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23-Mechatronics-A2 Circuits and Electronics · December 2019

Question 1 of 11: Ladder network — equivalent resistance and branch voltage

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 1: Ladder network — equivalent resistance and branch voltage [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. A DC ladder network between terminals A and B (Figure-1).

ElementValue
Series resistor at A12 Ω
Parallel pair 1 (10 Ω & 5 Ω)recombine at a node
Series resistor (mid)10 Ω
Parallel pair 2 (20 Ω & 25 Ω), $V_1$ across itrecombine at B
Source when switched50 V dc, A(+) to B(−)

Find. $R_{AB}$, and the voltage $V_1$ across the 20 Ω / 25 Ω pair once the 50 V source is applied.

A12ΩI+−Vdc=50VB10Ω5Ω10Ω20Ω25ΩLRV1
Figure-1: 12 Ω in series from A, feeding a parallel pair (10 Ω, 5 Ω), then a 10 Ω series link, then a second parallel pair (20 Ω, 25 Ω) to B. $V_1$ is measured across the second pair.

Approach. Reduce the ladder pair-by-pair from the terminals inward, then use the reduced network to find the total current and back-substitute for $V_1$.

  1. Reduce the first parallel pair. $$10\|5=\frac{10\times5}{10+5}=3.333\ \Omega.$$
  2. Reduce the second parallel pair. $$20\|25=\frac{20\times25}{20+25}=11.111\ \Omega.$$
  3. Add the series links. $$R_{AB}=12+3.333+10+11.111=\boxed{36.444\ \Omega}.$$
  4. Total current once 50 V is applied. $$I=\frac{V_s}{R_{AB}}=\frac{50}{36.444}=\boxed{1.3720\ \text{A}}.$$
  5. Voltage across the second parallel pair. The full current I flows through the 10 Ω link before reaching this pair, so $$V_1=I\times(20\|25)=1.3720\times11.111=\boxed{15.244\ \text{V}}.$$
QuantityResult
$R_{AB}$$\boxed{36.444\ \Omega}$
$V_1$$\boxed{15.244\ \text{V}}$
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