Thévenin equivalent and maximum power (AC): 22-Elec-A1 Circuits, December 2013, Q4

Question 4: Thévenin equivalent and maximum power (AC) [12 + 2 + 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 AC bridge with the source across the left diagonal:

Element	Value	Position
Source	$100\angle30^{\circ}$ V (rms)	node P (+) to node N
Top branch	$6\ \Omega$ then $j10\ \Omega$	P → A → R
Bottom branch	$-j5\ \Omega$ then $5\ \Omega$	N → B → R
Ports	A (top mid), B (bottom mid)	open

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

Figure 4 — With A–B open the network is a single series loop P→A→R→B→N→P; A and B are the tap points whose difference is the open-circuit (Thévenin) voltage.

Approach. With A–B open, one loop current circulates; $V_{th}=V_A-V_B$ follows from the taps. Deactivating the source gives $Z_{th}$ as the parallel of the two half-branches; conjugate matching gives $Z_{load}$ and $P_{max}$.

Loop current (A–B open). The whole bridge is one series loop: $$I=\frac{100\angle30^{\circ}}{(6+j10)+(5-j5)}=\frac{100\angle30^{\circ}}{11+j5}=8.27\angle5.56^{\circ}\ \text{A}.$$
Thévenin voltage. From the taps, $V_A-V_B$ spans the $j10$ and $5\,\Omega$ that lie between A, R and B: $$V_{th}=V_A-V_B=(5+j10)\,I=\boxed{92.5\angle69.0^{\circ}\ \text{V (rms)}}.$$
Thévenin impedance. Short the source (P≡N). From A–B two paths appear in parallel — $(6-j5)$ through the left node and $(5+j10)$ through R: $$Z_{th}=(6-j5)\parallel(5+j10)=\frac{(6-j5)(5+j10)}{11+j5}=\frac{80+j35}{11+j5}=\boxed{7.23-j0.10\ \Omega}.$$
Load for maximum power. Conjugate match: $$Z_{load}=Z_{th}^{*}=\boxed{7.23+j0.10\ \Omega}.$$
Maximum power. With the reactances cancelled the load resistance is $R_{th}=7.23\ \Omega$ and (rms phasors) $$P_{max}=\frac{|V_{th}|^{2}}{4R_{th}}=\frac{(92.5)^{2}}{4(7.23)}=\boxed{296\ \text{W}}.$$

Quantity	Result
Thévenin voltage	$V_{th}=92.5\angle69.0^{\circ}$ V (rms)
Thévenin impedance	$Z_{th}=7.23-j0.10\ \Omega$
Load for max power	$Z_{load}=7.23+j0.10\ \Omega$
Maximum power	$\boxed{P_{max}=296\ \text{W}}$

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-back (voltage sources shorted); matching $Z_{load}=Z_{th}^{*}$; $P_{max}=|V_{th,\text{rms}}|^{2}/(4R_{th})$.

Why this works. For a fixed rms source, delivered power is $P=|V_{th}|^{2}R_L/|Z_{th}+Z_L|^{2}$. Choosing $X_L=-X_{th}$ cancels the reactance, then $dP/dR_L=0$ gives $R_L=R_{th}$; substituting leaves one-quarter of $|V_{th}|^{2}/R_{th}$. Because the source drives a single open-circuit loop, $V_{th}$ is just the impedance-divider voltage between the two taps.

Common pitfall. Using $|V_{th}|^{2}/(8R_{th})$ (that form is for a peak-amplitude phasor, not rms), or matching to $Z_{th}$ instead of its conjugate.

Source. Sadiku §4.5–4.6 and §11.5 (maximum average power transfer in AC circuits).

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

Question 4: Thévenin equivalent and maximum power (AC) [12 + 2 + 6]