Thévenin Equivalent and Maximum Power Transfer: 04-BS-4, May 2014, Q2

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. $V_{s1}$ ties node X directly to ground (top rail), so $V_X=V_{s1}$ is fixed regardless of anything else on X. The ideal current source $I_s$ (in series with $R_6$) forces exactly 20 A from X into node W independent of every resistance in the circuit; $R_7$ and the load $R_L$ hang from W to ground.

Given data
Quantity	Value
$R_1\ldots R_5$	50, 100, 50, 100, 100 Ω
$R_6$, $R_7$	20 Ω, 80 Ω
$V_{s1}$	20 V
$I_s$	20 A
$V_{s2}$	5 V

Find. $V_{th}$, $R_{th}$ at the load terminals; $R_L$ and $P_{max}$ for maximum power transfer; and $P_{R_L}$ at $R_L=100\,\Omega$.

[Figure not reproduced: Figure 2 — redrawn from the printed figure. The $V_{s2}$/$R_1$/$R_2$/$R_3$/$R_4$/$R_5$ sub-network only ever touches node X, whose voltage is already pinned by the ideal source $V_{s1}$ — it cannot influence the load terminals (W–ground) and is a deliberate distractor. See the official exam paper.]

Approach. Recognize the two ideal sources: $V_{s1}$ fixes $V_X$ outright, and $I_s$ forces a fixed 20 A into node W through the series $R_6$ branch regardless of load. Everything left of X (the $V_{s2}$ ladder) and $R_6$ itself therefore drop out of the Thevenin calculation at W.

(a) Thevenin voltage $V_{th}$. With the load removed, ALL of the fixed 20 A from $I_s$ must return to ground through $R_7$ (the only path from W): $$\boxed{V_{th}=I_sR_7=(20)(80)=1600\text{ V}}$$
(b) Thevenin resistance $R_{th}$. Deactivate sources: $V_{s1}\to$ short, $I_s\to$ open. Opening $I_s$ disconnects $R_6$ from node X entirely, leaving $R_6$ a dead-end resistor hanging off W (carries no current, contributes nothing). The only path left from W to ground is $R_7$: $$\boxed{R_{th}=R_7=80\,\Omega}$$
(c) Maximum power transfer. Matched-load condition $R_L=R_{th}$: $$\boxed{R_{L,mp}=R_{th}=80\,\Omega}\qquad \boxed{P_{max}=\dfrac{V_{th}^2}{4R_{th}}=\dfrac{1600^2}{4(80)}=8000\text{ W}}$$
(d) Power at $R_L=100\,\Omega$. Voltage-divider power into the mismatched load: $$P_{R_L}=\left(\dfrac{V_{th}}{R_{th}+R_L}\right)^2R_L =\left(\dfrac{1600}{180}\right)^2(100)$$ $$\boxed{P_{R_L}=7901.23\text{ W}}$$

Final results — Question 2
Quantity	Value
$V_{th}$	1600 V
$R_{th}$	80 Ω
$R_L$ for max transfer	80 Ω
$P_{max}$	8000 W
$P_{R_L}$ at 100 Ω	7901.23 W

Topic: Thévenin equivalent circuits; ideal sources fixing node voltages/branch currents; maximum power transfer theorem.

Key relations: $R_{th}$ from deactivated-source network; $V_{th}=$ open-circuit voltage; $R_{L,mp}=R_{th}$; $P_{max}=V_{th}^2/(4R_{th})$.

Why this works: An ideal voltage source clamps its node's potential no matter what else is attached, and an ideal current source clamps its branch current no matter what impedance it drives into — both facts let large parts of a network be recognized as irrelevant to a specific pair of terminals before doing any algebra. Here that recognition collapses a seven-resistor, two-source network to a single series R7 path.

Common pitfall: Grinding through nodal equations for the entire $V_{s2}$ ladder before noticing it cannot affect the load — a huge waste of exam time. Always check for "dead" branches (fixed nodes, dangling resistors after source deactivation) before setting up simultaneous equations.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, Ch. 4 (Thévenin's Theorem, Maximum Power Transfer).

Question 2 of 7: Thévenin Equivalent and Maximum Power Transfer

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