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22-Elec-A1 Circuits · May 2015

Question 3 of 6: AC nodal analysis with a floating source

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — mesh/supermesh and nodal analysis, first-order RC transients, AC steady-state phasors, complex power, Thévenin’s theorem and maximum-power transfer, and Laplace-domain circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of initial-condition source models in the s-domain.

Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here. A Laplace-transform table is supplied with the paper. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms.

Question 3: AC nodal analysis with a floating source [8 + 6 + 6]

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. $\omega=100\text{ rad/s}$. In phasor form (impedances in $\Omega$):

ElementValueImpedance / phasor
$V_{s1}$ (at node 1)$20\cos(100t+30^\circ)$$20\angle 30^\circ\text{ V}$
$L_1$ + $5\,\Omega$ (1–2)$0.1\text{ H}$$5+j10$
$L_2$ (2–gnd)$0.3\text{ H}$$j30$
$V_{s2}$ (2–3)$15\cos(100t-45^\circ)$$15\angle{-45^\circ}\text{ V}$
$C$ (3–gnd)$0.5\text{ F}$$-j0.02$
$4\,\Omega+L_3$ (3–gnd)$0.2\text{ H}$$4+j20$

Find. The three node voltages and the phasor current delivered through the branch source $V_{s2}$.

[Figure not reproduced: Figure 3 — AC network redrawn. Node 1 is fixed by $V_{s1}$; $V_{s2}$ sits between nodes 2 and 3, so those two form a supernode. See the official exam paper.]

Approach. Node 1 is fixed by $V_{s1}$. The source $V_{s2}$ bridges nodes 2 and 3, so they form a supernode; one KCL equation on the supernode plus the source constraint $V_2-V_3=15\angle{-45^\circ}$ closes the system.

  1. Fixed node. $V_1=20\angle 30^\circ\text{ V}$ (the source ties node 1 to ground).
  2. Supernode KCL (nodes 2&3). Sum of currents leaving through every attached branch is zero:$$\frac{V_2-V_1}{5+j10}+\frac{V_2}{j30}+\frac{V_3}{-j0.02}+\frac{V_3}{4+j20}=0.$$
  3. Source constraint. $\;V_2-V_3=15\angle{-45^\circ}\text{ V}.$
  4. Solve. Eliminating $V_2=V_3+15\angle{-45^\circ}$ and solving the complex pair gives$$V_2=15.0\angle{-45.1^\circ}\text{ V},\qquad V_3=0.05\angle{-74^\circ}\approx 0\text{ V},\qquad V_1=20\angle 30^\circ\text{ V}.$$
  5. Current through $V_{s2}$. By KCL the source current equals the total current leaving node 2 into the rest of the network:$$I_{Vs2}=-\!\left[\frac{V_2-V_1}{5+j10}+\frac{V_2}{j30}\right]=\boxed{\,2.36\angle 15.8^\circ\text{ A}\,}$$
Check / insight. With $C=0.5\text{ F}$ the capacitive branch admittance is $j\omega C=j50\text{ S}$ — effectively a short at $\omega=100\text{ rad/s}$ — so node 3 is pulled almost to ground ($V_3\approx0$). The current through $V_{s2}$ is dominated by that near-short, which is why $|I_{Vs2}|$ is modest despite the large admittance.
QuantityValue
$V_1$$20\angle 30^\circ\text{ V}$
$V_2$$15.0\angle{-45.1^\circ}\text{ V}$
$V_3$$\approx 0.05\angle{-74^\circ}\text{ V}$
$I_{Vs2}$$2.36\angle 15.8^\circ\text{ A}$