Thévenin equivalent with a controlled source, and maximum power transfer: 22-Elec-A1 Circuits, Undated paper, Q4

Question 4: Thévenin equivalent with a controlled source, 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.

Figure 4 (rms phasors) — $I_s=5\angle{-30^\circ}\text{ A}$ in parallel with a $5\,\Omega$ (voltage $V_o$ across it); a series $-j2\,\Omega$ and $V_s=10\angle0^\circ\text{ V}$ ($+$ toward A) reach terminal A; the load side carries a $j4\,\Omega$, a dependent current source $4V_o$ (upward, into A) and $Z_L$, all between A and B.

Given (rms phasors). $I_s=5\angle{-30^\circ}\text{ A}$; the $5\,\Omega$ carries $V_o$ ($+$ at the top node, call it $M$, so $V_o=V_M$); series branch $M\!\to\!-j2\,\Omega\!\to\!V_s\!\to\!A$ with $V_s=10\angle0^\circ$ ($+$ at A); at A the branches $j4\,\Omega$, the VCCS $4V_o$ (arrow up = current injected into A) and $Z_L$ all return to B.

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

Check — the controlled source makes this one-port active. With the figure taken literally (VCCS current $4V_o$ injected into node A, $V_o$ measured $+$ at the top of the $5\,\Omega$), the transconductance of $4\text{ S}$ is large enough that the Thévenin resistance comes out with a negative real part: the network can source power. The classical maximum-power-transfer result ($Z_L=Z_{th}^{*}$, $P_{\max}=|V_{th}|^2/4\,\mathrm{Re}\,Z_{th}$ for rms) assumes a passive Thévenin source ($\mathrm{Re}\,Z_{th}>0$); it is reported below with that caveat. Had the controlled-source reference been the opposite sense, one would instead get $Z_{th}=0.249-j0.082\,\Omega$, $Z_L=0.249+j0.082\,\Omega$ and $P_{\max}\approx323\text{ W}$.

Approach. Solve the two-node phasor network for the open-circuit voltage ($V_{th}=V_A$), then short A–B for $I_{sc}$; $Z_{th}=V_{th}/I_{sc}$ (a test source confirms it, since a dependent source is present).

Node equations (A open, $Z_L$ removed). Writing the series-branch current as $\big(V_M-(V_A-V_s)\big)/(-j2)$ and using $V_o=V_M$:$$M:\ 5\angle{-30^\circ}=\frac{V_M}{5}+\frac{V_M-(V_A-10)}{-j2},$$$$A:\ \frac{V_M-(V_A-10)}{-j2}+4V_M=\frac{V_A}{j4}.$$
Open-circuit voltage. Solving the pair gives $V_M=0.64\angle{-153.7^\circ}$ and $$V_{th}=V_A=\boxed{16.70\angle31.0^\circ\text{ V (rms)}}=14.31+j8.61\text{ V}.$$
Short-circuit current. Setting $V_A=0$, $V_{M,sc}=(I_s-10/(-j2))/(1/5+1/(-j2))$ and $I_{sc}=\dfrac{V_{M,sc}+10}{-j2}+4V_{M,sc}=60.6\angle{-123.5^\circ}\text{ A}.$
Thévenin impedance. $$Z_{th}=\frac{V_{th}}{I_{sc}}=\boxed{-0.249+j0.119\,\Omega.}$$ An independent driving-point test (kill $I_s$, short $V_s$, inject $1\text{ A}$ at A) returns the identical value. The negative real part flags an active one-port.
Matched load (part c). The conjugate-match condition is $$Z_L=Z_{th}^{*}=\boxed{-0.249-j0.119\,\Omega.}$$ Note this is a negative resistance: with an active source no passive $Z_L$ attains a finite maximum (see the check note); the conjugate value is the formal optimum of the transfer formula.
Maximum power (part b). The rms transfer formula $P_{L,\max}=\dfrac{|V_{th}|^2}{4\,\mathrm{Re}\,Z_{th}}=\dfrac{16.70^2}{4(-0.249)}$ evaluates to a negative number, i.e. it does not represent power delivered to a passive load — the theorem’s precondition $\mathrm{Re}\,Z_{th}>0$ is violated. The physically meaningful statement is that this terminal pair, as drawn, is a source rather than a sink; a bounded maximum exists only under the alternative sign convention noted above ($\approx323\text{ W}$).

Quantity	Result
Thévenin voltage $V_{th}$	$16.70\angle31.0^\circ\text{ V (rms)}$
Thévenin impedance $Z_{th}$	$-0.249+j0.119\,\Omega$ (active)
Matched load $Z_L=Z_{th}^{*}$	$-0.249-j0.119\,\Omega$
Maximum power to a passive $Z_L$	unbounded / N.A. (active source — see note)

Concept & Theory

Topic: Thévenin reduction of a network containing a dependent source, and the maximum-power-transfer theorem in the phasor domain.

Key relations: $Z_{th}=V_{oc}/I_{sc}$ (or a $1\text{ A}$ test source with independent sources dead) when a controlled source is present; conjugate match $Z_L=Z_{th}^{*}$; for rms phasors $P_{L,\max}=|V_{th}|^2/(4\,\mathrm{Re}\,Z_{th})$ (use $8$ in the denominator for peak-amplitude phasors).

Why this works: a dependent source cannot be “turned off,” so $Z_{th}$ must be found from the terminal behaviour (open/short or a test excitation) rather than by inspection. Maximum power transfer follows from maximising $P=|I_L|^2\mathrm{Re}\,Z_L$ over $Z_L$; the stationary point is the conjugate match — provided $\mathrm{Re}\,Z_{th}>0$ so the source is passive.

Common pitfall: mechanically applying $Z_L=Z_{th}^{*}$ and the power formula without checking the sign of $\mathrm{Re}\,Z_{th}$; a strong controlled source can make the one-port active, where the passive-load maximum is undefined. Source: Hayt, Engineering Circuit Analysis, 9th ed., Ch. 5 (Thévenin with dependent sources) and Ch. 11 (AC maximum power transfer).

Question 4 of 6: Thévenin equivalent with a controlled source, and maximum power transfer

Question 4: Thévenin equivalent with a controlled source, and maximum power transfer [6 + 6 + 2 + 6]