Thévenin Equivalent and Maximum Power Transfer: 04-BS-4, May 2013, Q2

Question 2: Thévenin Equivalent and Maximum Power Transfer (equal value)

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.

[Figure not reproduced: Figure 2 — redrawn: node R sits directly across the ideal source Vs, so R is a fixed 20 V node regardless of Is or the R1–R4 network to its left. R5 & R6 form a divider from R down to ground, and R7 leads on to the (open) load terminal T. See the official exam paper.]

Given.

Given data
R1	12.5 MΩ	R2	22.5 kΩ
R3	300 kΩ	R4	100 kΩ
R5	10 kΩ	R6	10 kΩ
R7	5 kΩ	Is	2 A
Vs	20 V	RL (part c)	100 Ω

Find. $R_{th}$, $V_{th}$ at the load terminal; power to $R_L=100\ \Omega$; the load resistance and power for maximum power transfer.

Approach. Remove the load and reduce everything else to a Thévenin source. The key observation is that node R (where Vs connects) is pinned at 20 V by the ideal source no matter what current Is or the R1–R2–R3–R4 sub-network draws through it — so that entire left-hand sub-network (and Is) has zero effect on the Thévenin equivalent seen from the load. Then reduce the source-deactivated network for $R_{th}$ and the open-circuit condition for $V_{th}$.

Thévenin resistance. Deactivate the sources: Is → open circuit (breaks the only path from the R1–R4 side to node R, so that whole sub-network drops out entirely), Vs → short circuit (ties node R directly to ground). Looking into the load terminal T with RL removed: R7 is in series with (R5 ∥ R6), since R5 now returns to ground-via-shorted-Vs exactly like R6 does. $$R_{th} = R_7 + \dfrac{R_5 R_6}{R_5+R_6} = 5\text{k} + \dfrac{(10\text{k})(10\text{k})}{20\text{k}} = 5\text{k}+5\text{k} = \boxed{10\ \text{k}\Omega}$$
Thévenin voltage. With the load open, no current can flow through R7 (it dead-ends at the open terminal), so $V_{th}$ equals the voltage at the R5–R6 junction, a simple divider fed from the fixed 20 V at node R: $$V_{th} = V_s\cdot\dfrac{R_6}{R_5+R_6} = 20\cdot\dfrac{10\text{k}}{20\text{k}} = \boxed{10\ \text{V}}$$
Power to RL = 100 Ω. Since $R_{th}=10\text{k}\Omega \gg R_L$, only a small fraction of $V_{th}$ appears across the load: $$P_{R_L} = \left(\dfrac{V_{th}}{R_{th}+R_L}\right)^2 R_L = \left(\dfrac{10}{10100}\right)^2(100) = \boxed{9.80\times10^{-5}\ \text{W}}\ (98.0\ \mu\text{W})$$
Maximum power transfer. Occurs when $R_L=R_{th}$: $$R_{L,mp} = \boxed{10\ \text{k}\Omega},\qquad P_{max} = \dfrac{V_{th}^2}{4R_{th}} = \dfrac{100}{40000} = \boxed{2.5\ \text{mW}}$$

Question 2 — final results
Quantity	Value
R_th	10 kΩ
V_th	10 V
P (R_L = 100 Ω)	9.80×10−5 W
R_L for max transfer	10 kΩ
P_max	2.5 mW

Topic: Thévenin equivalent circuits and the maximum power transfer theorem.

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

Why this works: Any linear resistive one-port reduces to a single voltage source in series with a single resistance (Thévenin's theorem), so the load "sees" only two numbers no matter how complicated the internal network is. Recognizing that an ideal source fixes its own terminal voltage independent of everything else connected to it (here, node R) is what collapses this seven-resistor, two-source network down to a trivial two-resistor divider.

Common pitfall: Assuming every resistor in the schematic must enter the Thévenin calculation. Students often grind through R1–R4 and Is unnecessarily; the giveaway is that R1–R4 and Is only connect to the rest of the circuit through the ideal voltage source Vs, which isolates their effect completely from the load terminal.

Source: Sadiku, 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 (equal value)