AC bridge — Thévenin equivalent and maximum power transfer: 23-Mechatronics-A2 Circuits and Electronics, December 2019, Q4

Question 4: AC bridge — Thévenin equivalent and maximum power transfer [6+6, 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 (Figure-4), 50 Hz-class rms phasors.

Quantity	Value
Source	$100\angle25^\circ$ V (rms)
Top-left arm (L–B)	$j10\ \Omega$
Top-right arm (B–R)	$6\ \Omega$
Bottom-left arm (L–A)	$2\ \Omega$
Bottom-right arm (A–R)	$-j5\ \Omega$

Find. $V_{th}$ and $Z_{th}$ at terminals A–B; $Z_{load}$ for maximum power transfer; $P_{max}$.

Figure-4: AC bridge, source $100\angle25^\circ$ V (rms) across nodes L–R; arms $j10\,\Omega$ (L–B), $6\,\Omega$ (B–R), $2\,\Omega$ (L–A), $-j5\,\Omega$ (A–R); output terminals A (bottom) and B (top).

Approach. With A–B open, each side of the bridge is a simple impedance divider from L to R; with the source killed (shorted), L and R merge and $Z_{th}$ is the series combination of the two resulting parallel pairs.

Open-circuit voltage. Treating L–R as the divider ends, $$V_B=V_s\frac{Z_{BR}}{Z_{LB}+Z_{BR}},\qquad V_A=V_s\frac{Z_{AR}}{Z_{LA}+Z_{AR}}.$$ Evaluating, $V_B=51.450\angle-34.04^\circ$ V and $V_A=92.848\angle3.20^\circ$ V, so $$V_{th}=V_B-V_A=\boxed{60.51\angle-145.84^\circ\ \text{V (rms)}}.$$
Thévenin impedance. Shorting the source merges L and R, putting $Z_{LB}\|Z_{BR}$ in series with $Z_{LA}\|Z_{AR}$ between A and B: $$Z_{th}=\frac{j10\times6}{6+j10}+\frac{2\times(-j5)}{2-j5}=\boxed{6.136+j1.957\ \Omega}.$$
Matched load. Maximum power transfer requires the conjugate match: $$Z_{load}=Z_{th}^{*}=\boxed{6.136-j1.957\ \Omega}.$$
Maximum power. $$P_{max}=\frac{|V_{th}|^2}{4R_{th}}=\frac{60.51^2}{4\times6.136}=\boxed{149.17\ \text{W}}.$$

Quantity	Result
$V_{th}$	$\boxed{60.51\angle-145.84^\circ\ \text{V}}$
$Z_{th}$	$\boxed{6.136+j1.957\ \Omega}$
$Z_{load}$ (matched)	$\boxed{6.136-j1.957\ \Omega}$
$P_{max}$	$\boxed{149.17\ \text{W}}$

Concept & Theory

Topic. Thévenin equivalent of an AC bridge and conjugate-match maximum power transfer.

Key relations. Divider form $V=V_s\,Z_2/(Z_1+Z_2)$; parallel impedance $Z_1\|Z_2$; conjugate match $Z_L=Z_{th}^*$; $P_{max}=|V_{th}|^2/(4R_{th})$.

Why this works. With no dependent sources, each side of the bridge is an independent voltage divider from the source, so the open-circuit terminal voltage is just the difference of two divider outputs; killing the (independent) source and looking into the terminals gives the network’s own impedance directly, and complex-power theory shows a source delivers maximum average power to a load exactly when the load impedance is the conjugate of the source’s Thévenin impedance.

Common pitfall. Matching $Z_L=Z_{th}$ (the resistive/dc rule) instead of the conjugate $Z_{th}^{*}$ for an AC network with reactance; mixing up which node (A or B) is the divider reference when subtracting.

Source. Sadiku, Fundamentals of Electric Circuits §9.4, §11.5 (AC Thévenin equivalents, maximum average power transfer).

Question 4 of 11: AC bridge — Thévenin equivalent and maximum power transfer

Part A — Circuits

Question 4: AC bridge — Thévenin equivalent and maximum power transfer [6+6, 2, 6]