Thévenin equivalent and maximum power transfer: 22-Elec-A1 Circuits, May 2017, Q5

Question 5: Thévenin equivalent and maximum power transfer [8 + 4 + 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. A $25\angle10^\circ\text{ V rms}$ source feeds a $5\,\Omega$ to node $1$; from node $1$ a capacitor $-j4\,\Omega$ to b and an inductor $j6\,\Omega$ to node a. At a–b: a $5\angle0^\circ\text{ A rms}$ source (into a) in parallel with a $2\,\Omega$ resistor. All sources rms.

Find. $V_{th}$, $Z_{th}$, the matched $Z_L$, and $P_{max}$.

[Figure not reproduced: Figure 5 — redrawn. Open-circuit terminals a–b (load removed). Node 1 sits between the $5\,\Omega$, the $-j4\,\Omega$ (to b) and the $j6\,\Omega$ (to a); at a–b the $5\angle0^\circ$ current source and $2\,\Omega$ are in parallel. See the official exam paper.]

Approach. For $Z_{th}$, deactivate both sources (voltage→short, current→open) and reduce the impedance seen at a–b. For $V_{th}$, solve the open-circuit node voltages (nodes $1$ and a) with both sources active; $\overline{V_{th}}=\overline{V_a}$. Then $Z_L=Z_{th}^{*}$ and $P_{max}=|V_{th}|^2/(4R_{th})$ for rms.

(a) Thévenin impedance. Short the voltage source (so the $5\,\Omega$ runs from node 1 to b) and open the current source. Looking in at a–b: $$Z_{th}=2\,\big\|\,\big[\,j6+(5\,\|\,(-j4))\,\big].$$ With $5\,\|\,(-j4)=\dfrac{5(-j4)}{5-j4}=1.951-j2.439\,\Omega$, the bracket is $1.951+j3.561\,\Omega$, and $$\boxed{Z_{th}=1.44+j0.50\,\Omega=1.53\angle19.3^\circ\,\Omega.}$$
(a) Open-circuit node equations. Ground b. Node 1 and node a (= $V_{th}$), with the $5\angle0^\circ$ source feeding node a: $$\frac{V_1-25\angle10^\circ}{5}+\frac{V_1}{-j4}+\frac{V_1-V_a}{j6}=0,\qquad \frac{V_a-V_1}{j6}+\frac{V_a}{2}=5\angle0^\circ.$$
(a) Solve for $V_{th}=V_a$. The $2\times2$ complex solve gives $$\boxed{\overline{V_{th}}=8.55\angle-22.81^\circ\text{ V (rms)}.}$$
(b) Matched load. Maximum power to a complex load requires the conjugate match: $$\boxed{Z_L=Z_{th}^{*}=1.44-j0.50\,\Omega.}$$
(c) Maximum power. With the match, the reactances cancel and the load sees $R_{th}$; for rms $\overline{V_{th}}$, $$P_{max}=\frac{|V_{th}|^2}{4R_{th}}=\frac{(8.55)^2}{4(1.44)}=\boxed{12.69\text{ W}.}$$

Quantity	Value
$Z_{th}$	$1.44+j0.50\,\Omega$
$V_{th}$ (rms)	$8.55\angle-22.81^\circ\text{ V}$
Matched load $Z_L$	$1.44-j0.50\,\Omega$
$P_{max}$	$12.69\text{ W}$

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

Key relations: $Z_{th}$ = impedance at the terminals with sources deactivated; $V_{th}$ = open-circuit terminal voltage; match $Z_L=Z_{th}^{*}$; $P_{max}=|V_{th,\text{rms}}|^2/(4R_{th})$.

Why this works: Any linear two-terminal AC network looks, from its terminals, like one source $V_{th}$ behind one impedance $Z_{th}$. Choosing $Z_L=Z_{th}^{*}$ cancels the series reactance (resonating it out) so the current is maximised and purely limited by $2R_{th}$; the load then absorbs the same real power that $R_{th}$ does. Using rms phasors, average power is $|V|^2/R$ without a factor of $\tfrac12$, which is why $P_{max}=|V_{th}|^2/(4R_{th})$.

Common pitfall: matching to $Z_{th}$ instead of its conjugate; forgetting the sources here are rms (a peak value would add a spurious factor of $\tfrac12$); and mishandling the deactivation — the current source becomes an open, the voltage source a short.

Source: Sadiku, Fundamentals of Electric Circuits, §10.4 (Thévenin in AC) and §11.5 (maximum average power transfer).

Question 5 of 6: Thévenin equivalent and maximum power transfer

Question 5: Thévenin equivalent and maximum power transfer [8 + 4 + 2 + 6]