AC nodal analysis and power supplied by a source: 23-Mechatronics-A2 Circuits and Electronics, December 2018, Q5

Question 5: AC nodal analysis and power supplied by a source [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.

Given. $f=60\ \text{Hz}$ ($\omega=2\pi f=377\ \text{rad/s}$); phasors below use rms magnitudes.

Element	Value
Source $e$	$40\sqrt2\cos(\omega t+30^\circ)$ V ⇒ $E=40\angle30^\circ$ V (rms)
Current source $i$	$5\sqrt2\cos\omega t$ A ⇒ $I=5\angle0^\circ$ A (rms)
Capacitor	0.00133 F, in series feeding node 1
Series $R$–$L$ (node 1→2)	8 Ω + 0.016 H
Shunt resistors	3 Ω at node 1, 10 Ω at node 2

Find. The node-voltage equations at nodes 1, 2; the phasor node voltages; and the real power supplied by $e$.

Figure-5: e=40√2 cos(ωt+30°) V through a 0.00133 F capacitor into node 1; 8 Ω+0.016 H in series from node 1 to node 2; 3 Ω shunt at node 1, 10 Ω shunt at node 2, and a 5√2 cosωt A current source into node 2; bottom rail is the reference (60 Hz).

Approach. Convert every element to its impedance at $\omega=377\ \text{rad/s}$, write nodal KCL in phasor form, solve the linear complex system, then use $P=\operatorname{Re}\{E\,I_e^{*}\}$ (rms phasors) for the source power.

Impedances. $$Z_C=\frac{1}{j\omega(0.00133)}=-j1.994\ \Omega,\qquad Z_{RL}=8+j\omega(0.016)=8+j6.03\ \Omega.$$
Node-voltage equations (part a). With $V_1,V_2$ the phasor node voltages, $$\text{Node 1:}\quad \frac{E-V_1}{Z_C}=\frac{V_1}{3}+\frac{V_1-V_2}{Z_{RL}},$$ $$\text{Node 2:}\quad \frac{V_1-V_2}{Z_{RL}}+I=\frac{V_2}{10}.$$
Solve the complex system. Substituting the impedances and clearing to standard form gives $$V_1=29.40\angle62.83^\circ\ \text{V},\qquad V_2=40.89\angle27.98^\circ\ \text{V}.$$
Current supplied by $e$. $$I_e=\frac{E-V_1}{Z_C}=\boxed{11.08\angle73.82^\circ\ \text{A}}.$$
Power supplied by $e$. Using rms phasors directly, $$P_e=\operatorname{Re}\{E\,I_e^{*}\}=\operatorname{Re}\{40\angle30^\circ\times11.08\angle{-73.82}^\circ\}=\boxed{319.7\ \text{W}}.$$

Quantity	Result
Node 1 voltage	$V_1=29.40\angle62.83^\circ\ \text{V}$
Node 2 voltage	$V_2=40.89\angle27.98^\circ\ \text{V}$
Power supplied by $e$	$\boxed{P_e=319.7\ \text{W}}$

Concept & Theory

Topic. Phasor (ac steady-state) nodal analysis and instantaneous-to-average power.

Key relations. $Z_C=1/(j\omega C)$, $Z_L=j\omega L$; nodal KCL in phasor form; average power $P=\operatorname{Re}\{VI^*\}$ using rms phasors (no extra factor of ½ needed once amplitudes are already rms).

Why this works. For sinusoidal steady state, every voltage/current can be represented by a complex phasor and every element by a frequency-dependent impedance, turning the differential circuit equations into an algebraic linear system identical in form to a dc nodal solve, just with complex arithmetic.

Common pitfall. Mixing peak and rms magnitudes inside the same phasor calculation (the source is specified as $\ldots\sqrt2\cos(\cdot)$, i.e. peak $=$ amplitude$\times\sqrt2$, so the rms phasor drops the $\sqrt2$); forgetting that $Z_{RL}$ is a single series impedance, not two separate branches, in the node equation.

Source. Sadiku, Fundamentals of Electric Circuits §9.6–10.2 (phasor nodal analysis, ac power).

Question 5 of 11: AC nodal analysis and power supplied by a source

Question 5: AC nodal analysis and power supplied by a source [8, 6+6]