Thévenin Equivalent and Maximum Power Transfer: 04-BS-4, May 2017, 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. Figure 2: R₁ is in parallel with the ideal source V_s (both fix node 1); an ideal current source I_s sits in series feeding node 2; R₂ carries that current into node 3, where R₄ shunts to ground; R₃ then leads to the open load terminals.

Given data
Quantity	Value
R₁	7 kΩ
R₂	3 kΩ
R₃	150 Ω
R₄	250 Ω
I_s	8 mA
V_s	20 V
R_L (part c)	600 Ω

Find. R_th, V_th at the load terminals; the power delivered to a 600 Ω load; and the load resistance and maximum power at matched conditions.

Figure 2: Circuit for Thévenin / maximum power transfer

Approach. Deactivating the sources for R_th (V_s shorted, I_s opened) strands R₁, V_s, R₂ on a dead-end branch, so only R₃ and R₄ reach the load terminals. For V_th, the open load forces the entire I_s current through R₄ alone (R₃ carries zero current), so no drop appears across R₃.

Thevenin resistance (part a). Short V_s and open I_s: R₁ is now shorted to ground (irrelevant) and the current source's open circuit strands R₂ on a dead end (no closed path back to ground through it). Only R₃ and R₄ remain in series looking into the load terminals: $$ R_{th} = R_3 + R_4 = 150 + 250 = \boxed{400\ \Omega}. $$
Thevenin voltage (part b). With the load removed, no current can flow through R₃ (dead end at the open terminals), so the full I_s = 8 mA flows through R₂ and then entirely through R₄ to ground; since I(R₃)=0 there is no drop across R₃, so the open-circuit voltage equals the R₄ voltage: $$ V_{th} = I_s R_4 = (8\times10^{-3})(250) = \boxed{2\text{ V}}. $$
Power to a 600 Ω load (part c). $$ P_{load} = \frac{V_{th}^2 R_L}{(R_{th}+R_L)^2} = \frac{(2)^2(600)}{(400+600)^2} = \frac{2400}{1{,}000{,}000} = \boxed{2.4\text{ mW}}. $$
Maximum power transfer (part d). Matching the load to the source, $$ R_{L,mp} = R_{th} = \boxed{400\ \Omega}, \qquad P_{max} = \frac{V_{th}^2}{4R_{th}} = \frac{(2)^2}{4(400)} = \boxed{2.5\text{ mW}}. $$

Final results
Quantity	Value
R_th	400 Ω
V_th	2 V
P (R_L=600Ω)	2.4 mW
R_L for max power	400 Ω
P_max	2.5 mW

Topic: Thévenin equivalents and the maximum-power-transfer theorem with mixed independent sources.

Key relations: $R_{th}$ from source-deactivated resistance, $V_{th}$ = open-circuit terminal voltage, $P=\dfrac{V_{th}^2R_L}{(R_{th}+R_L)^2}$, matched-load condition $R_L=R_{th}\Rightarrow P_{max}=\dfrac{V_{th}^2}{4R_{th}}$.

Why this works: an ideal current source in series forces its exact current into whatever follows it, regardless of what precedes it; opening that source for the resistance calculation removes any path back through the upstream elements, so they cannot influence R_th or V_th at all — a deliberate distractor built into the topology.

Common pitfall: folding R₁, V_s, and R₂ into the Thevenin resistance or voltage calculation; because they sit upstream of the series current source, none of them can affect the load-terminal behaviour.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. 4 (Thévenin/Norton equivalents, maximum power transfer).

Question 2 of 7: Thévenin Equivalent and Maximum Power Transfer

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