Thevenin's Theorem: AC Circuits with Different Frequencies

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Calculating the Thevenin equivalent for AC circuits with different frequencies involves using superposition, where each voltage source is analyzed separately. For two sources, v1(t) and v2(t), the frequency of interest dictates which source is active while the other is shorted. This results in distinct circuits, each yielding unique open circuit voltages corresponding to their respective frequencies. It is crucial to note that different frequencies cannot be combined, as they produce separate voltage expressions. The frequency notation should also be understood in terms of radians per second, not hertz.
Vishera
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My textbook has this excerpt:

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I am not sure why it refers to Example 10.6 as that is unrelated. I don't understand to be honest. How can you calculate the Thevenin equivalent at different frequencies? Let us say that you have two voltage source: v1(t)=10cos(5t) and v2(t)=10cos(2t). So how do you calculate the Thevenin equivalent at the first frequency of 5 Hz? Do you change v2's frequency to 5 Hz?
 
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I think they mean when you use superposition, you use the voltage source you're intersted in and short the other one.

You will then have two different circuits, each with it's own voltage source and frequency.

Then when you go to find your open circuit voltage...you will end up with two different open circuit voltages with two different frequencies from two different sources. You cannot add different frequencies together.

One open circuit voltage will contain (5t) and the other open circuit voltage wil contain (2t).
You would never interchange these like you suggested and also the 5 in 5t is not hertz...it's 5 radians per second...

Just my first guess...or 2nd or third...
 
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I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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