Question 5 of 6: Thévenin equivalent and maximum power transfer (AC)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel and Δ–Y reduction, mesh/nodal analysis with dependent sources, first-order RC transients, AC phasor analysis, Thévenin equivalents, maximum-power transfer and Laplace (s-domain) analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of super-node bookkeeping and the initial-condition source models used in Q6.
Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here as a study resource. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages of the exam. Phasor magnitudes are written as magnitude $\angle$ angle with angles in degrees; a sine (peak) time reference is used in Q4 to match the source $100\sin(2\pi 60\,t)$, and the sources marked rms in Q5 are treated as rms.
Question 5: Thévenin equivalent and maximum power transfer (AC) [8 + 4 + 2 + 6]
Figure 5 — AC network with output terminals a–b and load $Z_L$; all phasors rms.
Given. A current source $\overline{I}_s=5\angle0^\circ\text{ A (rms)}$ in parallel with $8\,\Omega$; a series $5\,\Omega$ to node $D$; a capacitor $-j4\,\Omega$ from $D$ to ground; a series inductor $j6\,\Omega$ from $D$ to the output node $a$; from $a$ a $2\,\Omega$ to ground and a branch $1\,\Omega$ in series with $\overline{V}_s=25\angle10^\circ\text{ V (rms)}$ to ground. All phasors rms.
Approach. Get $Z_{th}$ by deactivating both sources and reducing from the terminals; get $V_{th}$ as the open-circuit voltage at $a$ by nodal analysis with both sources live; then apply the conjugate-match rule.
Thévenin impedance, part (a). Deactivate the sources: open $\overline{I}_s$ and short $\overline{V}_s$. Looking in from $D$ toward the left, the $5\,\Omega$ is in series with the now-grounded $8\,\Omega$, giving $13\,\Omega$, in parallel with the $-j4\,\Omega$:
$$13\,\|\,(-j4)=\frac{13(-j4)}{13-j4}=1.124-j3.654\,\Omega.$$
Add the series $j6$ to reach $a$: $j6+(1.124-j3.654)=1.124+j2.346\,\Omega$. This is in parallel with the $1\,\Omega$ (source shorted) and the $2\,\Omega$:
$$\boxed{Z_{th}=\big(1.124+j2.346\big)\,\|\,1\,\|\,2=0.575+j0.120\,\Omega.}$$
Open-circuit voltage, part (a). With the load removed, nodal analysis on nodes $P$ (top of $\overline{I}_s\|8\,\Omega$), $D$ and $a$ (the branch node is fixed at $\overline{V}_s$):
$$\Big(\tfrac15+\tfrac18\Big)V_P-\tfrac15 V_D=5;\quad -\tfrac15 V_P+\Big(\tfrac15+\tfrac1{-j4}+\tfrac1{j6}\Big)V_D-\tfrac1{j6}V_a=0;\quad -\tfrac1{j6}V_D+\Big(\tfrac1{j6}+1+\tfrac12\Big)V_a=\frac{25\angle10^\circ}{1}.$$
Solving, the open-circuit terminal voltage is
$$\boxed{V_{th}=V_a=12.54\angle15.18^\circ\text{ V (rms)}.}$$
Load for maximum power, part (b). Maximum power transfer to a complex load requires the conjugate match:
$$\boxed{Z_L=Z_{th}^{*}=0.575-j0.120\,\Omega.}$$
Maximum power, part (c). With the conjugate match the reactances cancel and the load resistance equals $R_{th}=0.575\,\Omega$; for rms phasors
$$\boxed{P_{max}=\frac{|V_{th}|^{2}}{4R_{th}}=\frac{(12.54)^{2}}{4(0.575)}=68.3\text{ W}.}$$
Check — source polarity. The paper does not print an explicit $\pm$ on $\overline{V}_s$; the natural reading (its labelled phasor $25\angle10^\circ$ with the $+$ terminal at the top of the branch, toward the $1\,\Omega$) is used. Flipping the assumed polarity would shift $V_{th}$ but leaves $Z_{th}$, $Z_L$ and the form of $P_{max}$ unchanged.
