Source voltage, Thevenin equivalent & maximum power: 22-Elec-A1 Circuits, May 2013, Q1

Question 1: Source voltage, Thevenin equivalent & maximum power [8 + 6 + 2 + 4]

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.

Part (A): Source voltage from a known branch current [8]

Given. A source $V_x$ in series with a 3 Ω resistor feeds a ladder: a 5 Ω series resistor to node A, two 10 Ω resistors from A to the bottom rail, a 5 Ω series resistor from A to node B, and a 10 Ω resistor from B to the rail. The current down the centre 10 Ω resistor is measured as $I = 2\ \text{A}$.

Find. The source voltage $V_x$.

Figure 1 — Ladder network for Q1(A). Node A is the common top node of both 10 Ω resistors and the two 5 Ω series arms; the centre branch current is fixed at 2 A.

Approach. The measured 2 A through the centre 10 Ω pins node A directly; then work outward to node B and back through the source branch by KCL and Ohm's law (bottom rail as reference).

Node A from the measured current. The 2 A flows down a 10 Ω resistor tied from A to the rail, so $$V_A = I\,(10) = (2)(10) = \boxed{20\ \text{V}}.$$
The parallel 10 Ω. The second 10 Ω sees the same $V_A$, so it also carries $I_{10}=V_A/10 = 20/10 = 2\ \text{A}$.
Node B (5 Ω–10 Ω divider). With B loaded only by its 10 Ω, KCL gives $\dfrac{V_A-V_B}{5}=\dfrac{V_B}{10}$, hence $V_B=\tfrac{2}{3}V_A = 13.33\ \text{V}$ and the right-arm current is $I_{5R}=\dfrac{V_A-V_B}{5}=\dfrac{20-13.33}{5}=1.33\ \text{A}$.
KCL at node A. The current delivered into A through the left 5 Ω equals the three currents leaving it: $$I_{5L}=I_{10}+I+I_{5R}=2+2+1.33=\boxed{5.33\ \text{A}}.$$ This is the whole current supplied by the source branch.
Climb back to the source. From A up through the left 5 Ω to the source terminal C, then across the internal 3 Ω: $$V_x = V_A + I_{5L}(5) + I_{5L}(3) = 20 + 5.33(5) + 5.33(3) = \boxed{62.7\ \text{V}}.$$

Quantity	Result
Node A voltage	$V_A = 20\ \text{V}$
Node B voltage	$V_B = 13.33\ \text{V}$
Source-branch current	$I_{5L} = 5.33\ \text{A}$
Source voltage	$\boxed{V_x = 62.7\ \text{V}}$

Concept & Theory

Topic. Ladder-network analysis working backward from a known branch current.

Key relations. Ohm's law $V=IR$; KCL $\sum I=0$ at a node; two-resistor current divider $V_B=V_A\,\dfrac{R_2}{R_1+R_2}$.

Why this works. A single measured branch current fixes the voltage of the node it hangs from, which removes one unknown and lets a ladder be solved by pure back-substitution — no simultaneous equations needed. Every subsequent node follows from Ohm's law and one KCL summation. The internal 3 Ω simply adds an $IR$ drop between the source EMF and its terminal.

Common pitfall. Forgetting that both 10 Ω resistors sit at node A (they are in parallel), or omitting the 3 Ω series drop and reporting the terminal voltage instead of the source EMF.

Source. Sadiku, Fundamentals of Electric Circuits §2 (Ohm's law, series/parallel) and §3 (nodal analysis).

Part (B): Thevenin equivalent and maximum power transfer [6 + 2 + 4]

Given. A 10 V source in series with a 10 Ω resistor sets node 1; a 5 A source injects into node 1; a 5 Ω resistor links node 1 to node 2; a 6 Ω resistor ties node 2 to the bottom rail (b); and a 2 Ω resistor connects node 2 to terminal a.

Find. $V_{th}$ and $R_{th}$ at a–b, the load $R_L$ for maximum power, and that maximum power.

Figure 2 — Network for Q1(B). Terminals a–b are open for the Thevenin voltage; the 2 Ω sits in series between node 2 and terminal a.

Approach. Get $V_{th}=V_{oc}$ by nodal analysis with a–b open (no current in the 2 Ω), then $R_{th}$ by deactivating both sources, and finally apply the maximum-power-transfer theorem.

Open-circuit node equations. With a–b open the 2 Ω carries no current, so $V_{oc}=V_2$. Taking the rail as reference (the source holds the far end of the 10 Ω at $+10\ \text{V}$): $$\text{node 1:}\ \ 5=\frac{V_1-10}{10}+\frac{V_1-V_2}{5},\qquad \text{node 2:}\ \ \frac{V_1-V_2}{5}=\frac{V_2}{6}.$$
Solve. The node-2 relation gives $6V_1=11V_2$; substituting into node 1 yields $\tfrac{21}{11}V_1=60$, so $V_1=\tfrac{220}{7}=31.43\ \text{V}$ and $$V_{th}=V_2=\frac{120}{7}=\boxed{17.14\ \text{V}}.$$
Thevenin resistance. Short the 10 V source and open the 5 A source. Looking in from a–b, the 2 Ω is in series with the parallel combination of the 6 Ω and the $(5+10)\ \Omega$ path: $$R_{th}=2+\bigl(6\parallel 15\bigr)=2+\frac{6\cdot 15}{21}=\boxed{6.29\ \Omega}.$$
Load for maximum power. The maximum-power-transfer theorem requires $$\boxed{R_L=R_{th}=6.29\ \Omega}.$$
Maximum power. With $R_L=R_{th}$ the load takes half the source EMF: $$P_{max}=\frac{V_{th}^{2}}{4R_{th}}=\frac{(17.14)^2}{4(6.29)}=\boxed{11.69\ \text{W}}.$$

Quantity	Result
Thevenin voltage	$V_{th}=17.14\ \text{V}$
Thevenin resistance	$R_{th}=6.29\ \Omega\ (=44/7)$
Load for max power	$R_L=6.29\ \Omega$
Maximum power	$\boxed{P_{max}=11.69\ \text{W}}$

Concept & Theory

Topic. Thevenin reduction and the maximum-power-transfer theorem.

Key relations. $V_{th}=V_{oc}$; $R_{th}$ from source-free look-back (voltage sources shorted, current sources opened); MPT: $R_L=R_{th}$ and $P_{max}=V_{th}^2/(4R_{th})$.

Why this works. Any linear two-terminal network is indistinguishable from a single EMF behind a single resistance. At the load, the delivered power $P=V_{th}^2R_L/(R_{th}+R_L)^2$ is maximised by $dP/dR_L=0$, which gives $R_L=R_{th}$; substituting back leaves exactly one-quarter of $V_{th}^2/R_{th}$.

Common pitfall. Deactivating sources incorrectly (opening a voltage source or shorting a current source), or forgetting the series 2 Ω when both finding $V_{oc}$ (it drops nothing when open) and $R_{th}$ (it adds in series).

Source. Sadiku, Fundamentals of Electric Circuits §4 (Thevenin's theorem, maximum power transfer).

Question 1 of 6: Source voltage, Thevenin equivalent & maximum power

Question 1: Source voltage, Thevenin equivalent & maximum power [8 + 6 + 2 + 4]

Part (A): Source voltage from a known branch current [8]

Part (B): Thevenin equivalent and maximum power transfer [6 + 2 + 4]