Question 4 of 7: AC Steady-State Phasor Analysis with Two Sources

Question 4: AC Steady-State Phasor Analysis with Two Sources (equal value)

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 not reproduced: Figure 4 — redrawn: vs1 in series with L1 forms the left branch to node V1; L2 shunts V1 to ground; C bridges V1 to V2; R and the ideal source vs2 both shunt V2 to ground (so V2 is fixed directly at vs2, independent of R). See the official exam paper.]

Given. $\omega=25$ rad/s (from the cosine arguments). $L_1=0.16$ H, $L_2=0.08$ H, $C=0.01$ F, $R=4\ \Omega$. $v_{s1}(t)=\sqrt2\cdot10\cos(25t+\tfrac{\pi}{4})$ V (peak amplitude $\sqrt2\cdot10=14.14$ V), $v_{s2}(t)=10\cos(25t)$ V (peak amplitude 10 V).

Find. $\underline{Z}_{L1},\underline{Z}_{L2},\underline{Z}_C$; the phasor $\underline{V}_1$; the phasors $\underline{I}_{L1},\underline{I}_{L2}$; and $i_R(t)$.

Approach. Convert each element to its impedance at $\omega=25$ rad/s and each source to a phasor (peak-amplitude convention, $v(t)=\text{Re}\{\underline{V}e^{j\omega t}\}$). Node V2 is fixed directly by the ideal source vs2 (R carries whatever current is needed and does not load V2). Node V1 is found from one KCL equation, since the vs1–L1 branch behaves as a practical source (EMF vs1 in series with impedance $Z_{L1}$) feeding V1 alongside L2 and C.

1. Impedances. $$\underline{Z}_{L1}=j\omega L_1 = j(25)(0.16) = \boxed{j4\ \Omega}\qquad \underline{Z}_{L2}=j\omega L_2 = j(25)(0.08)=\boxed{j2\ \Omega}$$ $$\underline{Z}_C = \dfrac{1}{j\omega C} = \dfrac{1}{j(25)(0.01)} = \boxed{-j4\ \Omega}$$
2. Source phasors. $$\underline{V}_{s1} = 14.14\angle45^\circ = 10+j10\ \text{V},\qquad \underline{V}_{s2}=\underline{V}_2 = 10\angle0^\circ\ \text{V}$$ ($\underline{V}_2$ is fixed by the ideal source vs2 directly, with no series impedance in that branch.)
3. KCL at node V1. Current arriving from V2 through C equals current leaving through the vs1–L1 branch plus through L2: $$\dfrac{\underline{V}_2-\underline{V}_1}{\underline{Z}_C} = \dfrac{\underline{V}_1-\underline{V}_{s1}}{\underline{Z}_{L1}} + \dfrac{\underline{V}_1}{\underline{Z}_{L2}}$$ Solving for $\underline{V}_1$: $$\underline{V}_1 = \dfrac{\dfrac{\underline{V}_2}{\underline{Z}_C}+\dfrac{\underline{V}_{s1}}{\underline{Z}_{L1}}}{\dfrac{1}{\underline{Z}_C}+\dfrac{1}{\underline{Z}_{L1}}+\dfrac{1}{\underline{Z}_{L2}}} = \boxed{5\angle90^\circ\ \text{V}}\ \ (=j5\ \text{V})$$
4. Current phasors. $$\underline{I}_{L1} = \dfrac{\underline{V}_1-\underline{V}_{s1}}{\underline{Z}_{L1}} = \dfrac{j5-(10+j10)}{j4} = \boxed{2.80\angle116.6^\circ\ \text{A}}$$ $$\underline{I}_{L2} = \dfrac{\underline{V}_1}{\underline{Z}_{L2}} = \dfrac{j5}{j2} = \boxed{2.5\angle0^\circ\ \text{A}}$$ Check via KCL: $\underline{I}_C=(\underline{V}_2-\underline{V}_1)/\underline{Z}_C = 1.25+j2.5\ \text{A} = \underline{I}_{L1}+\underline{I}_{L2}$ ✓
5. Resistor current. R shunts the fixed node V2 directly, so $$\underline{I}_R = \dfrac{\underline{V}_2}{R} = \dfrac{10\angle0^\circ}{4} = 2.5\angle0^\circ\ \text{A} \quad\Rightarrow\quad i_R(t) = \boxed{2.5\cos(25t)\ \text{A}}$$