Thévenin equivalent and maximum power transfer: 22-Elec-A1 Circuits, December 2016, Q5

Question 5: Thévenin equivalent and maximum power transfer [(6+6) + 4 + 4]

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 network: $30\angle0^\circ\text{ V}$ (rms) in series with $5\,\Omega$ to node P; a $j10\,\Omega$ inductor from P to ground; $-j6\,\Omega$ from P to node Q; a $5\angle0^\circ\text{ A}$ (rms) current source injecting into Q; and a $5\,\Omega$ from Q to ground. Terminal a is node Q, b is the bottom rail.

Element	Value (rms sources)
Voltage source	$30\angle0^\circ\text{ V}$
Series R	$5\,\Omega$
Inductor (P–gnd)	$j10\,\Omega$
Capacitor (P–Q)	$-j6\,\Omega$
Current source (into Q)	$5\angle0^\circ\text{ A}$
Shunt R at Q	$5\,\Omega$

Find. $V_{th},Z_{th}$ at a–b; the matched $Z_L$; and $P_{max}$.

[Figure not reproduced: Figure 5 — redrawn. With the load removed at a–b, $V_{th}$ is the open-circuit voltage at Q; $Z_{th}$ is found by killing both sources (short the $30\text{ V}$, open the $5\text{ A}$) and looking in from a–b. See the official exam paper.]

Approach. Find $Z_{th}$ by deactivating the independent sources and reducing to a–b; find $V_{th}$ as the open-circuit node-Q voltage by nodal analysis at P and Q. Then $Z_L=Z_{th}^{*}$ and $P_{max}=|V_{th}|^2/(4R_{th})$ (rms).

$Z_{th}$ — deactivate sources. Short the $30\text{ V}$ (so the $5\,\Omega$ runs P to ground) and open the $5\text{ A}$. Looking from Q, the left path is $-j6$ in series with $5\,\|\,j10$: $$5\,\|\,j10=\frac{5\cdot j10}{5+j10}=4+j2\,\Omega,\qquad -j6+(4+j2)=4-j4\,\Omega.$$
Combine with the $5\,\Omega$ at Q. $$Z_{th}=5\,\|\,(4-j4)=\frac{5(4-j4)}{9-j4}=\boxed{2.68-j1.03\,\Omega}\;(=2.87\angle-21.0^\circ).$$
$V_{th}$ — open-circuit nodal analysis. With a–b open the $5\,\Omega$ at Q carries all of node Q’s current. KCL at P and Q: $$\frac{V_P-30}{5}+\frac{V_P}{j10}+\frac{V_P-V_Q}{-j6}=0,\qquad\frac{V_Q-V_P}{-j6}+\frac{V_Q}{5}=5.$$
Solve for $V_Q=V_{th}$. The complex solve gives $$\boxed{V_{th}=22.70\angle13.66^\circ\text{ V (rms)}}\;=22.06+j5.36\text{ V}.$$
(b) Matched load. Maximum power to a complex load requires the conjugate match $$\boxed{Z_L=Z_{th}^{*}=2.68+j1.03\,\Omega}.$$
(c) Maximum power. The reactances cancel, leaving $R_{th}$ across $R_L=R_{th}$; with rms $V_{th}$ $$P_{max}=\frac{|V_{th}|^2}{4R_{th}}=\frac{(22.70)^2}{4(2.68)}=\boxed{48.08\text{ W}}.$$

Quantity	Value
$Z_{th}$	$2.68-j1.03\,\Omega$
$V_{th}$ (rms)	$22.70\angle13.66^\circ\text{ V}$
Matched load $Z_L$	$2.68+j1.03\,\Omega$
Maximum power $P_{max}$	$48.08\text{ W}$

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

Key relations: $Z_{th}$ from deactivated sources; $V_{th}=V_{oc}$; conjugate match $Z_L=Z_{th}^{*}$; $P_{max}=|V_{th,rms}|^2/(4R_{th})$.

Why this works: Any linear two-terminal AC network looks like one source $V_{th}$ behind one impedance $Z_{th}$. A conjugate load cancels the source reactance so the circuit is purely resistive at the load frequency, maximising real power; the load then sees half of $V_{th}$, giving the $|V_{th}|^2/4R_{th}$ formula. Because the sources are rms, no extra factor of two is needed.

Common pitfall: using $Z_L=Z_{th}$ instead of the conjugate, or inserting a $\tfrac12$ factor that belongs to peak phasors; here the values are rms, so $P_{max}=|V_{th}|^2/4R_{th}$ directly. Also, the current source must be opened (not shorted) when finding $Z_{th}$.

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 [(6+6) + 4 + 4]