Thévenin equivalent and maximum power transfer (AC): 22-Elec-A1 Circuits, December 2019, Q4

Question 4: Thévenin equivalent and maximum power transfer (AC) [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.

[Figure not reproduced: Figure‑4 — redrawn AC bridge. Arms: $j10\,\Omega$ (L–A), $6\,\Omega$ (A–R), $2\,\Omega$ (L–B), $-j5\,\Omega$ (B–R). The $100\angle25^\circ$ V (rms) source drives the horizontal diagonal L–R; the load hangs on the vertical diagonal A–B. See the official exam paper.]

Given. A four‑arm bridge fed on one diagonal by $V_s=100\angle25^\circ$ V (rms). The arms are $Z_{LA}=j10\,\Omega$, $Z_{AR}=6\,\Omega$, $Z_{LB}=2\,\Omega$, $Z_{BR}=-j5\,\Omega$; the output terminals A (top) and B (bottom) form the other diagonal.

Arm	Impedance	Arm	Impedance
L–A	$j10\,\Omega$	L–B	$2\,\Omega$
A–R	$6\,\Omega$	B–R	$-j5\,\Omega$
Source	$100\angle25^\circ$ V (rms) across L–R

Find. $V_{th}$ and $Z_{th}$ at A–B; the matched $Z_{load}$; and $P_{max}$.

Approach. Each side of the bridge is an independent voltage divider of the source, so $V_{th}=V_A-V_B$ follows by two dividers. Killing the source shorts L to R, after which $Z_{th}$ is two parallel pairs in series. The matched load is $Z_{th}^{*}$ and $P_{max}=|V_{th}|^2/(4R_{th})$ with $V_{th}$ in rms.

Open‑circuit voltage at A. Take $V_L=100\angle25^\circ$, $V_R=0$. The top branch L‑A‑R is a divider: $$V_A=V_L\frac{Z_{AR}}{Z_{LA}+Z_{AR}}=100\angle25^\circ\cdot\frac{6}{6+j10}.$$
Open‑circuit voltage at B. The bottom branch L‑B‑R divider gives $$V_B=V_L\frac{Z_{BR}}{Z_{LB}+Z_{BR}}=100\angle25^\circ\cdot\frac{-j5}{2-j5}.$$
Thévenin voltage. $$V_{th}=V_A-V_B=\boxed{60.51\angle{-145.84^\circ}\ \text{V (rms)}}\;=\;-50.07-j33.98\ \text{V}.$$
Thévenin impedance. Short the source (L ≡ R). Then A sees $j10\,\|\,6$ and B sees $2\,\|\,(-j5)$, in series between A and B: $$Z_{th}=\frac{j10\cdot6}{6+j10}+\frac{2\cdot(-j5)}{2-j5}=\boxed{6.14+j1.96\ \Omega.}$$
Matched load. Maximum power requires the conjugate match $$\boxed{Z_{load}=Z_{th}^{*}=6.14-j1.96\ \Omega.}$$
Maximum power. With $V_{th}$ in rms and the reactances cancelling, $$P_{max}=\frac{|V_{th}|^2}{4R_{th}}=\frac{(60.51)^2}{4(6.14)}=\boxed{149.2\ \text{W}.}$$

Quantity	Result
$V_{th}$	$60.51\angle{-145.84^\circ}\,\text{V (rms)}$
$Z_{th}$	$6.14+j1.96\,\Omega$
$Z_{load}$ (matched)	$6.14-j1.96\,\Omega$
$P_{max}$	$149.2\,\text{W}$

Concept & Theory

Topic: AC Thévenin equivalent and the conjugate‑match maximum‑power theorem.

Key relations: voltage divider $V_x=V\dfrac{Z_2}{Z_1+Z_2}$; $Z_{th}$ from the dead network; $Z_{load}=Z_{th}^{*}$; $P_{max}=\dfrac{|V_{th,\text{rms}}|^2}{4R_{th}}$.

Why this works: the two half‑bridges do not interact when A–B is open, so each output node is a simple divider. Conjugate matching cancels the source reactance so all delivered power is real and dissipated in $R_{load}=R_{th}$, which the theorem shows is the optimum split.

Common pitfall: using $|V_{th}|^2/(8R_{th})$ (that is the peak‑amplitude form); with an rms phasor the denominator is $4R_{th}$. Also, match to the conjugate, not to $Z_{th}$ itself.

Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., §10.4 & §11.5.

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

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