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Here is the circuit:

I have to find maximum average power. By $$Pmax= |Vth|^2 /8 Rth $$ For that I need load impedence Z

I correctly found it as,

$$ Z= (j10- j4 + 8)|| 5 =3.421 + j 0.737$$

But I'm unable figure out How to find Vth?

Is this solution correct: open circuit the capacitor then the left side of circuit Is at 8*2=16 Volts . We have to find Vth so by voltage division Vth=16 (5/5+j10)

enter image description here

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    \$\begingroup\$ Note that this circuit is only driven by DC. Think about how that simplifies things, especially the capacitor and inductor. \$\endgroup\$ – Olin Lathrop Nov 1 '16 at 11:05
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    \$\begingroup\$ You just edited the question to make the schematic less readable!!? What the ...? \$\endgroup\$ – Olin Lathrop Nov 1 '16 at 12:31
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    \$\begingroup\$ Oh but why in the heaven's name should this be a DC circuit? What frequency gives a capacitor -j4 ohm of impedance? What about inductance impedance? I'd rather avoid adding confusion to the OP's attempt! BTW to find Vth I'd rather try with a current divider to find currnt thru 5ohm resistor ;) \$\endgroup\$ – carloc Nov 1 '16 at 15:31
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    \$\begingroup\$ @carloc, yes, it is a very misleading question \$\endgroup\$ – Chu Nov 1 '16 at 15:35
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    \$\begingroup\$ Well basically the only misleading thing is Olin's first comment. It is clearly a phasor domain problem. Complex impedances given leave no other chance. I shall say I really can't imagine how something like DC may pop up in mind. \$\endgroup\$ – carloc Nov 2 '16 at 7:23
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OK Maya, we are on the same page now. To get the Thevenin voltage means we need the voltage across the 5 Ohm with no load Because a voltage is desired and the circuit has a current source it seems like KCL is the way to go. Only problem is that a lot of complex calculations, having 3 equations and three unknowns. So let's use the fact that we can get the Thevenin voltage from the impedance and Isc. If we short the 5 Ohm we can use a current divider to solve 2A*(8-j4)/((8-j4)+j10). I know you can do this because you already got found Z. So once you have this in the a+jb form just multiply it time your Z and you have Vth.

Now maybe you can help me. What did you use to write the equations in your question? I'm pretty new to stack exchange EE.

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  • \$\begingroup\$ Ok, but why did you short 5 ohm? \$\endgroup\$ – Maya Nov 3 '16 at 20:07
  • \$\begingroup\$ Just put "$$" signs around any equations you have to write. Like this: $$ equation $$ \$\endgroup\$ – Maya Nov 3 '16 at 20:16
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    \$\begingroup\$ To find the Thevenin and Norton equivalents, we need three things. Voc, Isc, and Zeq. If we have any two we can find the third. Voc is found by opening the circuit at the nodes where we are finding the equivalent. Isc is found by shorting the nodes where we are taking the equivalent. In this case Isc looks easier to find. That is why I shorted circuited it. It is easire to think of in DC circuits. If you graph the equivalent in the VI plane, Voc is the x intercept, Isc is the Y intercept, and the equivalent draws the line between them. \$\endgroup\$ – owg60 Nov 3 '16 at 20:16
  • \$\begingroup\$ If it was a capacitor, would we have shorted it? \$\endgroup\$ – Maya Nov 3 '16 at 20:30
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    \$\begingroup\$ If you change the circuit configuration you change everything. If from the arrow end of the current source the circuit went J10, 5, Zl then you will have a different Z. In this case you should see it would be easier to calculate Voc. Since in this case you just open the load Voc would be 2*(8-j4). Vth would be this Voc in series with the new Z you calculate. The Z would be easier if the 5Ohm was in series instead of parallel too. \$\endgroup\$ – owg60 Nov 3 '16 at 21:10
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Your proposed solution is wrong because you don't treat a capacitor as open when it is not in DC.

In solving for Vth, you could use any circuit analysis method (e.g. current division, voltage division, nodal analysis, etc). You need only to find the voltage across the 5ohm resistor

Remember, in using Thevenin's theorem, you solve for Rth by turning off all independent sources and getting the equivalent resistance as seen by point a and b (which is the 5 ohm resistor in your case). You've done this part right. In solving for Vth, you simply solve for it without turning off any independent sources. Literally, you just solve for Vth using any circuit analysis method.

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