Thévenin Equivalent and Maximum Power Transfer: 04-BS-4, May 2015, 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. A source network feeding a load through R3, with R4 shunting the mid-node. An ideal current source Is sits in series in the top branch, and R1 ∥ Vs form an isolated left island.

Given data
R1	R2	R3	R4	Is	Vs
2 kΩ	7 kΩ	50 Ω	150 Ω	2 A	20 V

Find. (a) R_Th and (b) V_Th at the load terminals; (c) load power for RL = 100 Ω; (d) the matched load and the maximum power.

Figure 2 — an ideal current source Is sits in series in the upper branch. It fixes that branch current at 2 A independently of everything to its left, so R1, R2 and Vs do not influence the load terminals.

Key observation. Is is an ideal current source in series with R2. It forces 2 A into the mid-node regardless of the left island, so R1, R2 and Vs are non-participating (distractor) elements for the Thévenin equivalent seen at the load. This is the crux the question tests.

Approach. Deactivate sources for R_Th (open the ideal current source); for V_Th, remove the load and note that the forced 2 A must flow entirely through R4 (no path through the open load), setting the open-circuit voltage.

a) Thévenin resistance. Open the ideal current source Is; the R2 branch then dangles (open) and drops out. Looking in from the load terminals: R3 in series with R4 to ground. $$\boxed{R_{Th}=R_3+R_4 = 50+150 = 200\ \Omega}$$
b) Thévenin (open-circuit) voltage. Remove RL; no current flows in R3 (its far end is open), so the forced $I_s = 2$ A flows entirely through R4: $$V_{Th}=V_{oc}=I_s R_4 = 2 \times 150$$ $$\boxed{V_{Th}=300\text{ V}}$$
c) Load power at RL = 100 Ω. $I_L=\dfrac{V_{Th}}{R_{Th}+R_L}=\dfrac{300}{200+100}=1\text{ A}$, hence $P_L=I_L^2 R_L = 1^2(100)$: $$\boxed{P_L = 100\text{ W}}$$
d) Maximum power transfer. Matching $R_L=R_{Th}=200\ \Omega$ gives $P_{max}=\dfrac{V_{Th}^2}{4R_{Th}}=\dfrac{300^2}{4(200)}$: $$\boxed{R_{L,\text{opt}}=200\ \Omega,\qquad P_{max}=112.5\text{ W}}$$

Question 2 — results
Quantity	Value
R_Th (a)	200 Ω
V_Th (b)	300 V
P_load at RL = 100 Ω (c)	100 W
Matched RL / P_max (d)	200 Ω / 112.5 W

Topic. Thévenin reduction and the maximum-power-transfer theorem, with an ideal-source isolation subtlety.

Key relations. $R_{Th}$ = resistance seen with sources deactivated (V-sources shorted, I-sources opened); $V_{Th}=V_{oc}$; matched-load $R_L=R_{Th}$ gives $P_{max}=V_{Th}^2/(4R_{Th})$.

Why this works. An ideal current source presents infinite impedance: opening it during the R_Th calculation severs the branch, and during the V_oc calculation it dictates its branch current no matter the surrounding voltages. Everything electrically "upstream" of a series ideal current source is invisible at the output — which is exactly why R1, R2 and Vs never appear in the answer.

Common pitfall. Trying to fold R1, R2 or Vs into R_Th or V_Th. They belong to a branch that the ideal current source decouples; including them is the classic error this problem screens for.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, ch. 4 (Thévenin, maximum power transfer).

Question 2 of 7: Thévenin Equivalent and Maximum Power Transfer

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