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eehelp150
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Homework Statement
Homework Equations
The Attempt at a Solution
How do I do Thevenin voltage and impedance with a dependent source? Any hints would be greatly appreciated.
At the "right side" of node V, is it (V+2V)/300 or (V-2V)/300?The Electrician said:You just did a network with a dependent source using nodal analysis. Solve for the voltage across the 300 ohm resistor in this network; that will be Vth.
eehelp150 said:At the "right side" of node V, is it (V+2V)/300 or (V-2V)/300?
Would the nodal equation be:
(V-9)/600 + V/(-j300) + (V+2V)/300 = 0?
No worries. That would just be 3V. Solve for V and multiply by 3The Electrician said:That equation is correct, but you need the voltage on the right side of the dependent source--that will be Vth.
Solving for V in that equation, I get V = 9/5 Volts. Is that correct? What would V300ohm be?gneill said:No worries. That would just be 3V. Solve for V and multiply by 3
eehelp150 said:Solving for V in that equation, I get V = 9/5 Volts. Is that correct? What would V300ohm be?
Interesting. I "re-pressed" the solve button on Wolfram alpha and got:gneill said:No, the voltage should be complex thanks to the capacitor in the circuit.
And that would be a much better result!eehelp150 said:Interesting. I "re-pressed" the solve button on Wolfram alpha and got:
V = (63-18i)/53.
So now I have the voltage across the capacitor and the value of the dependent source (which is simply 2*Vcapacitor). The voltage across the 300ohm resistor is Voc which would be: (Vcapacitor + 2Vcapacitor) for a total of 3 * Vcapacitor right?gneill said:And that would be a much better result!
Right.eehelp150 said:So now I have the voltage across the capacitor and the value of the dependent source (which is simply 2*Vcapacitor). The voltage across the 300ohm resistor is Voc which would be: (Vcapacitor + 2Vcapacitor) for a total of 3 * Vcapacitor right?
How would I find Zth?gneill said:Right.
One method is to find the short circuit current across the output (i.e. the Norton equivalent current). The impedance is the ratio of the Thevenin voltage to the Norton current.eehelp150 said:How would I find Zth?
I ended up getting I1 = -0.10975609756098 +0.01219512195122iThe Electrician said:Replace the 9 volt with a short and add a 1 amp current source at the output. Solve the network for V again. The voltage at the A-B terminals will be 3*V. The value of that voltage will be equal to the resistance at the A-B terminals which will be the desired Rth.
Since you already have an equation for V due to the 9 volt source, it will be a small change to add the 1 amp current source.
V/600 + V/(-j300) + 3V/300 - 1 =0The Electrician said:Neither is right. Let's see the equation you used.
Your original equations was (V-9)/600 + V/(-j300) + (V+2V)/300 = 0
What would it be if you short the 9 volt source and apply 1 amp to the A-B terminals?
redid with wolfram and got it. Thanks!The Electrician said:So is this the equation that gave you the values in post #17? If so, something's wrong with the solution because that's the right equation. Did you use MyAlgebra to solve it?
Here's what I get:
View attachment 107413
Thevenin voltage is the voltage at the output terminals of a linear electrical network, when all the sources and impedances within the network are replaced by an equivalent voltage source and series impedance.
Thevenin voltage can be calculated by short-circuiting the output terminals of the network and finding the voltage across them. This voltage is equivalent to the Thevenin voltage.
Thevenin impedance is the equivalent impedance of a linear electrical network, when all the sources within the network are replaced by an equivalent voltage source and series impedance.
Thevenin impedance can be calculated by first finding the Thevenin voltage and then calculating the ratio of the Thevenin voltage to the current flowing through the short-circuited output terminals.
Thevenin impedance is used in circuit analysis to simplify complex networks into a single equivalent circuit, making it easier to analyze and design circuits. It is also useful in determining maximum power transfer and stability of a circuit.