Maximum Average power Transferred to the load impedance

Thread Starter

Inawza

Joined Aug 30, 2010
2
(a) For the circuit shown in Figure Q1(a), determine the following:
(i) The maximum average power transferred to the load impedance.

V=20<0°




Solution:
I've find the ZL which equals to 5.76-j1.68.
To find the Pmax, I've used the (Vth)^2/4Re(ZL)
Vth=20
So,
Pmax= (20)^2 / 4(5.76)
=17.36 W.
Is Vth equals to the sinusoidal peak value? Or do we have to find it using the Thevenin's theorem?
Please verify whether my answer is correct or not.
 

t_n_k

Joined Mar 6, 2009
5,455
You appear to have correctly determined the Thevenin impedance at terminals a-b [5.76-j1.68].

For maximum power transfer one would place a load equivalent to the complex conjugate value ZL=5.76+j1.68

You then have to do a careful circuit analysis to find the load current with the given source voltage. And the Thevenin voltage isn't 20 V by the way. You would use nodal or mesh analysis to find the correct solution which I don't think you have at this stage.
 

t_n_k

Joined Mar 6, 2009
5,455
Don't give up yet. The fact that you found the Thevenin impedance was a good sign.

The Thevenin voltage is that voltage which would appear across terminals a-b with no load connected - basically the voltage across the capacitor with no load.

There are several ways of finding the Thevenin voltage with no load connected

1. Nodal analysis to find Vab
2. Mesh analysis to find the capacitor current and hence the capacitor voltage [=Vab]
3. Repeated application of circuit reduction [simplification] and use of the voltage divider rule to find Vab

Alternatively, you can bypass the derivation of the Thevenin voltage. That's not what the question is asking for. As suggested, set the load ZL to the complex conjugate of the Thevenin equivalent impedance and solve (using mesh analysis say) for the load current with the load connected. The load power will be (Iload)^2*R where R is the purely resistive part of the total complex load ZL - i.e. 5.76Ω.
 
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