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23-Mechatronics-A2 Circuits and Electronics · December 2018

Question 4 of 11: Thévenin equivalent with a dependent source, maximum power transfer

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

Notes on this paper

Paper: National Exams, December 2018 — 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 (Q7–Q11); 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-order and second-order transients, AC steady-state nodal 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.) — op-amp T-network feedback, diode rectifiers, MOSFET common-gate stages, NMOS inverters, and BJT bias-point analysis.

Reading the figures. The paper supplies the schematics but no separate figure data; every network below is redrawn element-by-element directly from the original drawing (several figures are small hand-style schematics). All node/reference choices and source polarities are stated with the solution. Q11 uses printed variable names (e.g. “VE”) exactly as labelled on the original circuit, even where the labelled terminal is physically the collector in one sub-part — this is called out explicitly where it occurs.

Question 4: Thévenin equivalent with a dependent source, maximum power transfer [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.

ElementValue
Source50 V dc
Series resistor2 Ω
Shunt resistor ($V_o$ across it)5 Ω
Feedback resistor4 Ω
Dependent source$0.5\,V_o$ (current-controlled by $V_o$)

Find. $R_L$ for maximum power transfer at terminals a–b, and $P_{max}$.

+−50V2Ω5Ω (Vo)+−4Ω0.5.Voab
Figure-4: 50 V source, 2 Ω series, node with a 5 Ω resistor ($V_o$ across it) and a 4 Ω resistor to the dependent current source $0.5\,V_o$ (arrow into the 5 Ω node); terminals a–b at the dependent-source side.

Approach. Because the network contains a dependent source, find $V_{oc}$ (a–b open) and $R_{th}$ by killing only the independent source and applying a test source at a–b (the dependent source stays active).

  1. Open-circuit voltage. Nodal analysis at the 5 Ω node ($V_1$) and terminal a ($V_2=V_{oc}$, open so no external current) gives the pair $$\frac{V_1-50}{2}+\frac{V_1}{5}+\frac{V_1-V_2}{4}-0.5V_1=0,\qquad \frac{V_2-V_1}{4}+0.5V_1=0.$$ Solving: $V_1=35.71\ \text{V}$, $$V_{oc}=V_2=\boxed{-35.71\ \text{V}}$$ (the negative sign only fixes the a–b reference polarity; magnitude is what feeds $P_{max}$).
  2. Thévenin resistance. Short the 50 V source (2 Ω now runs from the 5 Ω node to ground) and apply a 1 V test source at a–b; solving the same node equations with the test excitation gives a test current $I_t=0.3889\ \text{A}$, so $$R_{th}=\frac{1}{I_t}=\boxed{2.571\ \Omega}.$$
  3. Maximum power transfer condition. $$R_L=R_{th}=\boxed{2.571\ \Omega}.$$
  4. Maximum power. $$P_{max}=\frac{V_{oc}^2}{4R_{th}}=\frac{35.71^2}{4\times2.571}=\boxed{124.0\ \text{W}}.$$
QuantityResult
Open-circuit voltage$|V_{oc}|=35.71\ \text{V}$
Thévenin/load resistance$\boxed{R_L=2.571\ \Omega}$
Maximum power$\boxed{P_{max}=124.0\ \text{W}}$