AC node-voltage analysis and source power: 22-Elec-A1 Circuits, December 2018, Q5

Question 5: AC node-voltage analysis and source power [8 + 6 + 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 — 60 Hz ac network. Source e sets the left node; the series 8Ω+0.016 H links nodes 1 and 2; current source i injects into node 2. Reference is the bottom rail.

Given. $e=40\sqrt2\cos(\omega t+30^\circ)$ (rms phasor $\mathbf E=40\angle30^\circ$ V) drives a source node through $C=0.00133\text{ F}$ to node 1; a $3\,\Omega$ ties node 1 to ground; a series $8\,\Omega+L$ ($L=0.016\text{ H}$) links nodes 1 and 2; a $10\,\Omega$ ties node 2 to ground; and $i=5\sqrt2\cos\omega t$ (rms phasor $\mathbf I=5\angle0^\circ$ A) injects into node 2. At $f=60\text{ Hz}$, $\omega=2\pi(60)=377\text{ rad/s}$.

Quantity	Value	Impedance	Value
$\omega$	$377\text{ rad/s}$	$Z_C=1/(j\omega C)$	$-j1.99\,\Omega\;(\approx-j2)$
$L$	$0.016\text{ H}$	$Z_L=j\omega L$	$+j6.03\,\Omega\;(\approx+j6)$
Branch 1–2	$8\,\Omega+Z_L$	$Z_b$	$8+j6.03\,\Omega$

Find. the node equations, the phasor node voltages $\mathbf V_1,\mathbf V_2$, and the real power delivered by $e$.

Approach. Convert every element to an admittance, write KCL at nodes 1 and 2 (the source node voltage equals $\mathbf E$), solve the $2\times2$ complex system, then compute $\mathbf S=\mathbf E\,\mathbf I_e^{*}$ for the source.

(a) Node equations. With the bottom rail as reference and the source node held at $\mathbf E$, KCL at nodes 1 and 2 reads $$\frac{\mathbf V_1-\mathbf E}{Z_C}+\frac{\mathbf V_1}{3}+\frac{\mathbf V_1-\mathbf V_2}{Z_b}=0,\qquad \frac{\mathbf V_2-\mathbf V_1}{Z_b}+\frac{\mathbf V_2}{10}=\mathbf I,$$ with $Z_C=-j1.99\,\Omega$, $Z_b=8+j6.03\,\Omega$, $\mathbf E=40\angle30^\circ$ V and $\mathbf I=5\angle0^\circ$ A.
Admittance form. Using $Y_C=1/Z_C=j0.501\text{ S}$ and $Y_b=1/Z_b=0.0797-j0.0601\text{ S}$: $$(Y_C+\tfrac13+Y_b)\mathbf V_1-Y_b\mathbf V_2=\mathbf E\,Y_C,\qquad -Y_b\mathbf V_1+(Y_b+\tfrac1{10})\mathbf V_2=\mathbf I.$$
(b) Solve the node voltages. Solving the $2\times2$ complex system: $$\boxed{\mathbf V_1=29.4\angle62.8^\circ\text{ V},\qquad \mathbf V_2=40.9\angle28.0^\circ\text{ V}.}$$
Source current. All of the source current flows through $C$ into node 1: $$\mathbf I_e=\frac{\mathbf E-\mathbf V_1}{Z_C}=11.08\angle73.8^\circ\text{ A (rms)}.$$
Power supplied by $e$. The complex power delivered is $$\mathbf S=\mathbf E\,\mathbf I_e^{*}=(40\angle30^\circ)(11.08\angle{-73.8^\circ})=319.7-j306.8\text{ VA},$$ so the real power supplied is $$\boxed{P_e=\operatorname{Re}\{\mathbf S\}=319.7\text{ W}}$$ (with $306.8\text{ VAR}$ delivered capacitively/leading).

Quantity	Result
Node voltage $\mathbf V_1$	$29.4\angle62.8^\circ\text{ V}$
Node voltage $\mathbf V_2$	$40.9\angle28.0^\circ\text{ V}$
Source current $\mathbf I_e$	$11.08\angle73.8^\circ\text{ A}$
Real power supplied by $e$	$319.7\text{ W}$

Concept & Theory

Topic: phasor node analysis of an ac circuit and complex power.

Key relations: $Z_L=j\omega L$, $Z_C=1/(j\omega C)=-j/(\omega C)$; KCL in admittance form $\mathbf Y\mathbf V=\mathbf I$; complex power $\mathbf S=\mathbf V\mathbf I^{*}$ with rms phasors, $P=\operatorname{Re}\{\mathbf S\}$.

Why this works: in sinusoidal steady state every element becomes an impedance, so the same KCL machinery used for resistive circuits applies to complex admittances. Writing sources as rms phasors makes $\mathbf S=\mathbf V\mathbf I^{*}$ come out in watts and VARs directly.

Common pitfall: using amplitude ($\sqrt2\times$rms) phasors and forgetting the factor of $\tfrac12$ in power, or writing the capacitor admittance as $1/(j\omega C)$ instead of $j\omega C$.

Source: Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., §10.3–10.4 (nodal phasor analysis) and §11.5 (complex power).

Question 5 of 6: AC node-voltage analysis and source power

Question 5: AC node-voltage analysis and source power [8 + 6 + 6]