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22-Elec-A1 Circuits · December 2015

Question 2 of 6: AC nodal analysis with a floating source (supernode)

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) — nodal/mesh analysis, first- and second-order transients, AC steady-state phasors, complex power and power-factor correction, Thévenin’s theorem and maximum-power transfer, and Laplace-domain analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the source-free RLC circuit and s-domain transfer functions.

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 and a star–delta conversion table are supplied on the last two pages. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms and a common cosine time reference (ω = 377 rad/s) is used throughout Q2.

Question 2: AC nodal analysis with a floating source (supernode) [10 + 6 + 4]

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. Two nodes $V_1$ and $V_2$ at $\omega=377\text{ rad/s}$.

ElementValuePlacement
$v_{s1}$$20\cos(377t+30^\circ)$through $5\,\Omega$ to node $V_1$
$v_{s2}$$30\sin(377t+45^\circ)$floating, between $V_1(+)$ and $V_2(-)$
$L$$0.016\text{ H}$$V_1$ to ground
$R$$4\,\Omega$$V_2$ to ground
$C$$0.0013\text{ F}$$V_2$ to ground

Find. The node-voltage equations, the phasors $V_1,V_2$, and their time-domain expressions.

[Figure not reproduced: Figure 2 — redrawn. $v_{s1}$ drives $V_1$ through $5\,\Omega$; the $v_{s2}$ source floats between $V_1$ and $V_2$, so those two nodes form a supernode. See the official exam paper.]

Approach. Convert both sources to a common cosine reference, replace each element by its phasor impedance, and — because $v_{s2}$ floats between the two nodes — treat $V_1,V_2$ as a supernode (one KCL equation) closed by the source constraint $V_1-V_2=V_{s2}$.

  1. Reference the sources to cosine. Leaving $v_{s1}$ as is and shifting the sine by $-90^\circ$:$$V_{s1}=20\angle 30^\circ\text{ V},\qquad V_{s2}=30\sin(377t+45^\circ)=30\cos(377t-45^\circ)=30\angle{-45^\circ}\text{ V}.$$
  2. Element impedances at $\omega=377$.$$Z_L=j\omega L=j(377)(0.016)=j6.03\,\Omega,\qquad Z_C=\frac{1}{j\omega C}=\frac{1}{j(377)(0.0013)}=-j2.04\,\Omega.$$ Their admittances are $Y_L=1/Z_L=-j0.166\text{ S}$ and $Y_C=j\omega C=j0.490\text{ S}$.
  3. Supernode constraint (part i). The floating source ties the two node voltages:$$\boxed{\,V_1-V_2=V_{s2}=30\angle{-45^\circ}\,}$$
  4. Supernode KCL (part i). Sum the currents leaving the combined $V_1$–$V_2$ surface through the $5\,\Omega$, the inductor, the $4\,\Omega$ and the capacitor:$$\frac{V_1-V_{s1}}{5}+\frac{V_1}{j6.03}+\frac{V_2}{4}+\frac{V_2}{-j2.04}=0.$$ Grouping admittances gives the phasor node equations$$\left(0.2-j0.166\right)V_1+\left(0.25+j0.490\right)V_2=\tfrac{1}{5}V_{s1}=4\angle 30^\circ .$$
  5. Solve the 2×2 system (part ii). Substituting $V_1=V_2+30\angle{-45^\circ}$ into the KCL equation and solving the resulting complex linear equation:$$\boxed{\,V_1=36.84\angle{-15.47^\circ}\text{ V},\qquad V_2=18.27\angle{38.55^\circ}\text{ V}\,}$$ (verified by back-substitution — the KCL residual is zero).
  6. Time-domain form (part iii). With the cosine reference restored:$$v_1(t)=36.84\cos(377t-15.47^\circ)\text{ V},\qquad v_2(t)=18.27\cos(377t+38.55^\circ)\text{ V}.$$
QuantityValue
$Z_L,\;Z_C$$j6.03\,\Omega,\ -j2.04\,\Omega$
$V_1$$36.84\angle{-15.47^\circ}\text{ V}$
$V_2$$18.27\angle{38.55^\circ}\text{ V}$
$v_1(t)$$36.84\cos(377t-15.47^\circ)\text{ V}$
$v_2(t)$$18.27\cos(377t+38.55^\circ)\text{ V}$