Physics problem doubts - Circuits

[Anne Leite]

Homework Statement

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2. Homework Equations
Req= R1 + R2...Rn for resistors in series.
Req= (1/R1 + 1/R2 +... 1/Rn)^(-1) for resistors in paralel
And Ohm's Rules for circuits

The Attempt at a Solution

I did manage to solve it by using R3= 0 and thus finding R1= 2000 by using Ohm's Rules, then, using R3 going towards infinity, I find R23= R2, and from that I find R2 + R1= 6000. Since R1= 2000, R2= 4000, which is the correct answer. However, when I tried solving by using the point(5, 0.003) from the graph, I have the equation: R1 + (5R2/5+ R2) = 4000. Using R1 = 6000- R2 I get R2^2 -2000R2 - 10000=0. The positive root is 2000. That makes no sense since it inverts R1 and R2. Could someone tell me why it doesn't if I use that point?

 

[mfb]

The problem statement is inconsistent. As you figured out correctly, with the given numbers R1 and R2 are much larger than the scale of R3, but then R3 cannot have such a large influence with a few Ohm.

 

[SammyS]

Homework Statement

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View attachment 110878

2. Homework Equations
Req= R1 + R2...Rn for resistors in series.
Req= (1/R1 + 1/R2 +... 1/Rn)^(-1) for resistors in paralel
And Ohm's Rules for circuits

The Attempt at a Solution

I did manage to solve it by using R3= 0 and thus finding R1= 2000 by using Ohm's Rules, then, using R3 going towards infinity, I find R23= R2, and from that I find R2 + R1= 6000. Since R1= 2000, R2= 4000, which is the correct answer.

That is a correct analysis.

However, There are a number of errors in the following.

However, when I tried solving by using the point(5, 0.003) from the graph, I have the equation: R1 + (5R2/5+ R2) = 4000. Using R1 = 6000- R2 I get R2^2 -2000R2 - 10000=0. The positive root is 2000. That makes no sense since it inverts R1 and R2. Could someone tell me why it doesn't if I use that point?

You should include enough parentheses to correct the equation R1 + (5R2/5+ R2) = 4000.

It should be R₁ + (5⋅R₂/(5+ R₂)) = 4000 , which is the equivalent resistance for the three resistors and which you apparently got from (12 V) / (0.003 A) .

Solving this should not result in a quadratic equation, but it does give a negative value for R₂ .

As @mfb pointed out, there must be a problem with the statement of the problem.

It seems the problem is that R_3S must be 20kΩ rather than the 20Ω that is stated.

You will also find that the current for R₃ = 5kΩ is a bit less than 3mA .

 

[Anne Leite]

I do believe it is a quadratic, based on this calculation. So, does that mean problem is physically impossible if R3 is 20 ohms? Thanks for the huge help btw

 

[SammyS]

I do believe it is a quadratic, based on this calculation on my DS hehe. So, does that mean problem is physically impossible if R3 is 20 ohms? Thanks for the huge help btw

The values in the graph are physically impossible if r_3s = 20 Ω .

Using your approximate values from reading the graph, i = 3mA and R₃ = 5 Ω, then the parallel combination of R₂ and R₃ would have to have an effective resistance of 2000Ω. The effective resistance of two resistors in parallel is less than the smaller of the two resistors involved. 2000 Ω is impossible if R₃ is 5 Ω.

5kΩ is more reasonable a value for R₃.

You will find that with the other two resistance values you found, that R₃ = 4kΩ will give a current of exactly 3mA .

 

[mfb]

Use the current at R3=0 to find R1, then you don't get a quadratic equation.