04-BS-4, December 2016, Q2 | Nivaar Exam Prep

Question 2 (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 ladder network of Figure 2: an ideal source Vs1 fixes node n1; a side branch (Vs2 with R1, R2 in parallel, and R3) hangs off n1; from n1 the ladder continues through R4 to node n2 (R5 to ground), then through R6 to node n3 (R7 to ground), where the load RL connects.

Given data
Vs1	90 V	Vs2	5 V
R1	50 Ω	R2	100 Ω
R3	50 Ω	R4	30 Ω
R5	60 Ω	R6	10 Ω
R7	30 Ω	RL (part d)	45 Ω

Find. (a) Thevenin voltage at the load; (b) Thevenin resistance at the load; (c) the load resistance for maximum power transfer and that maximum power; (d) the power delivered to RL = 45 Ω.

Figure 2 — ladder network: Vs1 fixes n1 directly; the Vs2/R1/R2/R3 branch is a distractor with nowhere else to go; the load sees the network through R4–n2(R5)–R6–n3(R7).

Approach. Because Vs1 is an ideal source connected directly across n1 and ground, it fixes $V_{n1}=90\text{ V}$ regardless of anything else attached to n1 — so the Vs2/R1/R2/R3 branch cannot change $V_{n1}$ and, having no other connection, is a distractor for this problem. Find the Thevenin equivalent at the load terminals (n3–ground) of the remaining R4–R5–R6–R7 ladder, then apply the maximum-power-transfer theorem and a loaded nodal solve for part (d).

(a) Open-circuit voltage at the load. With RL removed, nodal equations at n2 and n3 give $$ \frac{V_{n1}-V_{n2}}{R_4}=\frac{V_{n2}}{R_5}+\frac{V_{n2}-V_{n3}}{R_6},\qquad \frac{V_{n2}-V_{n3}}{R_6}=\frac{V_{n3}}{R_7} $$ Solving with $V_{n1}=90$ gives $V_{n2}=40.0\text{ V}$ and $$ V_{th}=V_{n3}=\boxed{30.0\text{ V}} $$
(b) Thevenin resistance. Zero the source (short Vs1 to ground); the distractor branch then dangles off a grounded node and drops out entirely. Looking into the load terminals: R4 and R5 meet at n2 ($R_4\Vert R_5$), that combination is in series with R6, and the result is in parallel with R7: $$ R_{th}=R_7\left\Vert\Big(R_6+\big(R_4\Vert R_5\big)\Big)\right.=30\Vert(10+20)=\boxed{15.0\text{ }\Omega} $$
(c) Maximum power transfer. Maximum power transfer occurs when $R_L=R_{th}$: $$ R_{L,\text{opt}}=\boxed{15.0\text{ }\Omega},\qquad P_{max}=\frac{V_{th}^2}{4R_{th}}=\frac{30.0^2}{4(15.0)}=\boxed{15.0\text{ W}} $$
(d) Power at RL = 45 Ω. Re-solving the loaded nodal system with RL in place at n3 (in parallel with R7) gives $V_{n3}=22.5\text{ V}$, so $$ P_{R_L}=\frac{V_{n3}^2}{R_L}=\frac{22.5^2}{45}=\boxed{11.25\text{ W}} $$ (equivalently, from the Thevenin model: $P_{R_L}=\left(\dfrac{V_{th}}{R_{th}+R_L}\right)^2R_L=\left(\dfrac{30.0}{60}\right)^2(45)=11.25\text{ W}$, confirming the two methods agree.)

Question 2 — final results
Quantity	Value
Thevenin voltage $V_{th}$	30.0 V
Thevenin resistance $R_{th}$	15.0 Ω
Optimal $R_L$ / max power	15.0 Ω / 15.0 W
Power at $R_L=45\,\Omega$	11.25 W

Topic: Thevenin equivalent circuits and the maximum-power-transfer theorem.
Key relations: $V_{th}=$ open-circuit voltage at the terminals; $R_{th}=$ resistance seen with independent sources zeroed; $P_{max}=V_{th}^2/(4R_{th})$ at $R_L=R_{th}$; general loaded power $P_{R_L}=\big(V_{th}/(R_{th}+R_L)\big)^2R_L$.
Why this works: Any linear resistive one-port reduces to a single source and series resistance at a chosen pair of terminals, which is why an ideal source elsewhere in the network (here Vs1) can fully decouple an unrelated branch (Vs2/R1/R2/R3) from the terminals of interest.
Common pitfall: Including the Vs2/R1/R2/R3 branch in the Thevenin reduction — since it only connects back to the already-fixed node n1, it carries current but cannot influence $V_{n1}$, $V_{th}$, or $R_{th}$ at the load.
Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., Ch. 4 (Thevenin's Theorem, Maximum Power Transfer).

Question 2 of 7

Question 2 (20 marks)