Question 4 of 7: AC Steady-State Phasor Analysis

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. Node 1 (v1) joins the vs1-L1 branch, L2, and the left plate of C; node 2 joins the right plate of C, R, and the vs2 branch (vs2's "+" facing the ground-return leg). Both sources share ω=25 rad/s.

Find. Z_L1, Z_L2, Z_C; the node-1 phasor V₁; branch phasors I_L1, I_L2; and i_R(t).

Approach. vs2 is an ideal source with no series impedance in its own branch and its "+" facing the ground-return leg, so it fixes node 2's phasor directly; one KCL equation at node 1 then solves V1, after which every branch current follows directly.

1. Impedances (part a). $$ Z_{L1}=j\omega L_1 = j(25)(0.16) = j4\ \Omega, \qquad Z_{L2}=j\omega L_2 = j(25)(0.08) = j2\ \Omega, $$ $$ Z_C = \frac1{j\omega C} = \frac1{j(25)(0.02)} = \boxed{-j2\ \Omega}. $$
2. Source phasors (amplitude form). $$ \underline{V}_{s1} = 10\sqrt2\angle45^\circ\text{ V}, \qquad \underline{V}_{s2}=10\angle0^\circ\text{ V}. $$ Because vs2's "+" terminal faces the ground-return leg (no impedance in that branch), it fixes node 2 directly: $$ \underline{V}_2 = -\underline{V}_{s2} = 10\angle180^\circ\text{ V}. $$
3. Voltage phasor V1 (part b). KCL at node 1 (currents leaving node 1 via L1 back to the source, L2 to ground, and C toward node 2): $$ \frac{\underline{V}_1-\underline{V}_{s1}}{Z_{L1}} = \frac{\underline{V}_1}{Z_{L2}} + \frac{\underline{V}_1-\underline{V}_2}{Z_C}. $$ Substituting the numeric impedances and solving: $$ \underline{V}_1 = \boxed{-10+j10 = 14.14\angle135^\circ\text{ V}}. $$
4. Current phasors (part c). $$ \underline{I}_{L1} = \frac{\underline{V}_1-\underline{V}_{s1}}{Z_{L1}} = \boxed{j5 = 5\angle90^\circ\text{ A}}, $$ $$ \underline{I}_{L2} = \frac{\underline{V}_1}{Z_{L2}} = \boxed{5+j5 = 7.07\angle45^\circ\text{ A}}. $$
5. Resistor current, time domain (part d). $$ \underline{I}_R = \frac{\underline{V}_2}{R} = \frac{10\angle180^\circ}{2} = 5\angle180^\circ\text{ A} \quad\Rightarrow\quad i_R(t) = \boxed{5\cos(25t+180^\circ)\text{ A} = -5\cos(25t)\text{ A}}. $$