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23-Mechatronics-A2 Circuits and Electronics · December 2019

Question 2 of 11: Node-voltage equations with a dependent source

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

Notes on this paper

Paper: National Exams, December 2019 — 16-Mex-A2 Circuits and Electronics. Closed-book, 3-hour paper (approved Casio/Sharp calculator only). Two parts: Part A — Circuits (Q1–Q6) and Part B — Electronics, Questions (1)–(5); candidates normally answer 5 of the 11 questions (3+2 or 2+3 split). Full worked solutions to all eleven questions are given below so the set can be used for study regardless of which five a candidate chose.

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed.) — resistive networks, superposition, Thévenin/max-power transfer, first- and second-order transients, AC phasor analysis; W. H. Hayt et al., Engineering Circuit Analysis (9th ed.) — Laplace-domain circuit models; A. S. Sedra & K. C. Smith, Microelectronic Circuits (8th ed.) — diode limiters, MOSFET biasing and small-signal gain, op-amp slew rate, and CMOS inverter VTC.

Reading the figures. Every network below is redrawn element-by-element from the printed figures. Part B Question 1’s diode/zener network is analysed as the standard bidirectional bridge limiter (current is routed through the same pair of forward diodes for either output polarity, with the zener idle in the direct path); this is flagged as an engineering assumption below since the printed figure does not resolve current directions unambiguously. Part B Question 4 (CMOS inverter VTC) is answered symbolically in terms of the given threshold voltages, since the question supplies no device transconductance parameters.

Part A — Circuits

Question 2: Node-voltage equations with a dependent source [10 + 10]

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.

QuantityValue
Current source into $V_1$5 A
Shunt resistor at $V_1$5 Ω
Link $V_1$–$V_2$ (defines $V_x$, + at $V_1$)6 Ω
Independent source $V_2$ to ground15 V, + at $V_2$
Dependent source $V_2$–$V_3$$2V_x$, + at $V_2$
Shunt resistor at $V_3$5 Ω
Current source $V_3\to V_4$5 A
Shunt resistor at $V_4$2 Ω

Find. The node-voltage equations for $V_1,V_2,V_3,V_4$, and their solved values.

5A5ΩV16Ω+−VxV2+−15V+−2VxV35Ω5AV42Ω
Figure-2: 5 A current source and 5 Ω shunt at $V_1$; 6 Ω link to $V_2$ (control voltage $V_x$); 15 V source fixes $V_2$; dependent source $2V_x$ links $V_2$ to $V_3$; 5 Ω shunt at $V_3$; 5 A source carries current from $V_3$ to $V_4$; 2 Ω shunt at $V_4$.

Approach. Use the two ideal sources directly (one fixes a node voltage, the other fixes a voltage difference) and write ordinary KCL only where an unknown current genuinely needs one.

  1. Node $V_2$ is fixed directly. The 15 V source connects $V_2$ to ground, so $$V_2=\boxed{15\ \text{V}}$$ with no KCL needed there — its branch current is whatever the rest of the network demands.
  2. Node $V_1$ KCL. The 5 A source is the only current entering $V_1$; it leaves via the 5 Ω shunt and the 6 Ω link to $V_2$: $$5=\frac{V_1}{5}+\frac{V_1-V_2}{6}.$$
  3. Solve for $V_1$. Substituting $V_2=15$ and clearing denominators (×30): $150=6V_1+5(V_1-15)=11V_1-75$, so $$V_1=\boxed{20.455\ \text{V}}.$$
  4. Control voltage. $$V_x=V_1-V_2=20.455-15=\boxed{5.455\ \text{V}}.$$
  5. Node $V_3$ from the dependent source. The dependent voltage source directly fixes $V_2-V_3=2V_x$ (no KCL needed to find its value): $$V_3=V_2-2V_x=15-2(5.455)=\boxed{4.091\ \text{V}}.$$
  6. Node $V_4$ KCL. The 5 A source fixes the current arriving at $V_4$, which leaves entirely through the 2 Ω shunt: $$5=\frac{V_4}{2}\ \Rightarrow\ V_4=\boxed{10.0\ \text{V}}.$$
QuantityResult
$V_1$$\boxed{20.455\ \text{V}}$
$V_2$$\boxed{15.0\ \text{V}}$
$V_3$$\boxed{4.091\ \text{V}}$
$V_4$$\boxed{10.0\ \text{V}}$