Two-Source AC Phasor Network: 04-BS-4, May 2014, Q4

Question 4: Two-Source AC Phasor Network (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. $v_{s1}$ drives node 1 (top of $L_2$/left of $C$) through $L_1$; node 1 also feeds $L_2$ to ground. $v_{s2}$ is an ideal source landing directly on node 2 (right of $C$, top of $R$), so $V_2=v_{s2}$ is fixed outright.

Given data
Quantity	Value
$L_1$, $L_2$	160 mH, 80 mH
$R$	4 Ω
$C$	10 mF
$v_{s1}(t)$	$\sqrt2\,10\cos(25t+\pi/4)$ V
$v_{s2}(t)$	$10\cos(25t)$ V
$\omega$	25 rad/s

Find. $Z_{L1},Z_{L2},Z_C$; the node-1 phasor $V_1$; the inductor current phasors $I_{L1},I_{L2}$; and $i_R(t)$.

[Figure not reproduced: Figure 4 — redrawn from the printed figure, confirming the topology in the original AI-vision caption: $v_{s1}$–$L_1$ feed node 1, which also carries $L_2$ (to ground) and $C$ (to node 2); node 2 carries $v_{s2}$ directly and $R$ (to ground). See the official exam paper.]

Approach. Use the literal peak-phasor convention already printed in the source ($v_{s1}$ carries an explicit $\sqrt2$ factor on its peak amplitude). Node 2 is fixed by the ideal source $v_{s2}$, so only node 1 needs a KCL equation.

(a) Impedances. $\omega=25$ rad/s: $$Z_{L1}=j\omega L_1=j(25)(0.16)=j4\,\Omega\qquad Z_{L2}=j\omega L_2=j(25)(0.08)=j2\,\Omega$$ $$Z_C=\dfrac{1}{j\omega C}=\dfrac{1}{j(25)(0.01)}=-j4\,\Omega$$
(b) Node-1 voltage phasor. Peak phasors: $V_{s1}=\sqrt2(10)\angle45^\circ=10{+}j10$, $V_2=V_{s2}=10\angle0^\circ$ (fixed by the ideal source). KCL at node 1 ($L_1$ in, $L_2$ and $C$ out): $$\dfrac{V_{s1}-V_1}{Z_{L1}}=\dfrac{V_1}{Z_{L2}}+\dfrac{V_1-V_2}{Z_C}$$ Solving, $$\boxed{V_1=5.0\angle90^\circ\text{ V}=j5.0\text{ V (peak)}}$$
(c) Inductor currents. $$I_{L1}=\dfrac{V_{s1}-V_1}{Z_{L1}}=\dfrac{(10{+}j10)-j5}{j4} \;\Rightarrow\;\boxed{I_{L1}=2.795\angle{-63.43^\circ}\text{ A}}$$ $$I_{L2}=\dfrac{V_1}{Z_{L2}}=\dfrac{j5}{j2} \;\Rightarrow\;\boxed{I_{L2}=2.500\angle0^\circ\text{ A}}$$
(d) Resistor current in time domain. Because $V_2=V_{s2}$ exactly, $R$ simply sees the source voltage directly: $$i_R(t)=\dfrac{v_{s2}(t)}{R}=\dfrac{10\cos(25t)}{4} \;\Rightarrow\;\boxed{i_R(t)=2.5\cos(25t)\text{ A}}$$

Final results — Question 4
Quantity	Value
$Z_{L1},Z_{L2},Z_C$	$j4,\ j2,\ {-}j4\ \Omega$
$V_1$	$5.0\angle90^\circ$ V
$I_{L1}$	$2.795\angle{-63.43^\circ}$ A
$I_{L2}$	$2.500\angle0^\circ$ A
$i_R(t)$	$2.5\cos(25t)$ A

Topic: Phasor (frequency-domain) analysis of a two-source AC network at a single frequency.

Key relations: $Z_L=j\omega L$, $Z_C=1/(j\omega C)$; KCL in phasor form at each unknown node; time-domain recovery $i(t)=|I|\cos(\omega t+ \angle I)$.

Why this works: Because $v_{s2}$ lands directly on node 2 with no series impedance, node 2's phasor is not an unknown at all — it equals the source phasor outright. This reduces the KCL system from two unknown node voltages to one, exactly the same "ideal-source-fixes-a-node" idea used in Questions 1 and 2.

Common pitfall: Mixing peak and rms phasor conventions — the $\sqrt2$ printed explicitly in $v_{s1}(t)$ signals its underlying magnitude is already meant as an rms-style "10", scaled up to a 10√2 peak; taking phasor magnitudes directly from the literal $\cos(\cdot)$ amplitudes (peak convention, consistently for both sources) avoids an inconsistent mixed-convention error.

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

Question 4 of 7: Two-Source AC Phasor Network

Question 4: Two-Source AC Phasor Network (20 marks)