AC Steady-State Phasor Analysis: 04-BS-4, December 2013, Q3

Question 3: 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 two-source AC network of Figure 3 at $\omega=25$ rad/s, in sinusoidal steady state. Peak-phasor (amplitude) convention is used throughout.

Given data
$L_1$	$L_2$	$R$	$C$	$v_{s1}(t)$	$v_{s2}(t)$
160 mH	80 mH	2 Ω	20 mF	$\sqrt2\,10\cos(25t+\tfrac{\pi}{4})$ V	$10\cos(25t)$ V

Find. (a) the impedances $Z_{L1},Z_{L2},Z_C$; (b) the node phasor $V_1$; (c) the inductor currents $I_{L1},I_{L2}$; (d) the resistor current $i_R(t)$ in the time domain.

Figure 3 — Two-source AC network for Question 3 (ω = 25 rad/s).

With $v_{s2}$ an ideal source tied (through its polarity) to node 2, that node voltage is fixed and node 1 is the single unknown. One nodal equation there closes the problem; the branch currents then follow by Ohm's law in phasor form.

(a) Impedances. $$Z_{L1}=j\omega L_1=j4,\quad Z_{L2}=j\omega L_2=j2,\quad Z_C=\frac{1}{j\omega C}=-j2.$$
Source phasors and node 2. $V_{s1}=10+j10$ V and $V_{s2}=10$ V (peak). The drawn polarity of $v_{s2}$ puts node 2 at $V_2=-V_{s2}=-10$ V.
(b) Node 1 phasor. KCL at node 1 ($\,(V_1-V_{s1})/Z_{L1}+V_1/Z_{L2}+(V_1-V_2)/Z_C=0\,$) gives $$\boxed{V_1=30+j10=31.623\,\angle\,18.435^\circ\text{ V}.}$$
(c) Inductor currents. $$I_{L1}=\frac{V_{s1}-V_1}{Z_{L1}}=5\,\angle\,90^\circ\text{ A},\qquad I_{L2}=\frac{V_1}{Z_{L2}}=15.811\,\angle\,-71.565^\circ\text{ A}.$$
(d) Resistor current in time. $I_R=V_2/R=5\,\angle\,180^\circ$ A, so $$i_R(t)=5\cos\!\left(25t+180^\circ\right)\text{ A}.$$

Final results (peak phasors)
Quantity	Value
$Z_{L1},Z_{L2},Z_C$	$j4$, $j2$, $-j2$
$V_1$	$31.623\,\angle\,18.435^\circ$ V
$I_{L1}$	$5\,\angle\,90^\circ$ A
$I_{L2}$	$15.811\,\angle\,-71.565^\circ$ A
$i_R(t)$	$5\cos(25t+180^\circ)$ A

Topic: AC steady-state (phasor) analysis of an RLC network.

Key relations: Impedances $Z_L=j\omega L$, $Z_C=1/(j\omega C)$; phasor KCL $\sum \underline{I}=0$; Ohm's law $\underline{I}=\underline{V}/Z$; time recovery $x(t)=|X|\cos(\omega t+\angle X)$.

Why this works: In steady state every element becomes a complex impedance and the algebra is identical to a DC nodal solve over the complex field. An ideal voltage source pins its node, collapsing the unknowns to one, so a single complex equation delivers $V_1$ and hence every branch current.

Common pitfall: Dropping the source polarity when writing $V_2$ (a sign error there propagates into every current); and confusing RMS with peak amplitude when returning to the time domain.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. 9–10 (phasors, sinusoidal steady-state analysis).

Question 3 of 7: AC Steady-State Phasor Analysis

Question 3: AC Steady-State Phasor Analysis (20 marks)