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The given question : enter image description here

I went on to get Rth by adding a voltage of 1v to control the dependent source from which I obtain the current in the branches and get the total current , as follow then later decide it by 1 since 1/I =v by ohms law to get Rth:

enter image description here

But in the the answer of the text book I’m using ((fundamentals of electric circuits 5th ed chapter 4 practical problem 13)) , no 1v voltage (or current ) were introduced and he solved as : enter image description here enter image description here

Which I found very confusing.

So the bottom line is : When and when not are voltage sources introduced in the analysis to get Rth when there’s a dependent source ?

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2 Answers 2

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So the bottom line is : When and when not are voltage sources introduced in the analysis to get Rth when there’s a dependent source ?

Both should work. As a summary, you can find the \$R_{TH}\$ by doing one of the following:

Option 1: If all the sources in your circuit are independent (not in this case since there is a dependent one), you just deactivate them and find the equivalent resistance of the network you are left with. By deactivating, voltage sources become shorts, and current sources become open circuits.

Option 2: You can find the short circuit current, (\$I_{sc}\$), flowing through the terminals of interest and to find \$R_{TH}\$, you compute \$R_{TH}=\dfrac{V_{TH}}{I_{sc}}\$. In this case, you don't deactivate any of the sources.

Option 3: This is what you are doing, but you are making a mistake. You add a test source across the terminals of interest, but you have to deactivate the independent sources, and leave the dependent ones untouched. So, in your method, you need to make the 9V source a short circuit (0V).

Both option 2 and 3 should give you the right result.

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People often forget that superposition theorem also applies with dependent sources as demonstrated in several papers. It truly helps in some cases like in this one with a controlled source. The first thing is to start with the control variable, \$v_x\$ in this example. Using the labels I associate in the below sketch, a few lines get you there:

\$v_x=R_1\frac{V_1-3v_x}{R_1+R_2}\$ and solving for \$v_x\$ leads to \$v_x=\frac{R_1V_1}{4R_1+R_2}=2\;V\$ with \$V_1=9\;V\$. Now that you have the control variable, apply superposition by letting \$V_1\$ be 0 V (replace it by a short) and you are left with the \$3v_x\$ source with the \$v_x\$ we have determined. You calculate \$V_{o1}\$. Then, reduce to 0 V the \$3v_x\$ source and calculate \$V_{o2}\$. Summing up both results lead you to \$V_th\$. If everything goes well, you should find \$V_{th}=\frac{V_1(3R_1+R_2)}{4R_1+R_2}=7\;V\$. A quick sim confirms this result:

enter image description here

For the Thévenin resistance, short the output and determine the short-circuit current. You may calculate the 3 currents and determine in this configuration the voltage \$v_x\$ which is equal to \$v_x=\frac{R_1V_o(R_2+R_3)}{3R_1R_3+R_2R_3}\$. \$V_0\$ is the voltage at the resistances junction and is defined as \$V_0=V_1-v_x\$. Replacing \$v_x\$ by its definition and solving for \$V_o\$ gives the final short-circuit current equal to \$I_{sc}=\frac{V_o}{R_3}=1.658\;A\$. The Thévenin resistance is then defined as \$R_{th}=\frac{V_{th}}{I_{sc}}=\frac{7}{1.658}=4.222\;\Omega\$. A quick sim will let you know if this final value is ok; install a 1-A current source and ask the simulator to determine the transfer function from node 4 to the 1-A source:

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The simulator gives you:

***** SMALL SIGNAL DC TRANSFER FUNCTION

output_impedance_at_V(4)        4.222222e+000
i1#Input_impedance      4.222222e+000
Transfer_function       4.222222e+000

which is what we found. Trying to simplify these problems by breaking them into small pieces is the divide and conquer method promoted by the FACTs. Learn how they work and solve transfer functions in a swift and efficient manner. The below Mathcad sheet summarizes the calculations:

enter image description here

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