Computation of Thevenin Equivalent

In summary, the conversation is about a student seeking help with a circuit problem on a forum. The problem involves finding the absolute value of impedance, the maximum power dissipated on the load, and the current amplitude through the load. The student initially struggles with the second and third questions, but with the help of other forum members and the maximum power transfer theorem, they are able to find the correct solutions.
  • #1
peroAlex
35
4
Hello!
Recently I was going through some old exams and upon encountering this problem (which seemed pretty easy) I got stuck. Exams at my university are composed of individual tasks, each having three subquestions with four plausible answers respectively. Solution sheet gives results only, so there’s no way for me to check where my attempt at solution went wrong. I ask members of this forum for help, maybe someone will see where I made a mistake. Your help is very appreciated, so thank you in advance.

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Homework Statement



Task gives a circuit with following data: ## u(t) = 60 \cos{\omega t} ##, ## R = 10 \Omega ##, ##X_C = -3j \Omega ##, ## X_{L1} = 10j \Omega ## and ##X_{L2} = 20j \Omega ##.

First question asks for absolute value of impedance ##|Z_b|##.
Second question asks for maximum power dissipated on the load ##Z_b##.
Third question asks for current amplitude through load ##Z_b##.

Homework Equations



My university provides a formula sheet, so for this question I believe only ##P_max = \frac{|U_{Th}|^2}{8 \cdot R_{Th}} ## is relevant here. If anyone thinks I’m missing some equation, please mention it in reply.

The Attempt at a Solution



So first of all, computing Thevenin resistance (first question) was pretty simple, just use $$ R_{Th} = jX_{L2} + \frac{ (R+jX_{L1}) (jX_C) }{(R+jX_{L1}) + (jX_C)} = 0.604 + 16.577j \Omega$$ and for the final touch compute absolute value of complex number ## |R_{Th}| = 16.588 \Omega ##. Solution sheet agrees with me.

Now for second and third question I’ve been banging my head for the past 6 hours. Correct result should be that power dissipated on the load is ##P_{max} = 45 W ## and that current amplitude through the load should be ##I_0 = 12.2 A##. My very first attempt was finding current in each loop.

Solving system of equations ## I_1 (R + jX_{L1} + jX_C) + I_2 (jX_C) = -60 ## and ##I_2 (16.588 + jX_{L2} + jX_C) + I_1(X_C) = 0 ## yields two complex currents ##I_1 = -4.088 + 2.680i ## and ##I_2 = -0.606 – 0.118i ##. So using second current (in absolute) times previously computed resistance yields ##U_Th = 10.241 U## and ##P_{max} = 0.790 W##.
I don’t know how to solve this task, been through videos on YouTube, visited sites like Hyper Physics (which helped me compute Thevenin impedance. Could someone please help me? At least a hint or by pointing me in the right direction.

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By the way, I really hope you’re having a lovely Monday!
 

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  • #3
At first the voltage- in rms[root mean square]- has to be 60/sqrt(2) since u(t)=60*cos(wt) 60 it is maximum value Um=sqrt(2)*U.
Second, according to the article presented by cnh1995 the maximum power will be when Xload=-Xth and Rload=Rth.
Iload=Vth/(Zth+Zload)
Iload=-4.95-j7.07 A
Iload =sqrt(4.95^2+7,07^2)= 8.63 A [RMS] or
Iload max=sqrt(2)*8,63=12.20 maxim.
Pload max=0.604*8.63^2=45 W
 
  • Like
Likes peroAlex
  • #4
Thank you so much, to the both of you! I cannot express my gratitude enough! It's been bugging me for a long time... and now as I see how easy it actually is. Thanks again!
 

1. What is the Thevenin equivalent circuit?

The Thevenin equivalent circuit is a simplified circuit model that represents a complex network of resistors, voltage sources, and current sources. It is used to simplify circuit analysis and is based on the principle that any linear circuit can be represented by a single voltage source and a single resistor in series.

2. How is the Thevenin equivalent calculated?

The Thevenin equivalent is calculated by finding the open circuit voltage, which is the voltage across the terminals of the circuit when no load is connected, and the equivalent resistance, which is the resistance seen by the load when all voltage sources are removed and replaced with their internal resistances. The open circuit voltage is the Thevenin voltage, and the equivalent resistance is the Thevenin resistance.

3. When is it useful to use the Thevenin equivalent circuit?

The Thevenin equivalent circuit is useful in circuit analysis when a complex network needs to be simplified to a single voltage source and a single resistor. It is also useful for finding the maximum power transfer theorem and determining the behavior of a circuit at different loads.

4. What are the assumptions made when using the Thevenin equivalent circuit?

The assumptions made when using the Thevenin equivalent circuit are that the circuit is linear, meaning that the voltage and current relationships are constant, and that the circuit is in a steady-state, meaning that the voltages and currents have reached their steady-state values. Additionally, the Thevenin equivalent assumes that the load is resistive.

5. How does the Thevenin equivalent circuit differ from the Norton equivalent circuit?

The Thevenin equivalent circuit is a voltage source in series with a resistor, while the Norton equivalent circuit is a current source in parallel with a resistor. Both models are equivalent and can be used interchangeably in circuit analysis, but the Thevenin equivalent is more commonly used in practical applications.

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