AC Steady-State Phasor Analysis: 04-BS-4, December 2015, Q4

Question 4: AC Steady-State Phasor Analysis (20 marks)

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. The source v_s(t), R₁, and L₁ are in series, feeding node V₁, where R₂ in series with C forms one parallel branch to ground, and L₂ forms the other. Return current i₁(t) equals the source (series-branch) current.

Given data
L1	L2	R1	R2	C	v_s(t)	ω
160 mH	80 mH	5 Ω	2 Ω	20 mF	√2·10cos(100t) V	100 rad/s

Find. Z_L1, Z_L2, Z_C; the phasors V₁ and I₁ (peak convention, matching the given v_s amplitude); i_C(t).

[Figure not reproduced: Figure 4 (redrawn): source-R1-L1 series feed into the R2+C || L2 parallel pair at node v1. See the official exam paper.]

Approach. Convert every element to its impedance at ω=100 rad/s, reduce the parallel combination (R₂+Z_C) ∥ Z_L2, then apply Ohm's law and the voltage/current divider relations in the phasor domain, using peak-value phasors to match the given v_s(t) amplitude.

(a) Impedances.
$$Z_{L1}=j\omega L_1 = j(100)(0.160) = \boxed{j16\ \Omega}$$$$Z_{L2}=j\omega L_2 = j(100)(0.080) = \boxed{j8\ \Omega}$$$$Z_C = \frac{1}{j\omega C} = \frac{1}{j(100)(0.020)} = \boxed{-j0.5\ \Omega}$$
Reduce the parallel section. R₂+Z_C = 2−j0.5 Ω is in parallel with Z_L2=j8 Ω:
$$Z_{par} = \frac{(2-j0.5)(j8)}{(2-j0.5)+j8} = 2.124+j0.033\ \Omega$$Total impedance seen by the source: $Z_{tot}=R_1+Z_{L1}+Z_{par} = 7.12+j16.03\ \Omega$.
(c) Current phasor I₁. Using the peak source phasor $V_m=10\sqrt{2}\angle0^\circ$ V:
$$I_1 = \frac{V_m}{Z_{tot}} = \frac{14.142\angle0^\circ}{17.55\angle66.04^\circ} = \boxed{0.806\angle{-66.04^\circ}\text{ A}}$$
(b) Voltage phasor V₁.
$$V_1 = I_1\,Z_{par} = (0.806\angle{-66.04^\circ})(2.125\angle0.89^\circ) = \boxed{1.713\angle{-65.15^\circ}\text{ V}}$$
(d) Capacitor branch current and time-domain form. The R₂+C branch carries $I_C=V_1/(R_2+Z_C)$:
$$I_C = \frac{1.713\angle{-65.15^\circ}}{2.062\angle{-14.04^\circ}} = 0.831\angle{-51.11^\circ}\text{ A}$$$$\boxed{i_C(t) = 0.831\cos(100t - 51.11^\circ)\text{ A}}$$(A KCL check confirms $I_1 = I_C + V_1/Z_{L2}$ to within rounding.)

Final results — Question 4 (peak-value phasors)
Quantity	Value
Z_L1	j16 Ω
Z_L2	j8 Ω
Z_C	−j0.5 Ω
I₁	0.806∠−66.04° A
V₁	1.713∠−65.15° V
i_C(t)	0.831cos(100t−51.11°) A

Topic: AC steady-state analysis by phasors and impedance.

Key relations: $Z_L=j\omega L$, $Z_C=1/(j\omega C)$; series/parallel impedance combination exactly mirrors resistive circuits; a phasor $I\angle\theta$ corresponds to $i(t)=I\cos(\omega t+\theta)$ under the peak-value convention.

Why this works: for a linear circuit driven by a single-frequency sinusoid at steady state, every voltage and current is a sinusoid at the same frequency; representing each as a complex phasor converts the differential equations of L and C into ordinary complex algebra (Ohm's law with complex impedance), which is then solved exactly as a resistive network.

Common pitfall: mixing peak and RMS phasor conventions — since v_s(t) here is given directly with its peak amplitude ($\sqrt{2}\cdot 10$), all derived phasors (I₁, V₁, I_C) are consistently peak values too, and converting back to i_C(t) requires no additional √2 factor.

Source: Sadiku, Fundamentals of Electric Circuits, Ch. 9 (Sinusoids and Phasors).

Question 4 of 7: AC Steady-State Phasor Analysis

Question 4: AC Steady-State Phasor Analysis (20 marks)