AC Steady-State Phasors, Two Sources: 04-BS-4, December 2014, Q4

Question 4: AC Steady-State Phasors, Two Sources (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. $\omega=25\text{ rad/s}$; $v_{s1}$ is IN SERIES with $L_1$ (confirmed against the source drawing – not in parallel, as a coarse reading might suggest), forming the left branch to node 1; $L_2$ hangs from node 1 to ground; $C$ connects node 1 to node 2; $R$ and the ideal source $v_{s2}$ both hang from node 2 to ground (so $v_{s2}$ fixes node 2 directly).

Given data
Quantity	Value
$L_1,\,L_2$	$160\text{ mH},\ 80\text{ mH}$
$R,\,C$	$2\,\Omega,\ 5\text{ mF}$
$v_{s1}(t)$	$\sqrt2\cdot10\cos(25t+45^\circ)$ V
$v_{s2}(t)$	$\sqrt2\cdot8\cos(25t)$ V

[Figure not reproduced. See the official exam paper.]

Find. (a) $\underline Z_{L1},\underline Z_{L2},\underline Z_C$; (b) $\underline V_1$; (c) $\underline I_{L1},\underline I_{L2}$; (d) $i_R(t)$.

Approach. Compute the three phasor impedances at $\omega=25$; note node 2 is fixed by the ideal source $v_{s2}$, so a single KCL equation at node 1 (with the $L_1$-$v_{s1}$ branch modelled as a Thévenin source $v_{s1}$ behind $Z_{L1}$) solves $\underline V_1$ directly.

a) Impedances. $$\underline{Z}_{L1}=j\omega L_1=j(25)(0.16)=\boxed{j4\ \Omega}, \qquad \underline{Z}_{L2}=j\omega L_2=j(25)(0.08)=\boxed{j2\ \Omega}$$ $$\underline{Z}_C=\frac{1}{j\omega C}=\frac{1}{j(25)(0.005)}=\boxed{-j8\ \Omega}$$
b) Node-1 voltage phasor. Node 2 is fixed: $\underline V_2=\underline V_{s2}=8\angle0^\circ$ V. Writing KCL at node 1 with $\underline V_{s1}=10\angle45^\circ$ V (current from the $L_1$ branch entering node 1, plus current from $C$, balancing current leaving through $L_2$): $$\frac{\underline V_{s2}-\underline V_1}{\underline Z_C}=\frac{\underline V_1-\underline V_{s1}}{\underline Z_{L1}}+\frac{\underline V_1}{\underline Z_{L2}}$$ Solving for $\underline V_1$ (collecting all source terms on the left): $$\underline V_1=\frac{\dfrac{\underline V_{s2}}{\underline Z_C}+\dfrac{\underline V_{s1}}{\underline Z_{L1}}}{\dfrac{1}{\underline Z_C}+\dfrac{1}{\underline Z_{L1}}+\dfrac{1}{\underline Z_{L2}}}=\boxed{3.08\angle66.5^\circ\text{ V}}$$
c) Branch current phasors. $$\underline I_{L1}=\frac{\underline V_1-\underline V_{s1}}{\underline Z_{L1}}=\boxed{1.81\angle126.0^\circ\text{ A}}, \qquad \underline I_{L2}=\frac{\underline V_1}{\underline Z_{L2}}=\boxed{1.54\angle{-}23.5^\circ\text{ A}}$$ As a check, $\underline I_C=(\underline V_{s2}-\underline V_1)/\underline Z_C$ equals $\underline I_{L1}+\underline I_{L2}$ exactly (KCL at node 1 closes).
d) Resistor current in time domain. Since node 2 is fixed directly by the ideal source $v_{s2}(t)$, $R$'s current follows Ohm's law with no phasor algebra needed: $$i_R(t)=\frac{v_{s2}(t)}{R}=\frac{\sqrt2\cdot8\cos(25t)}{2}=\boxed{\sqrt2\cdot4\cos(25t)\text{ A}}$$

Final Results – Question 4
Quantity	Value
$\underline Z_{L1},\underline Z_{L2},\underline Z_C$	$j4,\ j2,\ -j8\ \Omega$
$\underline V_1$	$3.08\angle66.5^\circ$ V
$\underline I_{L1}$	$1.81\angle126.0^\circ$ A
$\underline I_{L2}$	$1.54\angle{-}23.5^\circ$ A
$i_R(t)$	$\sqrt2\cdot4\cos(25t)$ A

Topic: Phasor (nodal) analysis of an AC steady-state circuit with two independent sources.

Key relations: $\underline Z_L=j\omega L$, $\underline Z_C=1/(j\omega C)$; KCL in the phasor domain; an ideal voltage source fixes the phasor voltage of the node it terminates on.

Why this works: once every element is replaced by its phasor impedance and every source by its phasor value, the circuit reduces to ordinary (complex) linear algebra — here a single unknown node voltage, because the second node is pinned by an ideal source.

Common pitfall: mis-reading $v_{s1}$ as being in parallel with $L_1$ rather than in series (which would incorrectly fix node 1's voltage directly and eliminate the need for any nodal solve at all); always verify branch series/parallel structure against the original figure before writing the governing equation.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, Ch. 9–10 (Sinusoids and Phasors, Sinusoidal Steady-State Analysis).

Question 4 of 7: AC Steady-State Phasors, Two Sources

Question 4: AC Steady-State Phasors, Two Sources (20 marks)