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I have a circuit with a variable resistor R which I need to find the value of. I am told that the power dissipated in the resistor is 250W.

The circuit:

enter image description here

I used the mesh current method naming the loops from left to right as Ia, Ib, and Ic.

These are my equations thus far:

25Ia + 100(Ia - Ib) - 200V = 0

10Ib + 20(Ib - Ic) + 30ix + 100(Ib - Ia) = 0

RIc - 30ix + 20(Ic-Ib) = 0

At this point I realized I needed a restraint equation for both ix and R. My ix restraint equation is:

ix - Ib = 0

For R's restraint equation, I related it to the power dissipated by the resistor which is 250W. I know that p = i^2 * R, so I can say that

250W = Ic^2 * R

However, when setting up my matrix I run into the problem of what to do with Ic^2. How would I set up the matrix to solve the system of equations using these equations? Or are my equations incorrect?

P.S. I cannot use any Thevenin stuff.

EDIT: The answer key I am looking at states that the two values or R should be 2.5 Ohms and 22.5 Ohms. I am just confused at how to arrive at those values.

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  • \$\begingroup\$ As a separate equation or rewrite the 250W = Ic^2 * R equation? \$\endgroup\$
    – Zearia
    Nov 4, 2015 at 9:00

2 Answers 2

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Your textbook answer is correct, and so are your equations (@Arka Sadhu probably just typed them wrong in Matlab). Anyway, here's how to cheat-solve this (first):

enter image description here

The transfer function of your circuit with the output considered the voltage over \$R_o\$ is

$$ V_o = \frac{200 R_o}{2 R_o + 15}$$

Since power over this resistor is \$V_o^2/R_o = 250\$, solve this equation to get exactly what your textbook said we should.

enter image description here


Now if we solve your system of equations (which is correct), but with R as a parameter, we get

enter image description here

Again if we use the value of the last current to solve for R given the power, we again get the same result:

enter image description here

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From what I can see, your equations are correct. So all you need to do is set up the matrix using the first four equations, and get a Ia, Ib, Ic as a function of R0.

Then simply substitute the value of Ic(R0) into the last equation to get the value of R0.

I tried it on Matlab, and got R0 = 40 or 40/49.

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  • \$\begingroup\$ Hm, are you sure that works? I think the values should be 2.5 and 22.5 Ohms. \$\endgroup\$
    – Zearia
    Nov 4, 2015 at 9:22
  • \$\begingroup\$ Looks like you typed the equations in wrong or something like that. \$\endgroup\$ Nov 5, 2015 at 22:51

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