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A linear circuit containing two voltage dependent sources, as shown. The elements in this circuit have the following values: \$R_1 = 7.2k\$ Ohms, \$R_2 = 12k\$ Ohms, \$R_3 = 3.3k\$ Ohms, \$\alpha = 0.025\$ A/V and \$\beta = 5\$.

Calculate the numerical value for the Thevenin equivalent resistance for terminal A-B.

I am having a hard time solving this. Letting \$i\$ be the current flowing into \$R_3\$, \$i_1\$ the current flowing through \$R_1\$ and \$i_2\$ the current through \$R_1\$—and the node between R3 and R1 be e_1;

Using KVL and KCL, I've managed to find that

\$i + I = i_1 + i_2\$ Hence \$\frac{e_1-\beta v_2}{R_3} + \alpha v_1 = \frac{v_2}{R_2} + \frac{v_1}{R_1}\$

also from one of the loops around V, R3 and R2, I have \$e_1 - 2\beta v_2 + v_2 = 0\$.

I have 3 unknowns so far, so I'm trying to find another equation, but I'm not sure what to do from here.

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  • \$\begingroup\$ Are \$V_1\$ or/and \$V_2\$ given parameters? Otherwise, everything is equal to \$0\$. \$\endgroup\$ – Jan Jan 13 at 16:37
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Homework - no ready to copy solutions will be written, but some guidance.

Connect an ideal voltage source Us to AB and calculate how much current your circuit takes from it. You'll see that the current taken from Us is (QCF) x Us, where (QCF) is a quite complex formula which contains alpha, beta and the resistances R1, R2, R3.

This means that your circuit behaves like a resistor 1/(QCF). That's the Thevenin equivalent resistance. The Thevenin equivalent voltage of your circuit is zero.

This circuit has controlled sources. You cannot use solution methods which remove any of the controlled sources or any of the branches which control the sources. Make straight Kirchoff's current law equation where v2 is the unknown voltage. You can express v1 with Us and v2.

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  • \$\begingroup\$ I hope this isn't too stupid of a question, but when you said "Connect an ideal voltage source Us to AB and calculate how much current your circuit takes from it.", I was wondering how to go about it. Using V/R = I, I will need to find the resistance, but to me, it seems that this resistance I have to find is itself the R_TH I'm looking for \$\endgroup\$ – singularity Jan 12 at 16:56
  • \$\begingroup\$ If you do not know general network solution methods such as making an equation where the sum of all currents which come to the connection node of R1,R2 and R3 from different directions is written to be zero, you are out of luck.You must learn the well known general network laws and how to make equations with them at first. Then you solve v2 as a function of Us. Then you write current =(Us-v2)/R1 and see what I said: The current =(QCF) x Us. \$\endgroup\$ – user287001 Jan 12 at 17:07
  • \$\begingroup\$ Ah, I think you mean the KCL equations, then. From your help I was able to get that v2 = 0.956568... and hence I = 0.000006. Since I let Us = 1, I have R_TH = 1/QCF = 165776 Ohms, or 165.78 kilo Ohms. Is this line of thinking right? \$\endgroup\$ – singularity Jan 12 at 23:42
  • \$\begingroup\$ 165,78 kOhm is OK. You did the calculations with fixed Us=1V and assumed the resulted current from Us is proportional to Us. That's a logical weakness. Full proof needs symbolic Us and showing that I = (QCF) x Us with every Us. \$\endgroup\$ – user287001 Jan 13 at 0:02
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To find thevenin equivalent resistance just short the voltage source and open the current source. You will Get Rth=R3||R2+R1

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  • \$\begingroup\$ so from what I understand, it should be 7.2 + 1/(1/12+ 1/3.3) = 9.78 k ohms, but apparently that is wrong... Did I understand you the wrong way? \$\endgroup\$ – singularity Jan 12 at 15:56
  • \$\begingroup\$ also, I'm not sure if I can just short and open circuits since they're dependent sources.. \$\endgroup\$ – singularity Jan 12 at 16:28

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