Maximum average power for a purely resistive load

[zealeth]

Homework Statement

Consider the circuit shown in the figure below. Suppose that R = 46Ω and Z=j14Ω. Determine the maximum average power that can be delivered to the load if the load is pure resistance. Note that the voltage source magnitude is given as V_max, not V_RMS

Homework Equations

j = \sqrt{-1}
V=IZ
I₁ = \frac{R_2*I_s}{R_1+R_2}
Z_t = Thevenin equivalent impedance = V_OC/I_SC
P = I_RMS²*R_L
Power is max when R_L = R_s, or Z_L = Z_s^*
Z^* = complex conjugate of Z

The Attempt at a Solution

Shorting the voltage source to find the thevenin equivalents, I combine the capacitor and inductor in parallel then combine that value with the resistor:

\frac{(-j10)(j14)}{-j10+j14} = -j35
-j35+46 = 46-j35 = Z_t

Going back to the original circuit and replacing the load resistance with a short circuit to find I_SC:

Current across capacitor = \frac{10}{-j10} = j
Using the current divider to find the current across the 46Ω resistor,
I_SC = \frac{j14*(j)}{j14+46} = -0.278+j*0.8477

V_t = I_SC*Z_t = (-0.278+j*0.0848)(46-j35) = -9.82+j13.63

For max power, R_L = |Z_t| = |46-j35| = sqrt(46^2+35^2) = 57.8

The current through the thevenin equivalent circuit with R_L attached is:

I = \frac{-9.82+j13.63}{46-j35+57.8} = -0.125+j*0.0893
I_RMS = \sqrt{0.125^2+0.0893^2}/\sqrt{2} = 0.1086

P = I_RMS²*R_L = 0.682 W which is incorrect. The correct answer is 2.95 W. Any ideas?

 

[gneill]

Your Thevenin impedance looks good, but I'm not liking your Thevenin voltage or the short circuit current. How did you conclude that the voltage across the capacitor is j when ZL is replaced by a short? Shorting RL places the resistor R in parallel with the inductor, it doesn't ground the top of the R.

 

[zealeth]

Thanks for the reply.

Your Thevenin impedance looks good, but I'm not liking your Thevenin voltage or the short circuit current.

Sorry, I meant to say that was the *current* across the capacitor, not the voltage.

Shorting RL places the resistor R in parallel with the inductor, it doesn't ground the top of the R.

So now repeating the same method starting from I_SC while combining R and Z:

R||Z = 46*j14/(46+j14) = 3.9+j12.8
R||Z + C = 3.9+j12.8 -j10 = 3.9+j2.8
I_SC = V/R = 10/(3.9+j2.8) = 1.692-j1.215
V_t = I_SC*Z_t = (1.692-j1.215)(46-35j) = 35.31-j115.1
R_L is still 57.8
Current through Thevenin equiv. circuit with R_L = I = (35.31-j115.1)/(46-j35+57.8) = 0.641-j*0.893
I_RMS = sqrt(0.641^2+0.893^2)/sqrt(2) = 0.777
P = 0.777^2*57.8 = 34.9W which is still incorrect.

 

[gneill]

Isc will be just the current through the resistor. You've taken it to be the total current produced by the source voltage driving the total impedance.

If I may suggest, since you've found the Thevenin Impedance easily enough (and it is correct), why not find the Thevenin Voltage as the open circuit voltage rather than by way of the short circuit current and impedance? With the load removed, the open circuit voltage at the terminals is the result of the action of the voltage divider created by the capacitor and inductor. Easy!

 

[zealeth]

Isc will be just the current through the resistor. You've taken it to be the total current produced by the source voltage driving the total impedance.

Isn't that what I did the first time, just with the mistake of me writing voltage instead of current?

If I may suggest, since you've found the Thevenin Impedance easily enough (and it is correct), why not find the Thevenin Voltage as the open circuit voltage rather than by way of the short circuit current and impedance? With the load removed, the open circuit voltage at the terminals is the result of the action of the voltage divider created by the capacitor and inductor. Easy!

Just tried that, V_OC = 35V which when plugged into the following equations gives me the right answer. However, I've noticed that this is very close to the real part of the V_t value I obtained in my first reply to you. Since it's asking for a purely resistive power, was I supposed to take just the real part of that voltage or is that just a coincidence that the values are similar? Nonetheless, thank you very much for your help.

 

[gneill]

Isn't that what I did the first time, just with the mistake of me writing voltage instead of current?

I don't think so. The source will see a total impedance of ZC + (Z||R), while you wrote that the capacitor current is 10/(-10j). That would only be true of the capacitor alone was across the source.

Just tried that, V_OC = 35V which when plugged into the following equations gives me the right answer. However, I've noticed that this is very close to the real part of the V_t value I obtained in my first reply to you. Since it's asking for a purely resistive power, was I supposed to take just the real part of that voltage or is that just a coincidence that the values are similar? Nonetheless, thank you very much for your help.

I think that's a coincidence.

 

[szynkasz]

...Just tried that, V_OC = 35V which when plugged into the following equations gives me the right answer...

Because it is 35\,V:
V_t=V_{oc}=10\cdot\frac{14j}{14j-10j}=35