Question

# a) Determine the load impedance for the circuit shown in Fig. that will result in maximum average power being transferred to the load if $\omega=5 \mathrm{krad} / \mathrm{s}$. b) Determine the maximum average power delivered to the load from part (a) if $v_g=$ $80 \cos 5000 t \mathrm{~V}$. c) Repeat part (a) when $Z_{\mathrm{L}}$. consists of two components from Appendix $\mathrm{H}$ whose values yield a maximum average power closest to the value calculated in part (b).

Solution

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In the frequency domain the reactances have the values:

\begin{align*} \mathbf{Z_L}&=j\omega L=j40\ \mathrm{\Omega}\\ \mathbf{Z_C}&=\cfrac{-j}{\omega C}=-j40\ \mathrm{\Omega} \end{align*}

The impedance necessary to reach maximum average power transfer is the conjugate of the Thevenin impedance seen from the terminals of $Z_{Load}$. This impedance is calculated by eliminating the voltage source, then the inductor stays in parallel with the resistance and this combination is in series with the capacitor:

\begin{align*} \mathbf{Z_{th}}&=\mathbf{Z_L}\|R+\mathbf{Z_C}\\ &=j40\|40-j40\\ &=20-j20\ \mathrm{\Omega} \end{align*}

And, the required load impedance is:

\begin{align*} \boxed{\mathbf{Z_{Load}}=20+j20\ \mathrm{\Omega}} \end{align*}

b) In the frequency domain $\mathbf{V_{g}}=80\angle{0^\circ}$. The Thevenin voltage at the terminals of $Z_{Load}$ (obtained by substituting the load impedance by an open circuit) is the voltage across the resistance because there isn't current flowing through $C$. This voltage could be calculated by applying voltage divisor:

\begin{align*} \mathbf{V_{th}}&=\mathbf{V_{g}}\frac{R}{\mathbf{Z_L}+R}\\ &=80\angle{0^\circ}\frac{40}{j40+40}\\ &=40-j40\ \mathrm{V}\\ &=56.57\angle{-45^\circ}\ \mathrm{V} \end{align*}