2} = 25$
Now if we look at point 10. We find that the current flowing in $E_2$ is 0.4 and upward
The current in $E_1$ is -0.2 which means 0.2 downward. We conclude that the current in $R_2$ is 0.2
$R_2 = \frac{E_2}{0.2} = \frac{10}{0.2}= 50$
Here is the inconsistency.

 

[TomHart]

If we look at the point where E2 is equal to 5 ( 5 , 0.1).

That looks right - that the I2 current is 0.1 A for E2 = 5V. So what does that tell you about the value of R2?

R2=E2/0.2=5/0.2=25

Now it looks like you have assigned a different current (0.2 A) through R2 for the same E2 voltage of 5 V. How can that be?

Edit: Corrected "E1" to "E2" in second to last sentence.

 

[haruspex]

and we can easily know R2
$R_2 = \frac{E_2}{0.2} = \frac{5}{0.2} = 25$

That assumes the currents through E₂ and R₂ are the same.

 

[TomHart]

Do you happen to have a better representation of the plot? It is very difficult to read, making me question my ability to read data points off of it accurately.

 

[Biker]

That assumes the currents through E₂ and R₂ are the same.

Well look at E2 0.4 A going to the node and 0.2 A coming out so the current in R2 must be equal to 0.2 A

That looks right - that the I2 current is 0.1 A for E2 = 5V. So what does that tell you about the value of R2?Now it looks like you have assigned a different current (0.2 A) through R2 for the same E2 voltage of 5 V. How can that be?

Edit: Corrected "E1" to "E2" in second to last sentence.

Not sure what you mean :/, There are two points with the same current going through R2 but different voltages how come that can be true?

Do you happen to have a better representation of the plot? It is very difficult to read, making me question my ability to read data points off of it accurately.

Nope, That is the only picture I got. :C sorry

 

[TomHart]

I'm sorry. Once again my failure to read the problem statement carefully has caused my misunderstanding. I thought I1 and I2 were currents through R1 and R2. My mistake. Disregard what I said above.

 

[Biker]

I'm sorry. Once again my failure to read the problem statement carefully has caused my misunderstanding. I thought I1 and I2 were currents through R1 and R2. My mistake. Disregard what I said above.

I blame the picture, It is just unclear. :D

 

[TomHart]

I blame the picture, It is just unclear. :D

Thank you for your kindness, but the text was pretty easy to read - if only I had read it.. :)

 

[TomHart]

R2=E2/0.2=5/0.2=25

Biker, how did you come up with this equation? Are you reading I1 = I2 = 0.1 for E2 = 5 V and adding them to get the R2 current?
If 5V is the E2 voltage where I1 and I2 intersect, it looks like it could be somewhat less than 0.1 Amps, but it's hard to tell for sure.

 

[Biker]

Biker, how did you come up with this equation? Are you reading I1 = I2 = 0.1 for E2 = 5 V and adding them to get the R2 current?
If 5V is the E2 voltage where I1 and I2 intersect, it looks like it could be somewhat less than 0.1 Amps, but it's hard to tell for sure.

Yes exactly, But the difference between 50 ohms and 25 ohms is pretty large, isn't it?

 

[TomHart]

Yes exactly, But the difference between 50 ohms and 25 ohms is pretty large, isn't it?

Yes, it is pretty large. But then again, if the intersection of I1 and I2 is really at E2 = 5V, I don't think it would be much of a stretch to say that those currents intersect at, say, 0.06 Amps, or even 0.05 Amps. Because if you look at the exact point where I2 is at 0.1 Amps, I1 appears to be some distance below it. The plot is just too vague to get very accurate information from it.

 

[haruspex]

Well look at E2 0.4 A going to the node and 0.2 A coming out so the current in R2 must be equal to 0.2 A

Sorry, not following your logic. What specific data points are you making use of? From your original post, looks like you are using E2=5V, but in that case where are you getting the 0.4A from?
The sketch of the graphs is rather hard to read numbers off. I took the following:
E2=2, I1=.2
E2=10, I1=-.2
E2=4, I2=0
E2=10, I2=.42
and got R2=8.

 

[TomHart]

Sorry, not following your logic. What specific data points are you making use of? From your original post, looks like you are using E2=5V, but in that case where are you getting the 0.4A from?

I'm pretty sure what he's doing there is reading the I₁ and I₂ currents at E₂ = 10 V.
At E₂ = 10 V, I₁ = -0.2 A and I₂ = +0.4 A (from the plot).
The current into R₂ has to be the sum of those two currents, or I_R2 = -0.2 + 0.4 = 0.2 Amps.
Since the voltage across R₂ is 10 V, R₂ is calculated to be (10 V)/(0.2A) = 50 ohm

Another data point I was looking at is where I₁ crosses the 0 line. I read E₂ to be approximately 5.8 V, and I₂ to be approximately 0.13 Amps. At that data point, I₂ is equal to the current through R₂. That would result in R₂ = (5.8)/(0.14) = 44.6 ohm

 

[haruspex]

's doing there is reading the I1 and I2 currents at E2 = 10 V.
At E2 = 10 V, I1 = -0.2 A and I2 = +0.4 A (from the plot).
The current into R2 has to be the sum of those two currents, or IR2 = -0.2 + 0.4 = 0.2 Amps.
Since the voltage across R2 is 10 V, R2 is calculated to be (10 V)/(0.2A) = 50 ohm

Ok, thanks.

 

[NascentOxygen]

If we look at the point where $E_2$ is equal to 5 ( 5 , 0.1). There is only one possibility of both current being upward so it can flow in R2
So we now know that the positive values must be upward.
and we can easily know R2
$R_2 = \frac{E_2}{0.2} = \frac{5}{0.2} = 25$
Now if we look at point 10. We find that the current flowing in $E_2$ is 0.4 and upward
The current in $E_1$ is -0.2 which means 0.2 downward. We conclude that the current in $R_2$ is 0.2
$R_2 = \frac{E_2}{0.2} = \frac{10}{0.2}= 50$
Here is the inconsistency.

The inconsistency lies in your use of a graphical aid. These are straight line plots, and you must not lose sight of that.

You decided the plots intersect at (5,0.1). Okay, we'll go with that.
You also decided another relevant point is at (10,0.4). Okay, we'll use that, too.

By those 2 points you have now fully defined these plots. If you need any more data points you must calculate them using geometry of a straight line. You should not attempt to read any further data off your scrappy thumbnail sketch because estimation errors in further readings will almost certainly give data not consistent with the plots defined by the 2 points you have already decided on.

If you had an accurate graph this "problem" would not arise. I put it in quotes because it is not really a problem at all. These are straight line plots, and you should appreciate the need to be rigorously consistent in treating the variables relationship at all times as linear.

 

[epenguin]

A rather hard thread to read. It is hard to read the diagram which led to me misreading the problem - please always use an app like DocScanHD to present a less murky one.

 

[gneill]

Seems to me the diagram has got to be wrong.
There is no way that the clockwise current through R2 can be negative. This current is just E2/R2, period. The current through R2 versus E2 graph must go through the origin.

Neither line represents the current through R2. They represent the currents through the sources.

 

[epenguin]

Neither line represents the current through R2. They represent the currents through the sources.

I have deleted most of my post as I misunderstood the data and problem.

It seems to me that significant points are where the graphs cross the horizontal axis. This gives you the ratio of the E's and the ratio of the R's.

As far as I can see the in this complicated thread the OP has not given any general equation for the circuit to base an easy-to-follow reasoning on.