Question 2 of 7: Thévenin Equivalent and Maximum Power Transfer
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2015 — 04-BS-4 Electric Circuits and Power, 3 hours, closed book (one aid sheet permitted, Casio/Sharp approved calculators). The paper instructs "any five questions constitute a complete paper" and marks only the first five answered; every question is solved so the set stands as a complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits (6th ed.) — DC/AC circuit analysis, Thevenin/Norton, first-order transients, phasors, magnetic circuits, diode rectifiers; Mano & Ciletti, Digital Design — combinational logic design (Question 7).
Check: several figure dimensions/polarities in this paper are read directly from the printed figure (vision-caption topology is a known failure mode on this subject); the specific reads are flagged inline where they affect a result (Q2 Thevenin distractor branch, Q5 mean magnetic path lengths, Q7 "desired temperature reached" logic).
Question 2: Thévenin Equivalent and Maximum Power Transfer (20 marks)
Given. The printed figure shows the ideal source Vs connected directly between the top rail (node Q) and ground, in parallel with R4 — so Vs pins VQ = 20 V regardless of R1, R2, R3, R4 (this left-hand branch, R1–R2–R3–R4, is a deliberate distractor for the Thévenin quantities). From Q, R5 leads to a mid-node M, where R6 returns to ground and R7 continues to the load terminals.
Given data
R1
R2
R3
R4
R5
R6
R7
Vs
RL
12.5 MΩ
22.5 kΩ
300 kΩ
100 kΩ
10 kΩ
10 kΩ
5 kΩ
20 V
100 Ω
Find. Rth, Vth at the load terminals; power delivered to RL=100 Ω; the load resistance and power at maximum power transfer.
[Figure not reproduced: Figure 2 (redrawn): R1/R2/R3/R4 form a distractor branch that does not affect the load-side Thévenin values. See the official exam paper.]
Check: R1, R2, R3 hang off node P, which connects to the load path only through R3 into node Q — once Vs is shorted (for Rth) or once R4 is left undisturbed (VQ already fixed for Vth), that whole branch carries no current relevant to the load side. the same "upstream distractor" pattern recurs across this subject's Thévenin questions.
Approach. Zero the source (short Vs) to get Rth by series/parallel reduction from the load terminals; find Vth as the open-circuit load voltage by voltage division (R7 carries no current when the load is open); then apply the maximum-power-transfer theorem.
(a) Rth — short Vs. With Vs replaced by a wire, node Q is shorted to ground, so R1, R2, R3, R4 all dangle off a grounded node and contribute nothing to the load path. Looking back from the load terminals: R6 is in parallel with R5 (both now run from the grounded node Q to the mid-node M), in series with R7: $$R_{th} = R_7 + (R_5 \parallel R_6) = 5\text{k} + \frac{(10\text{k})(10\text{k})}{10\text{k}+10\text{k}} = 5\text{k}+5\text{k} = \boxed{10.0\text{ k}\Omega}$$
(b) Vth — open-circuit load voltage. With the load removed, no current flows through R7, so node N (the open load terminal) sits at the same potential as M. R5 and R6 form a simple divider from VQ=20 V to ground: $$V_{th} = V_M = V_s\cdot\frac{R_6}{R_5+R_6} = 20\cdot\frac{10\text{k}}{20\text{k}} = \boxed{10.0\text{ V}}$$
Thévenin equivalent circuit at the load terminals.
(c) Power to the load, RL=100 Ω. $$I_L = \frac{V_{th}}{R_{th}+R_L} = \frac{10}{10{,}000+100} = 0.9901\text{ mA}$$$$P_{RL} = I_L^2 R_L = (0.9901\text{ mA})^2(100) = \boxed{98.03\ \mu\text{W}}$$
(d) Maximum power transfer. Maximum power is delivered when the load resistance equals the Thévenin resistance: $$R_{L,max} = R_{th} = \boxed{10.0\text{ k}\Omega}, \qquad P_{max} = \frac{V_{th}^2}{4R_{th}} = \frac{10^2}{4(10{,}000)} = \boxed{2.500\text{ mW}}$$
