Maximum average power for a purely resistive load

1. The problem statement, all variables and given/known data

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 Vmax, not VRMS

2. Relevant equations

j = [itex]\sqrt{-1}[/itex]
V=IZ
I1 = [itex]\frac{R_2*I_s}{R_1+R_2}[/itex]
Zt = Thevenin equivalent impedance = VOC/ISC
P = IRMS2*RL
Power is max when RL = Rs, or ZL = Zs*
Z* = complex conjugate of Z

3. 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:

[itex]\frac{(-j10)(j14)}{-j10+j14}[/itex] = -j35
-j35+46 = 46-j35 = Zt

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

Voltage across capacitor = [itex]\frac{10}{-j10}[/itex] = j
Using the current divider to find the current across the 46Ω resistor,
ISC = [itex]\frac{j14*(j)}{j14+46}[/itex] = -0.278+j*0.8477

Vt = ISC*Zt = (-0.278+j*0.0848)(46-j35) = -9.82+j13.63

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

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

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

P = IRMS2*RL = 0.682 W which is incorrect. The correct answer is 2.95 W. Any ideas?

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