Thévenin Equivalent and Maximum Power Transfer: 04-BS-4, December 2013, 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. The network of Figure 2. The current source $I_s$ sits in the lower rail, so it separates the left ground node from the load-side ground node.

Given data
$R_1$	$R_2$	$R_3$	$R_4$	$R_5$	$R_6$	$R_7$	$I_s$	$V_s$
12.5 kΩ	22 kΩ	50 Ω	350 Ω	10 kΩ	10 kΩ	5 kΩ	1 mA	20 V

Find. (a) $R_{Th}$ and (b) $V_{Th}$ at the load terminals; (c) power to $R_L=100 Ω$; (d) the $R_L$ for maximum transfer and that maximum power.

Figure 2 — Network for Question 2. The load (RL) sees the terminals inside the dashed box; Is sits in the lower rail.

The key observation is topological. Because $I_s$ is an ideal current source placed in series with the bottom conductor, deactivating it (open circuit) disconnects the entire left sub-network from the load reference node. The branch through $R_5$ then leads into an isolated island and carries no current, so it plays no part in the Thévenin resistance.

(a) Thévenin resistance. Kill the sources: short $V_s$, open $I_s$. Opening $I_s$ strands the $R_1,R_2,R_3,R_4$ island reached through $R_5$, so that branch is a dead end. Looking in from the load, only the series path $R_7$–$R_6$ survives:$$\boxed{R_{Th}=R_7+R_6=5 kΩ+10 kΩ=15 kΩ.}$$
(b) Thévenin voltage. With the load open, $R_7$ carries no current, so $V_{Th}=V_P$. Solving the active network (supernode across $V_s$, the $I_s$ injection, and KCL at $P$ giving $V_P=V_M/2$) yields $$V_{Th}=-10\text{ V}\quad(|V_{Th}|=10\text{ V}).$$The sign simply marks the polarity of the open‑circuit terminal.
(c) Power to $R_L=100 Ω$. With the equivalent source driving the load, $$P_L=\Big(\frac{V_{Th}}{R_{Th}+R_L}\Big)^2R_L=43.858\ \mu\text{W}.$$
(d) Maximum power transfer. Maximum power is delivered when $R_L=R_{Th}=15 kΩ$, giving $$P_{max}=\frac{V_{Th}^{\,2}}{4R_{Th}}=1.667\text{ mW}.$$

Final results
Quantity	Value
$R_{Th}$	15 kΩ
$\|V_{Th}\|$	10 V
$P_L$ at $R_L=100 Ω$	$43.858\ \mu\text{W}$
$R_L$ for max power	15 kΩ
$P_{max}$	$1.667\text{ mW}$

Check: the placement of $I_s$ in the bottom rail is load-bearing. If $I_s$ were instead a branch to ground on the source side, $R_5$ would parallel $R_6$ and $R_{Th}$ would fall to $R_7+R_5\Vert R_6$.

Topic: Thévenin equivalent and maximum power transfer.

Key relations: $R_{Th}$ from source-deactivated network; $V_{Th}$ = open‑circuit voltage; $P_L=[V_{Th}/(R_{Th}+R_L)]^2R_L$; maximum at $R_L=R_{Th}$ with $P_{max}=V_{Th}^2/4R_{Th}$.

Why this works: Reducing everything left of the load to a single source plus resistance turns a load-power sweep into one formula. An ideal current source in the return path is what isolates the bridge network, so most of the labelled resistors turn out to be distractors for $R_{Th}$ — they still shape $V_{Th}$.

Common pitfall: Assuming $R_5\Vert R_6$ without checking how $I_s$ sits in the circuit; the in-rail source opens that parallel path.

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

Question 2 of 7: Thévenin Equivalent and Maximum Power Transfer

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