Thévenin equivalent and maximum power transfer: 22-Elec-A1 Circuits, May 2018, Q5

Question 5: Thévenin equivalent 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‑5 — AC circuit for Thévenin analysis at terminals a–b (all sources in rms phasors).

Given. All phasors are rms. A $50\angle45^\circ\text{ V}$ source in series with a capacitor $-j4\,\Omega$ feeds central node $M$ (source $+$ taken at the $M$ side). Between $M$ and terminal a sit a $4\,\Omega$ resistor and a $4\angle0^\circ\text{ A}$ source in parallel (arrow toward a). A $6\,\Omega$ runs $M\!\to\!N$, an inductor $j2\,\Omega$ runs $N\!\to\!$b, and an $8\,\Omega$ runs b$\to$reference.

Element	Value (rms)	Element	Value
Source	$50\angle45^\circ\text{ V}$	Series cap	$-j4\,\Omega$
Current source	$4\angle0^\circ\text{ A}$	$R$ ($M$–a)	$4\,\Omega$
$M$–$N$	$6\,\Omega$	$N$–b	$j2\,\Omega$
b–ref	$8\,\Omega$

Find. (a) $V_{th},Z_{th}$; (b) $Z_L$; (c) $P_{max}$.

Check: source polarity. The $50\angle45^\circ$ source’s $+$ terminal is taken at the cap/$M$ side (the source driving the network). a reversed marking would flip the sign of $V_{th}$ but leave $Z_{th}$, $Z_L$ and the magnitude pattern of the method unchanged.

Approach. Get $Z_{th}$ by killing both sources and reducing; get $V_{th}$ from the open-circuit node voltages; then $Z_L=Z_{th}^{*}$ and $P_{max}=|V_{th}|^{2}/(4R_{th})$.

Thévenin impedance (part a). Short the voltage source and open the current source. Looking in from a–b, the $4\,\Omega$ ($a\!-\!M$) is in series with the parallel combination of the two paths from $M$ to b: $(-j4+8)$ through the shorted-source/8 $\Omega$ branch and $(6+j2)$ through the $6\,\Omega$/inductor branch. Since $(8-j4)\,\|\,(6+j2)=\dfrac{(8-j4)(6+j2)}{14-j2}=\dfrac{56-j8}{14-j2}=4\,\Omega$, $$\boxed{Z_{th}=4+4=8\,\Omega\ (\text{purely resistive}).}$$
Node $M$ on open circuit (part a). With a–b open, the $4\,\Omega$ and the $4\angle0^\circ$ source form a closed loop that draws no net current from $M$, so $M$ only feeds the $6\,\Omega\!-\!j2\!-\!8\,\Omega$ string. KCL at $M$: $\dfrac{50\angle45^\circ-V_M}{-j4}=\dfrac{V_M}{14+j2}$, giving $V_M=50\angle45^\circ\cdot\dfrac{14+j2}{14-j2}=50\angle61.26^\circ\text{ V}.$
Terminal voltages (part a). The current source pushes $4\angle0^\circ$ through the $4\,\Omega$, so $V_a=V_M+4\cdot4\angle0^\circ=V_M+16=59.37\angle47.59^\circ\text{ V}$. The string current $I=\dfrac{V_M}{14+j2}$ drops across the $8\,\Omega$ at b: $V_b=8I=28.28\angle53.13^\circ\text{ V}$.
Thévenin voltage (part a). $$\boxed{V_{th}=V_a-V_b=59.37\angle47.59^\circ-28.28\angle53.13^\circ=31.34\angle42.60^\circ\text{ V (rms)}.}$$
Load for maximum power (part b). Maximum power transfer needs the conjugate match; since $Z_{th}=8\,\Omega$ is real, $$\boxed{Z_L=Z_{th}^{*}=8\,\Omega.}$$
Maximum power (part c). With rms phasors and $R_{th}=8\,\Omega$, $$\boxed{P_{max}=\frac{|V_{th}|^{2}}{4R_{th}}=\frac{31.34^{2}}{4\cdot8}=30.7\text{ W}.}$$

Quantity	Result
$Z_{th}$	$8\,\Omega$
$V_{th}$	$31.34\angle42.60^\circ\text{ V (rms)}$
$Z_L$ for max power	$8\,\Omega$
$P_{max}$	$30.7\text{ W}$

Question 5 of 6: Thévenin equivalent and maximum power transfer

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