# 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 V_{max}, not V_{RMS}

**2. Relevant equations**

j = [itex]\sqrt{-1}[/itex]

V=IZ

I_{1} = [itex]\frac{R_2*I_s}{R_1+R_2}[/itex]

Z_{t} = Thevenin equivalent impedance = V_{OC}/I_{SC}

P = I_{RMS}^{2}*R_{L}

Power is max when R_{L} = R_{s}, or Z_{L} = Z_{s}^{*}

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 = Z_{t}

Going back to the original circuit and replacing the load resistance with a short circuit to find I_{SC}:

Voltage across capacitor = [itex]\frac{10}{-j10}[/itex] = j

Using the current divider to find the current across the 46Ω resistor,

I_{SC} = [itex]\frac{j14*(j)}{j14+46}[/itex] = -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 = [itex]\frac{-9.82+j13.63}{46-j35+57.8}[/itex] = -0.125+j*0.0893

I_{RMS} = [itex]\sqrt{0.125^2+0.0893^2}[/itex]/[itex]\sqrt{2}[/itex] = 0.1086

P = I_{RMS}^{2}*R_{L} = 0.682 W which is incorrect. The correct answer is 2.95 W. Any ideas?

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