Question 2 of 6: Node-voltage analysis with a dependent 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) — series/parallel reduction, node/mesh analysis with dependent sources, first-order RC transients, AC phasor analysis, Thévenin equivalents, maximum-power transfer and Laplace (s-domain) analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of super-node bookkeeping and the initial-condition source models used in Q6.
Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here as a study resource. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages of the exam. Phasor magnitudes are written as magnitude $\angle$ angle with angles in degrees; the cosine time reference is used throughout Q4 and Q5 (so $10\sin(377t+90^\circ)=10\cos 377t$ becomes $10\angle 0^\circ$), and the sources marked rms in Q5 are treated as rms.
Question 2: Node-voltage analysis with a dependent source [9 + 6 + 5]
Figure‑2 — Node‑voltage circuit with a voltage‑controlled dependent source 3V₀. Bottom rail is the reference node.
Given. Ground is the bottom rail. $V_o$ is the voltage across the $5\,\Omega$ at node 2, so $V_o=V_2$. The dependent source $3V_o$ has its $+$ terminal at node 2 and its $-$ terminal at the intermediate node $2'$ that feeds the $4\,\Omega$.
Approach. The $50\text{ V}$ source fixes $V_1$; the dependent source ties nodes $2$ and $2'$ into a super-node whose KCL, plus the control law and one KCL at node 3, closes the problem.
Fixed node and control law (part a). The $50\text{ V}$ source between node 1 and ground gives $V_1=50\text{ V}$, and $V_o=V_2$. The dependent source ($+$ at node 2) gives $V_2-V_{2'}=3V_o=3V_2$, i.e. $$V_{2'}=-2V_2.$$
Super-node KCL, nodes $\{2,2'\}$ (part a). Summing currents leaving the super-node through the $2\,\Omega$ (to node 1), the $5\,\Omega$ (to ground) and the $4\,\Omega$ (to node 3): $$\frac{V_2-V_1}{2}+\frac{V_2}{5}+\frac{V_{2'}-V_3}{4}=0.$$
KCL at node 3 (part a). The $5\text{ A}$ source injects into node 3: $$\frac{V_3-V_{2'}}{4}+\frac{V_3}{10}-5=0.$$
Reduce to two unknowns (part b). Put $V_1=50$ and $V_{2'}=-2V_2$ into the two KCLs and clear denominators. The super-node equation becomes $4V_2-5V_3=500$; the node-3 equation becomes $10V_2+7V_3=100$.
Output voltage (part c). $V_o$ is simply the node-2 voltage across the $5\,\Omega$: $$\boxed{V_o=V_2=51.28\text{ V}.}$$
Large magnitudes are correct. A dependent source with gain $3$ whose control $V_o$ sits on its own output node is positive feedback, so the node voltages ($V_3=-172.5\text{-scale}$ currents, $V_{2'}\approx-103\text{ V}$) run well above the $50\text{ V}$ supply. This is exact, not an error — the KCL residual is zero and the control law $V_2-V_{2'}=3V_2$ is satisfied.
