Hello, im trying to calculate \( V_o \) using Thevenin-Theorem, so first i calculated the equivalent resistance :
\( R_{eq} = 1+[ -j \Omega // [ 2 \Omega + ( j \Omega // (1 \Omega + 1 \Omega))]] \)

but im stuck in finding the thevenin voltage, any hints on how to do it?

 

The Thevenin resistance is done by setting the source to 0 V and calculating the resistance looking into the output.

You haven't quite done that.

Also, track your units properly!

To calculate the Thevenin voltage, you just find Vo for the circuit as shown -- not modification needed since you aren't treating any of it's components as a load.

 

The Thevenin resistance is done by setting the source to 0 V and calculating the resistance looking into the output.

You haven't quite done that.

Also, track your units properly!

To calculate the Thevenin voltage, you just find Vo for the circuit as shown -- not modification needed since you aren't treating any of it's components as a load.

the thevenin resistance is \( R_th = 1\Omega + [ (-j)\Omega // [2\Omega + (j \Omega // (1 \Omega))]] \\ \)
but \( V_o = \frac{1 \Omega }{1\Omega+R_th} V_th \)

 

the thevenin resistance is \( R_th = 1\Omega + [ (-j)\Omega // [2\Omega + (j \Omega // (1 \Omega))]] \\ \)
but \( V_o = \frac{1 \Omega }{1\Omega+R_th} V_th \)

What is it that you are finding the Thevenin equivalent for? The entire circuit? Or are you treating the right-most resistor as a load and only finding the Thevenin resistance of everything to the left of it?

You can do it either way, as long as you're consistent.

The equation you have here appears to be for the case where that right hand resistor is considered the load. But even then it isn't quite correct. You are saying that the (j1 Ω) impedance, which is the inductor, is in parallel with a (1 Ω) resistance. Which 1 Ω resistance is the inductor in parallel with?

Actually draw the circuit that is being used to determine Rth. That will greatly reduce the likelihood of making mistakes. Then I'd recommend assigning labels to each component, such as R1, R2, R3 to the resistors and Z1 and Z2 to the reactive components, and develop your equation using those. That makes it must easier to verify that your equation is actually correct. Then substitute the specific values when it comes time to compute a numerical result.

Yes, your equation for Vo in terms of Vth is correct (assuming that the right-most resistor is the load), but you still need to find Vth in order to apply it.

Vth is nothing more than the voltage that appears across Vo when the load is removed. So draw the circuit for that case and then solve it -- it's just a slightly simpler circuit that is analyzed just like any other circuit.

 

Okey im terribly sorry i cause you confusion, okey well im trying to find \( V_o \) which is the right most resistance, so the updated thevenin resistance which i gave is valid, all i need to do is find the thevenin voltage , so i can do source transformation?

 

View attachment 309295
Okey im terribly sorry i cause you confusion, okey well im trying to find \( V_o \) which is the right most resistance, so the updated thevenin resistance which i gave is valid, all i need to do is find the thevenin voltage , so i can do source transformation?

The last equation you gave for Rth (should really be Zth) was

\(
R_{th} \; = \; 1\Omega + \left[ \; \left(-j1 \Omega \right) \; || \left[ \; 2\Omega + \left( \; j1 \Omega \; || \; 1 \Omega \; \right) \; \right] \; \right]
\)

Replacing the values with the symbols you've defined above, this becomes (as best I can match up):


\(
Z_{th} \; = \; R_1 + \left[ \; Z_2 \; || \left( \; R_2 + \left(Z_1 || R_3\right) \; \right) \; \right]
\)

Look this equation over very carefully and see if you believe that it is correct.

Note, in particular, that it is claiming that Z1 is in parallel with R3. Do you agree with that?

Always remember that mistakes at this stage can't be fixed later and that any work after this point is wasted effort.

 

salhi, the Thevenin impedance is the impedance seen looking into the output of the network without the load connected (and with the 6 V source replaced with a short), not the impedance at the input of the network seen by the 6 V source.

The expression for Req given in post #1 is for the impedance seen by the 6 V source. The expression for RTh given in post #3 is almost identical to the Req expression with a small change.

 

salhi, the Thevenin impedance is the impedance seen looking into the output of the network without the load connected (and with the 6 V source replaced with a short), not the impedance at the input of the network seen by the 6 V source.

The expression for Req given in post #1 is for the impedance seen by the 6 V source. The expression for RTh given in post #3 is almost identical to the Req expression with a small change.

im sorry i dont really understand wym by impedance seen by the load, you mean V_o? that mean the equivalent impedance of all the components behind the load, no?

 

im sorry i dont really understand wym by impedance seen by the load, you mean V_o? that mean the equivalent impedance of all the components behind the load, no?

The load is the rightmost resistor. Vo is the voltage across that resistor. Vo is a voltage, not a resistor. With the rightmost resistor removed, what is the impedance that would be measured at the place where it was connected.

 

The load is the rightmost resistor. Vo is the voltage across that resistor. Vo is a voltage, not a resistor. With the rightmost resistor removed, what is the impedance that would be measured at the place where it was connected.

it would be \( R_{eq} = (R_1 + Z_1 || [(Z_2||R_3) +R_2] ) \) right?

 

it would be \( R_{eq} = (R_3 + Z_2 || (R_2 +Z_1) ) \) right?

You must replace the 6 volt source with a short circuit, so the first calculation is R3||Z2, then that is in series with R2. Can you continue on at that point?

 

it would be \( R_{eq} = (R_3 + Z_2 || (R_2 +Z_1) ) \) right?

You are still finding the impedance as seen by the supply voltage on the left.

But you looking for the Thevenin equivalent impedance as seen by the load.

I've suggested several times that you redraw the circuit with the source set to zero and the load removed. Then find the resistance that would be seen if a meter were places across the terminals where the load was previously connected.

This is what you would have if you did that:



If you were given this problem, could you find Zth?

 

You are still finding the impedance as seen by the supply voltage on the left.

But you looking for the Thevenin equivalent impedance as seen by the load.

I've suggested several times that you redraw the circuit with the source set to zero and the load removed. Then find the resistance that would be seen if a meter were places across the terminals where the load was previously connected.

This is what you would have if you did that:

View attachment 309735

If you were given this problem, could you find Zth?

look i've edited my answer previously, what about V_th now?

 

it would be \( R_{eq} = (R_1 + Z_1 || [(Z_2||R_3) +R_2] ) \) right?

This is correct. Now you need put the source back in place and calculate the voltage Vo at the terminals where the load resistor used to be (load resistor still not there).

 

look i've edited my answer previously, what about V_th now?

I would recommend against editing prior posts in such a way that it changes the meaning (grammar errors or typos is fine).

Because now, someone coming upon the thread is going to see your prior post and wonder why everyone is telling you that it's wrong when it's correct.

Just make a follow-up post with the corrected answer. Think of a thread as being a transcript of a conversation -- you can't go back and unsay something, you just correct yourself in the here-and-now and move on.

But, yes, now you have it correct.

The next step is to turn all of the sources back on and find the output voltage that appears across the terminals where the load used to be. This is just a straight up circuit analysis problem exactly like any other. Don't let yourself get confused because of words like Thevenin being thrown into the mix. Just analyze the circuit.