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 and Δ–Y reduction, mesh/nodal 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; a sine (peak) time reference is used in Q4 to match the source $100\sin(2\pi 60\,t)$, and the sources marked rms in Q5 are treated as rms.
Question 2: Node-voltage analysis with a dependent source [8 + 8 + 4]
Figure 2 — four-node circuit with the $2V_x$ voltage-controlled dependent source; ground is the bottom rail.
Given. Reference (ground) is the bottom rail. The controlling voltage is $V_x$, the voltage across the top $2\,\Omega$ resistor with $+$ at node 1, so $V_x=V_1-V_4$. Element list:
Element
Connection
Element
Connection
$10\text{ V}$ src ($+$ at 1)
node 1 → node 2
$1\,\Omega$
node 3 → gnd
$15\text{ V}$ src ($+$ at 2)
node 2 → gnd
$2\,\Omega$ (top, $V_x$)
node 1 → node 4
$4\,\Omega$
node 1 → gnd
$2Vx$ dep. src ($+$ at 3)
node 3 → node 4
$5\,\Omega$
node 2 → node 3
$2\,\Omega$
node 4 → gnd
Find. (a) the node equations; (b) $V_1\ldots V_4$; (c) $P_{5\Omega}$.
Approach. The two ideal sources on the left fix $V_1$ and $V_2$ outright; the dependent source ties nodes 3 and 4 into a super-node whose single KCL equation, together with the control law, closes the problem.
Fix the known nodes (part a). The $15\text{ V}$ source sits between node 2 and ground, so $V_2=15\text{ V}$. The $10\text{ V}$ source ($+$ at node 1) gives $V_1-V_2=10$, hence $V_1=25\text{ V}$. Both are determined without KCL.
Control law and super-node constraint (part a). $V_x=V_1-V_4=25-V_4$. The dependent source between nodes 3 ($+$) and 4 gives
$$V_3-V_4=2V_x=2(25-V_4)\;\Longrightarrow\;V_3=50-V_4.$$
Super-node KCL (part a). Summing the currents leaving the $\{3,4\}$ super-node through every resistor tied to it:
$$\frac{V_3-V_2}{5}+\frac{V_3}{1}+\frac{V_4-V_1}{2}+\frac{V_4}{2}=0.$$
Substituting the known $V_1,V_2$ gives the working equation $\dfrac{V_3-15}{5}+V_3+\dfrac{V_4-25}{2}+\dfrac{V_4}{2}=0.$
Solve (part b). Insert $V_3=50-V_4$ and clear denominators ($\times10$): $2(35-V_4)+500-10V_4+5(V_4-25)+5V_4=0$, i.e. $445-2V_4=0$, so
$$\boxed{V_4=222.5\text{ V},\quad V_3=50-V_4=-172.5\text{ V},\quad V_x=-197.5\text{ V},}$$
with $V_1=25\text{ V}$ and $V_2=15\text{ V}$ already fixed.
Power in the $5\,\Omega$ (part c). The current from node 2 to node 3 is $I_{5}=\dfrac{V_2-V_3}{5}=\dfrac{15-(-172.5)}{5}=37.5\text{ A}$, so
$$\boxed{P_{5\Omega}=I_5^{\,2}(5)=37.5^{2}\cdot5=7.03\times10^{3}\text{ W}.}$$
Check — large magnitudes are real. The controlled source has gain 2 and its control $V_x$ spans nodes 1 and 4, forming positive feedback; the node voltages and the resulting $5\,\Omega$ power are therefore far larger than the $10$–$15\text{ V}$ sources alone would suggest. This is the mathematically exact result of the stated network (check: $V_3-V_4=-395=2V_x$, super-node KCL residual $=0$). $V_x$ is taken across the top $2\,\Omega$ exactly as the figure’s “$+\,V_x\,-$” label shows.
