AC Thévenin equivalent and maximum power transfer: 22-Elec-A1 Circuits, December 2014, Q5

Question 5: AC Thévenin equivalent and maximum power transfer [10 + 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 source $V_s=100\angle30^{\circ}\text{ V}$ feeds two series ladders that meet the open terminals A and B:

Branch	To terminal A (left)	To terminal B (right)
from + rail	$10\,\Omega$ then $-j5\,\Omega$	$4\,\Omega$ then $j8\,\Omega$
from $-$ rail	$5\,\Omega$	$6\,\Omega$

Find. The Thévenin voltage $V_{th}$ and impedance $Z_{th}$ at A–B; the load $Z_{Load}$ for maximum power; and that maximum power.

[Figure not reproduced: Figure-5 (redrawn). Terminal A is the mid-point of the left ladder ($10\,\Omega,-j5\,\Omega$ above; $5\,\Omega$ below) and B the mid-point of the right ladder ($4\,\Omega,j8\,\Omega$ above; $6\,\Omega$ below). Each ladder is a voltage divider across the source. See the official exam paper.]

Approach. Each terminal is the tap of a voltage divider, so $V_{th}=V_A-V_B$ (open circuit). For $Z_{th}$, short the source (the two rails merge) and combine the parallel pairs seen from A and from B, in series. Maximum power to a complex load requires $Z_{Load}=Z_{th}^{*}$, and $P_{max}=|V_{th}|^2/(4R_{th})$ for rms phasors.

Open-circuit node voltages. With A and B open, each ladder is an undivided series divider from the $+$ rail to the $-$ rail. Taking the $-$ rail as reference and $V_s$ on the $+$ rail, the tap voltages (across the lower resistor of each ladder) are $$V_A=V_s\,\frac{5}{10-j5+5}=V_s\,\frac{5}{15-j5}=31.62\angle48.43^{\circ}\ \text{V},$$ $$V_B=V_s\,\frac{6}{4+j8+6}=V_s\,\frac{6}{10+j8}=46.85\angle{-8.66^{\circ}}\ \text{V}.$$
Thévenin voltage (i). $$V_{th}=V_A-V_B=31.62\angle48.43^{\circ}-46.85\angle{-8.66^{\circ}}=\boxed{39.82\angle129.5^{\circ}\ \text{V (rms)}}.$$
Thévenin impedance (i). Kill the source (short the two rails together). Looking into A, the upper arm $(10-j5)$ parallels the lower $5\,\Omega$; looking into B, the upper $(4+j8)$ parallels the lower $6\,\Omega$; the two results are in series between A and B: $$Z_{th}=\big[(10-j5)\|5\big]+\big[(4+j8)\|6\big]=(3.50-j0.50)+(3.80+j1.76),$$ $$Z_{th}=\boxed{7.30+j1.26\ \Omega}=7.41\angle9.76^{\circ}\ \Omega.$$
Load for maximum power (ii). Conjugate match: $$Z_{Load}=Z_{th}^{*}=\boxed{7.30-j1.26\ \Omega}\;(=7.41\angle{-9.76^{\circ}}\ \Omega).$$
Maximum power (iii). With the match, the load resistance equals $R_{th}=7.30\,\Omega$ and the reactances cancel; for rms phasors $$P_{max}=\frac{|V_{th}|^{2}}{4R_{th}}=\frac{39.82^{2}}{4(7.30)}=\boxed{54.26\ \text{W}}.$$

Quantity	Result
$V_{th}$	$\boxed{39.82\angle129.5^{\circ}\ \text{V (rms)}}$
$Z_{th}$	$\boxed{7.30+j1.26\ \Omega}$
$Z_{Load}=Z_{th}^{*}$	$\boxed{7.30-j1.26\ \Omega}$
$P_{max}$	$\boxed{54.26\ \text{W}}$

Concept & Theory

Topic. AC Thévenin equivalent and maximum-power transfer with a complex load.

Key relations. $V_{th}=V_{oc}$; $Z_{th}=$ impedance with sources killed; conjugate match $Z_{Load}=Z_{th}^{*}$; for rms phasors $P_{max}=|V_{th}|^2/(4R_{th})$ (use $|V_{th}|^2/(8R_{th})$ if $V_{th}$ is a peak amplitude).

Why this works. Any linear two-terminal AC network reduces to one source $V_{th}$ behind one impedance $Z_{th}$. Delivered power peaks when the load reactance cancels $X_{th}$ (series resonance, maximising current) and the load resistance equals $R_{th}$ (the standard resistive match). Both conditions together give the conjugate match.

Common pitfall. Using $/(8R_{th})$ when the source is already rms (the source here is explicitly rms, so use $/(4R_{th})$), or matching only the magnitude $|Z_{th}|$ instead of the conjugate. Also remember the open-circuit tap voltage is across the lower divider element.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, §10.6–10.7 and §11.5 (AC Thévenin, maximum average power transfer).

Question 5 of 6: AC Thévenin equivalent and maximum power transfer

Question 5: AC Thévenin equivalent and maximum power transfer [10 + 4 + 6]