Thevenin Equivalent and Maximum Power Transfer: 04-BS-4, December 2014, Q2

Question 2: Thevenin 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.

Given data
Quantity	Value
$R_1,\,R_2$	$12.5\text{ k}\Omega,\ 22\text{ k}\Omega$
$R_3,\,R_4$	$50\,\Omega,\ 350\,\Omega$
$R_5,\,R_6$	$40\,\Omega,\ 10\,\Omega$
$I_s,\,V_{s1},\,V_{s2}$	$1\text{ mA},\ 10\text{ V},\ 40\text{ V}$
$R_L$ (part c)	$72\,\Omega$

[Figure not reproduced. See the official exam paper.]

Find. (a) $R_{th}$; (b) $V_{th}$; (c) $P_L$ at $R_L=72\,\Omega$; (d) $R_L$ for maximum power transfer and $P_{max}$.

Approach. Recognize that the ideal source $V_{s1}$ fixes node M unconditionally, and the ideal source $V_{s2}$ (in series) then fixes node N – so the entire $R_1,R_2,R_3,I_s$ branch is electrically decoupled from the load terminals. Zero the independent sources to find $R_{th}$, then use the fixed node N together with the $R_5$–$R_6$ loop (closed even with the load removed) to find $V_{th}$.

a) Thevenin resistance. Zeroing $V_{s1}$ (short) ties node M to ground; zeroing $V_{s2}$ (short) then ties node N to ground as well, so looking into the open load terminals, $R_5$ and $R_6$ both run from N (≡ ground) to the other load terminal – i.e. they appear in parallel. The $R_1$-$R_2$-$R_3$-$I_s$ branch dangles off the (now grounded) node M and carries no current to the terminals: $$R_{th}=R_5\parallel R_6=\frac{(40)(10)}{40+10}=\boxed{8\ \Omega}$$
b) Thevenin voltage. With the load open, node M is fixed by $V_{s1}$: $V_M=10\text{ V}$. Node N follows through the series source $V_{s2}$: $V_N=V_M+V_{s2}=10+40=50\text{ V}$ (independent of any current, since both sources are ideal). Even with the load removed, a closed loop N–$R_5$–(bottom node G1)–$R_6$–ground–$V_{s1}$–M–$V_{s2}$–N still exists, driving a loop current $$I_{loop}=\frac{V_N}{R_5+R_6}=\frac{50}{40+10}=1\text{ A}$$ so the load's bottom terminal sits at $V_{G1}=I_{loop}R_6=(1)(10)=10\text{ V}$ above the deep ground, giving $$V_{th}=V_N-V_{G1}=50-10=\boxed{40\text{ V}}$$
c) Power to $R_L=72\,\Omega$. Using the Thevenin equivalent, $$I_L=\frac{V_{th}}{R_{th}+R_L}=\frac{40}{8+72}=0.5\text{ A}, \qquad P_L=I_L^2R_L=(0.5)^2(72)=\boxed{18\text{ W}}$$
d) Maximum power transfer. Maximum power transfer occurs when $R_L=R_{th}$: $$R_{L,opt}=R_{th}=\boxed{8\ \Omega}, \qquad P_{max}=\frac{V_{th}^2}{4R_{th}}=\frac{40^2}{4(8)}=\boxed{50\text{ W}}$$

Final Results – Question 2
Quantity	Value
$R_{th}$	8 Ω
$V_{th}$	40 V
$P_L$ ($R_L=72\,\Omega$)	18 W
$R_{L,opt}$	8 Ω
$P_{max}$	50 W

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

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

Why this works: Any two-terminal linear network reduces to a single source-resistance pair; once reduced, the load sees only $V_{th}$ and $R_{th}$, so the power-transfer optimization is a one-variable calculus result (maximize $I_L^2R_L=\left(\frac{V_{th}}{R_{th}+R_L}\right)^2R_L$ over $R_L$).

Common pitfall: Forgetting that zeroing a voltage source means short-circuiting it (not open-circuiting it) and a current source means open-circuiting it; here, an even subtler trap is missing that $R_6$ sits in the bottom (return) rail rather than in parallel with $R_5$, which is what actually splits the ground reference and makes $R_{th}=R_5\parallel R_6$ rather than $R_5+R_6$.

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

Question 2 of 7: Thevenin Equivalent and Maximum Power Transfer

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