a) Determine the load impedance for the circuit shown in Fig. that will result in maximum average power being transferred to the load if ω=5krad/s\omega=5 \mathrm{krad} / \mathrm{s}. b) Determine the maximum average power delivered to the load from part (a) if vg=v_g= 80cos5000t V80 \cos 5000 t \mathrm{~V}. c) Repeat part (a) when ZLZ_{\mathrm{L}}. consists of two components from Appendix H\mathrm{H} whose values yield a maximum average power closest to the value calculated in part (b).


Answered 2 years ago
Answered 2 years ago
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In the frequency domain the reactances have the values:

ZL=jωL=j40 ΩZC=jωC=j40 Ω\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 ZLoadZ_{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:

Zth=ZLR+ZC=j4040j40=20j20 Ω\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:

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

b) In the frequency domain Vg=800\mathbf{V_{g}}=80\angle{0^\circ}. The Thevenin voltage at the terminals of ZLoadZ_{Load} (obtained by substituting the load impedance by an open circuit) is the voltage across the resistance because there isn't current flowing through CC. This voltage could be calculated by applying voltage divisor:

Vth=VgRZL+R=80040j40+40=40j40 V=56.5745 V\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*}

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