04-BS-4 · Undated paper

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 Exam — 04-BS-4 Electric Circuits and Power — undated sitting (internal evidence places it at May 2019). 3 hours, closed book, one double-sided aid sheet, approved Casio/Sharp calculator only. The exam instructs "any five questions constitute a complete paper" — all seven are solved below as a complete study resource.

Reference texts: Sadiku, Fundamentals of Electric Circuits (7th ed.) — Ch. 2–4 (resistive circuits, node/mesh analysis, Thévenin/Norton, maximum power transfer), Ch. 7 (first-order RC/RL transients), Ch. 9–10 (sinusoidal steady state, phasors, AC power); Chapman, Electric Machinery Fundamentals (5th ed.) — Ch. 1 (magnetic circuits, reluctance, fringing, force on an armature); Sadiku, Elements of Electromagnetics — Ch. 5, 8 (magnetic energy and force); Mano & Ciletti, Digital Design — Ch. 2–3 (Boolean algebra, combinational logic design).

Question 2: Thévenin Equivalent and Maximum Power Transfer (20 marks)

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. The network of Figure 2, with the load $R_L$ removed between the terminals marked "Load."

Given data
Quantity	Value
$R_1$	$50\ \Omega$
$R_2$	$100\ \Omega$
$R_3$	$50\ \Omega$
$R_4$	$30\ \Omega$
$R_5$	$60\ \Omega$
$R_6$	$10\ \Omega$
$R_7$	$30\ \Omega$
$V_{s1}$	$90\text{ V}$
$V_{s2}$	$5\text{ V}$

Find. $V_{th}$, $R_{th}$, the load resistance and power at maximum power transfer, and the power delivered to a $45\ \Omega$ load.

Figure 2 — circuit for Question 2 (RL is the load seen at node T3–ground).

Approach. First recognize that $V_{s1}$ is an ideal source strapped directly across node T1 and ground, so $V_{T1}=V_{s1}$ no matter what current the $V_{s2}$/$R_1$/$R_2$/$R_3$ branch draws — that whole branch is a self-contained loop hanging off T1 and never appears in the equations for the load. What is left is the plain $R_4$–$R_5$–$R_6$–$R_7$ ladder that actually feeds the load, which is solved by open-circuit nodal analysis (for $V_{th}$) and by deactivating sources (for $R_{th}$).

Step 1 — identify the irrelevant sub-network. $V_{s1}$ connects node T1 directly to the ground rail with no series impedance, so $V_{T1}=V_{s1}=90\text{ V}$ is fixed regardless of $R_1,R_2,R_3,V_{s2}$ (that branch, in fact, is $V_{s2}$ directly in parallel with $R_3$, both bridging T1 to an isolated node M that only feeds $R_1\Vert R_2$ to ground — it carries its own closed-loop current and never touches the T2/T3/load side of the circuit). It is safely dropped from every later step.
Step 2 — open-circuit voltage $V_{th}$ (part a). Remove $R_L$ and solve the remaining ladder $V_{T1}(=90\text{ V})\to R_4\to T2\ (R_5\text{ to ground})\to R_6\to T3\ (R_7\text{ to ground, open at the load})$ by nodal analysis: $$\dfrac{V_{T2}-V_{T1}}{R_4}+\dfrac{V_{T2}}{R_5}+\dfrac{V_{T2}-V_{T3}}{R_6}=0,\qquad \dfrac{V_{T3}-V_{T2}}{R_6}+\dfrac{V_{T3}}{R_7}=0$$ Substituting values and solving gives $V_{T2}=40\text{ V}$ and $$V_{th}=V_{T3}=\boxed{30\text{ V}}$$
Step 3 — Thévenin resistance $R_{th}$ (part b). Deactivate both independent sources ($V_{s1}\to$ short, tying T1 to ground) and look into the load terminals: $R_4$ and $R_5$ are now both from T1(=0) to T2, i.e. in parallel, and that combination is in series with $R_6$ to T3, where $R_7$ is in parallel across the terminals: $$R_4\Vert R_5=\dfrac{30\cdot60}{30+60}=20\ \Omega\ \Rightarrow\ R_{th}=R_7\Vert\big(R_6+20\ \Omega\big)=\dfrac{30\cdot30}{30+30}=\boxed{15\ \Omega}$$
Step 4 — maximum power transfer (part c). Maximum power to a resistive load occurs when $R_L=R_{th}$: $$R_{L,\text{opt}}=R_{th}=\boxed{15\ \Omega},\qquad P_{\max}=\dfrac{V_{th}^2}{4R_{th}}=\dfrac{30^2}{4(15)}=\boxed{15\text{ W}}$$
Step 5 — power at $R_L=45\ \Omega$ (part d). With the load reattached, voltage-divide the Thévenin source between $R_{th}$ and $R_L$: $$V_{R_L}=V_{th}\cdot\dfrac{R_L}{R_L+R_{th}}=30\cdot\dfrac{45}{45+15}=22.5\text{ V}$$ $$P_{R_L}=\dfrac{V_{R_L}^2}{R_L}=\dfrac{22.5^2}{45}=\boxed{11.25\text{ W}}$$ As expected this is below the $15\text{ W}$ maximum, since $45\ \Omega\neq R_{th}$.

Final results — Question 2
Quantity	Value
$V_{th}$	$30\text{ V}$
$R_{th}$	$15\ \Omega$
$R_{L}$ for $P_{\max}$	$15\ \Omega$
$P_{\max}$	$15\text{ W}$
$P_{R_L}$ at $R_L=45\ \Omega$	$11.25\text{ W}$

Topic: Thévenin equivalents and the maximum power transfer theorem.

Key relations: $R_{th}$ = resistance at the terminals with all independent sources deactivated; $V_{th}$ = open-circuit terminal voltage; maximum power transfer at $R_L=R_{th}$, giving $P_{\max}=V_{th}^2/(4R_{th})$; general load power $P_{R_L}=\big(V_{th}R_L/(R_L+R_{th})\big)^2/R_L$.

Why this works: Any linear two-terminal network reduces to one voltage source in series with one resistance as seen from a chosen pair of terminals; this collapses a load-sizing question about an arbitrarily complicated network into a single-loop voltage-divider problem, and the $dP_{R_L}/dR_L=0$ condition on that reduced circuit always lands at $R_L=R_{th}$.

Common pitfall: Forgetting to strip out a sub-network that is not actually in the path to the load (an easy trap here, since $V_{s2}/R_1/R_2/R_3$ visually sits right at the front of the schematic); and deactivating $V_{s1}$ for the $V_{th}$ calculation as well as for $R_{th}$ (only $R_{th}$ deactivates the sources — $V_{th}$ needs them live).

Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., §4.5–4.7 (Thévenin/Norton equivalents, maximum power transfer).

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