MATH40007 · Practice Paper

MATH40007 Revised Historical Coverage Paper 3

Revision questions and worked solutions, presented read-only. This review interface was prepared after the recorded study period.

Status: Completed
Questions completed: 4 / 4
Suggested time: 120 minutes
Source date: 2026-08-08

Question 120 marks

Parameterized conductance network

Question 1 The circuit below has node 1 at unit voltage and node 3 grounded. Conductances are shown. 2 𝑐 1 1 1 3 𝑐 1 4 (a) Write the conductance-weighted Laplacian in the node order 1, 3, 2, 4. (2 marks) (b) Find the effective conductance in the limits 𝑐 → 0 and 𝑐 → ∞. (2 marks) (c) Find the voltages at nodes 2 and 4 for general 𝑐 > 0. (6 marks) (d) Find the effective conductance 𝐶 (𝑐) and check both limits. (4 marks) (e) Put a copy with parameter 𝑐 in series with a copy with parameter 1/𝑐. Find the effective conductance of the composite network. (6 marks) (Total: 20 marks) 9

Worked solution and marking guidance

Question 1 In the order 1, 3, 2, 4 the weighted Laplacian is 𝐿 = [ 𝑐 + 1 0 −𝑐 −1 0 𝑐 + 1 −1 −𝑐 −𝑐 −1 𝑐 + 2 −1 −1 −𝑐 −1 𝑐 + 2 ] . The limiting networks give 𝐶 (0) = 1/3 and 𝐶 (∞) = 3. KCL gives 𝑣 2 = 𝑐 + 1 𝑐 + 3 , 𝑣 4 = 2 𝑐 + 3 . The input current is 𝐶 (𝑐) = 𝑐(1 − 𝑣 2) + ( 1 − 𝑣 4) = 3𝑐 + 1 𝑐 + 3 . For a series connection with parameters 𝑐 and 1/𝑐, 𝐶series = (𝑐 + 3) (3𝑐 + 1) 2(5𝑐2 + 6𝑐 + 5) . Question difficulty allocation: 𝐴 = 8, 𝐵 = 4, 𝐶 = 4, 𝐷 = 4.

Question 220 marks

Ternary-tree resistance and voltage

Question 2 A ternary tree has 𝑛 generations. The root is held at unit voltage, all generation-𝑛 leaves are grounded, and every edge between generations 𝑗 − 1 and 𝑗 has conductance 𝑐 𝑗 > 0. root 𝑐 𝑐 𝑐 1 1 1 generation 2 (a) Find the numbers of vertices and edges and derive a formula for the effective resistance in terms of the 𝑐 𝑗 . (4 marks) (b) When 𝑐 𝑗 = 1, find the effective conductance and the voltage at every node in generation 𝑘. Find the limit as 𝑛 → ∞. (8 marks) (c) Find the infinite-tree effective conductance when 𝑐 𝑗 = 1/ 𝑗 . (4 marks) (d) Find it when 𝑐 𝑗 = 𝑗 . (4 marks) (Total: 20 marks) 10

Worked solution and marking guidance

Question 2 The vertex and edge counts are 3 𝑛+1 − 1 3 𝑛+1 − 3 𝑉 𝑛 = 1 + 3 + · · · + 3 𝑛 = , 𝐸 𝑛 = 3 + · · · + 3 𝑛 = . 2 2 Each generation- 𝑗 layer consists of 3 𝑗 equal conductors in parallel, so its resistance is 1/(3 𝑗 𝑐 𝑗 ). Hence 𝑛 ∑︁ 1 𝑅 𝑛 = 3 𝑗 𝑐 𝑗 , 𝐶 𝑛 = 𝑅 𝑛 −1. 𝑗=1 For 𝑐 𝑗 = 1, 2 𝐶 𝑛 = 1 − 3−𝑛 , 𝐶 ∞ = 2, and the voltage in generation 𝑘 is 3−𝑘 − 3−𝑛 𝑣 𝑘 = 1 − 3−𝑛 . For 𝑐 𝑗 = 1/ 𝑗, ∑︁ 𝑗 3 4 𝑅 = = , 𝐶 = . ∞ 3 𝑗 4 ∞ 3 𝑗≥1 For 𝑐 𝑗 = 𝑗, ∑︁ 1 3 1 𝑅 = = log , 𝐶 = . ∞ 𝑗3 𝑗 2 ∞ log(3/2) 𝑗≥1 Question difficulty allocation: 𝐴 = 4, 𝐵 = 6, 𝐶 = 4, 𝐷 = 6. 5

Question 320 marks

Complete-graph Laplacian and eigenbasis

Question 3 Consider the complete graph 𝐾5 with unit edge weights. (a) Write its graph Laplacian. (2 marks) (b) Find the effective conductance between nodes 1 and 2. (4 marks) (c) Give a real orthonormal eigenbasis and all eigenvalues. (6 marks) (d) A unit current enters at node 1 and leaves at node 2. In the gauge∑ 𝑖 𝑥𝑖 = 0, find the voltage vector by eigenvector expansion. (6 marks) (e) Use the voltage difference to check part (b). (2 marks) (Total: 20 marks) 11 MATH40007 Revised Historical-Coverage Practice Pack Questions

Worked solution and marking guidance

Question 3 For 𝐾5, 𝐿 = 5𝐼 − 𝐽. The effective resistance between any two nodes is 2 /5, hence 𝐶12 = 5/2. A convenient orthonormal eigenbasis is 𝑢0 = 1√ 5 (1, 1, 1, 1, 1)𝑇 , 𝜆 0 = 0, 𝑢1 = 1√ 2 (1, −1, 0, 0, 0)𝑇 , 𝑢 2 = 1√ 6 (1, 1, −2, 0, 0)𝑇 , 𝑢3 = 1√ 12 (1, 1, 1, −3, 0)𝑇 , 𝑢 4 = 1√ 20 (1, 1, 1, 1, −4)𝑇 , all with eigenvalue 5 except𝑢0. For 𝑓 = 𝑒1 − 𝑒2, only the 𝑢1 mode is present, so 𝑥 = 1 5 𝑓 = ( 1 5 , −1 5 , 0, 0, 0 )𝑇 . The voltage difference is 2/5, confirming 𝑅 = 2/5 and 𝐶 = 5/2. Question difficulty allocation: 𝐴 = 8, 𝐵 = 4, 𝐶 = 4, 𝐷 = 4.

Question 420 marks

Half-plane conductor and complex potential

Question 4 The right half-plane 𝑥 > 0 is a unit conductor. Let 0 < 𝑎 < 𝑏 and ℎ(𝑧) = − 𝑚 2𝜋 log ( (𝑧 − 𝑎) (𝑧 + 𝑏) (𝑧 + 𝑎) (𝑧 − 𝑏) ) , 𝜙 = ℜℎ(𝑧). (a) Show that the wall 𝑥 = 0 is grounded. (2 marks) (b) Identify all sources and sinks lying in the conductor. (4 marks) (c) Find 𝐽𝑥 − 𝑖𝐽 𝑦 = −ℎ′ (𝑧). (3 marks) (d) Show that 𝐽𝑦 = 0 on the wall. (1 marks) (e) Find 𝐽𝑥 (0, 𝑦) when 𝑎 = 1, 𝑏 = 3. (2 marks) (f) Calculate the total signed current through the wall. (4 marks) (g) Sketch 𝐽𝑥 (0, 𝑦), showing its zeros and the value at 𝑦 = 0. (4 marks) (Total: 20 marks) 12

Worked solution and marking guidance

Question 4 On 𝑧 = 𝑖𝑦, numerator and denominator in the logarithm have equal modulus, so 𝜙 = 0. The finite singularities in 𝑥 > 0 are a source +𝑚 at 𝑧 = 𝑎 and a sink −𝑚 at 𝑧 = 𝑏. 𝐽𝑥 − 𝑖𝐽 𝑦 = 𝑚 2𝜋 ( 1 𝑧 − 𝑎 − 1 𝑧 + 𝑎 − 1 𝑧 − 𝑏 + 1 𝑧 + 𝑏 ) . For 𝑎 = 1, 𝑏 = 3 and 𝑧 = 𝑖𝑦, 𝐽𝑥 (0, 𝑦) = 𝑚 𝜋 ( − 1 𝑦2 + 1 + 3 𝑦2 + 9 ) , 𝐽 𝑦 = 0. Its zeros are 𝑦 = ± √ 3, 𝐽𝑥 (0, 0) = −2𝑚/(3𝜋), and the tails are positive. The signed wall flux is zero because the source and sink strengths cancel: ∫ ∞ −∞ 𝐽𝑥 (0, 𝑦) 𝑑𝑦 = 0. Question difficulty allocation: 𝐴 = 12, 𝐵 = 6, 𝐶 = 0, 𝐷 = 2. 6