Thévenin equivalent and maximum power (AC): 22-Elec-A1 Circuits, May 2014, Q4

Question 4: Thévenin equivalent and maximum power (AC) [5 + 5 + 4 + 6]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An rms AC network with terminals A–B:

Element	Value	Position
Current source	$5\angle30^{\circ}$ A rms (↑)	left branch, node L to ground B
$R$ (top)	$10\ \Omega$	node L $\to$ terminal A
Series $C$+$R$	$-j4\ \Omega$ then $6\ \Omega$	node L $\to$ node N
Voltage source	$20\angle0^{\circ}$ V rms	A (+) to N (−)
Series $R$+$L$	$5\ \Omega$ then $j8\ \Omega$	node N $\to$ terminal B (ground)

Find. $V_{th}$, $Z_{th}$, $Z_{load}$ for maximum power, and $P_{max}$.

[Figure not reproduced: Figure 4 — Redrawn network. With A–B open, terminal A connects to the rest only through the 10 Ω and the $20$ V source; terminal B is the bottom rail (the current-source return). See the official exam paper.]

Approach. Take B (bottom rail) as reference. With A–B open, nodal analysis (a supernode across the $20$ V source) gives $V_{th}=V_A$. Deactivating both sources gives $Z_{th}$; conjugate matching gives $Z_{load}$ and $P_{max}$.

Open-circuit (Thévenin) voltage. Label node L (top of the current source), node N (− of the $20$ V source) and A; the $20$ V source sets $V_A-V_N=20$. Writing KCL at node L and at the $\{A,N\}$ supernode (terminal A draws no external current) and solving, $$V_{th}=V_A=\boxed{51.88\angle65.33^{\circ}\ \text{V (rms)}}.$$
Thévenin impedance. Open the current source and short the voltage source ($A\equiv N$). The 10 Ω and the $(6-j4)$ branch then dangle from node A to the now-floating node L (no path to B), so they carry no current; the only path from A to B is the $5+j8$ branch: $$Z_{th}=5+j8\ \Omega\;=\;\boxed{9.43\angle57.99^{\circ}\ \Omega}.$$
Load for maximum power. Conjugate match: $$Z_{load}=Z_{th}^{*}=\boxed{5-j8\ \Omega}.$$
Maximum power. The reactances cancel, leaving $R_{load}=R_{th}=5\ \Omega$; with rms phasors $$P_{max}=\frac{|V_{th}|^{2}}{4R_{th}}=\frac{(51.88)^{2}}{4(5)}=\boxed{134.6\ \text{W}}.$$

Quantity	Result
Thévenin voltage	$V_{th}=51.88\angle65.33^{\circ}$ V rms
Thévenin impedance	$Z_{th}=5+j8\ \Omega$
Load for max power	$Z_{load}=5-j8\ \Omega$
Maximum power	$\boxed{P_{max}=134.6\ \text{W}}$

Concept & Theory

Topic. Thévenin reduction of an AC network and the maximum-average-power-transfer theorem.

Key relations. $V_{th}=V_{oc}$; $Z_{th}$ from the source-free look-in (current sources opened, voltage sources shorted); match $Z_{load}=Z_{th}^{*}$; $P_{max}=|V_{th,\text{rms}}|^{2}/(4R_{th})$.

Why this works. For a fixed rms source, delivered power is $P=|V_{th}|^{2}R_L/|Z_{th}+Z_L|^{2}$; setting $X_L=-X_{th}$ cancels the reactance and $dP/dR_L=0$ gives $R_L=R_{th}$, leaving one-quarter of $|V_{th}|^{2}/R_{th}$. Here the whole left network hangs off node A through a single series source branch, so once that source is shorted only the $5+j8$ path reaches B.

Common pitfall. Using $|V_{th}|^{2}/(8R_{th})$ (that is for a peak-amplitude phasor, not rms), matching to $Z_{th}$ instead of its conjugate, or including the dangling 10 Ω/$(6-j4)$ stub in $Z_{th}$.

Source. Sadiku §4.5–4.6 (Thévenin) and §11.5 (maximum average power transfer).

Question 4 of 6: Thévenin equivalent and maximum power (AC)

Question 4: Thévenin equivalent and maximum power (AC) [5 + 5 + 4 + 6]