EEE U111 · Tutorial companion · Sheets 1–2

Tutorial sheets 1–2: hints and verified solutions

Every question from the two sheets, redrawn, with a hint to read before attempting and a full solution folded underneath. Read the hint, try the question, then open the solution only to compare.

Start here

Twenty-five questions, and they fall into just nine pattern families (table at the end). Sheet 2 is entirely Module 1 §1.3–1.5 material — KCL at one node, KVL round one loop, dependent sources — which the Module 1 lesson covers. Sheet 1 mixes that with three "fundamentals" families that are not in the Bobrow sections the handout lists (reading charge and energy off a graph, resistivity of a wire, nameplate ratings): they are Hughes-style and new to everyone in the room. The Module 1 extras lesson teaches exactly those three, nothing more.

Every answer here was recomputed independently. Three places where the sheet's printed answer needs a comment are marked with a rust Note box (T1 Q15, T2 Q3, T2 Q9).

How to work a sheet

Before any algebra: for a graph question, write the operation (area or slope?) and the units of the answer; for a circuit, letter the nodes and mark a polarity on every element you don't yet know. Half the wrong answers on these sheets are sign or unit slips made before the first equation.

Tutorial 1 · Fundamentals

Pages: Module 1 lesson §1, §3, §5 (power, sign convention, capacitor and inductor) and the extras lesson for Q1, Q5, Q9, Q11 (graphs and averages), Q8, Q10 (wire), Q2, Q3, Q6, Q7, Q12 (ratings).

Q1 · Charge from a current graphgraph → area · §1.1

The current through a capacitor is the piecewise-linear graph below. Charge acquired in the first 5 μs? (a) 5 μC (b) 10 μC (c) 15 μC (d) 20 μC

Hintq = ∫i dt is the area under the current graph. Cut the region 0–5 μs into a triangle and two trapezoids; A × μs = μC.
Solution
  1. 0–3 μs: triangle, ½·3·5 = 7.5.
  2. 3–4 μs: trapezoid from 5 down to 3, (5+3)/2·1 = 4.
  3. 4–5 μs: the segment from (4, 3) to (6, 5) has slope 1 A/μs, so i(5) = 4; trapezoid (3+4)/2·1 = 3.5.
  4. Total 7.5 + 4 + 3.5 = 15 μC — (c).
Q2 · Two coils in parallelratings · R = V²/P

Two coils in parallel across 100 V dc draw 10 A total; one coil dissipates 600 W. Resistance of each?

HintIn parallel every coil sees the full 100 V. Get total power, split it, then R = V²/P for each.
Solution
  1. Total P = VI = 100·10 = 1000 W, so the other coil takes 400 W.
  2. R₁ = 100²/600 = 16.67 Ω, R₂ = 100²/400 = 25 Ω.
  3. Check by currents: 6 A + 4 A = 10 A ✓.

Trap: R = 100/10 = 10 Ω is the combined resistance, not either coil's.

Q3 · Lamps in seriesratings · resistance is the fixed thing

How many 200 W/220 V lamps in series consume the same total power as one 100 W/220 V lamp (on 220 V)? (a) not possible (b) 4 (c) 3 (d) 2

HintA nameplate fixes the lamp's resistance, not its power. Work out R for a 200 W lamp, put n of them in series across 220 V, and set the power equal to 100 W.
Solution
  1. R = 220²/200 = 242 Ω per lamp.
  2. n in series: P = 220²/(n·242) = 200/n W.
  3. 200/n = 100 ⇒ n = 2 — (d). (Each lamp then runs at 110 V, 50 W, dimly.)
Q4 · Same R, current source vs voltage sourcepower puzzle · §1.2

A resistor absorbs 18 W from a current source, and 4.5 W from a voltage source of the same numerical magnitude. Source value and R?

HintWrite both powers: I²R = 18 and V²/R = 4.5 with V = I numerically. Divide one equation by the other and I disappears.
Solution
  1. I²R = 18, I²/R = 4.5.
  2. Divide: R² = 4 ⇒ R = 2 Ω; then I² = 9 ⇒ I = 3 A.
  3. 3 A and 2 Ω — (b). Check: 9·2 = 18 ✓, 9/2 = 4.5 ✓.
Q5 · Current from a charge graphgraph → slope · §1.1

Charge delivered by a source is the graph below. Current at (a) t = 1 s, (b) t = 3 s?

Hinti = dq/dt is the slope. Each straight piece has one constant slope; mC/s = mA.
Solution
  1. 0–2 s: slope 10 mC / 2 s = 5 mA at t = 1 s.
  2. 2–5 s: slope −10 mC / 3 s = −3.33 mA at t = 3 s. Negative: charge is flowing back into the source.
Q6 · Toaster on the wrong voltageratings · under-voltage

A 1000 W, 240 V toaster is connected to 220 V. Damaged? Rating affected?

HintThe rating tells you R; the supply tells you what actually flows. Compare the actual current with the rated current.
Solution
  1. R = 240²/1000 = 57.6 Ω; rated current 1000/240 = 4.17 A.
  2. At 220 V: I = 220/57.6 = 3.82 A — below rating, so not damaged.
  3. Power actually drawn 220²/57.6 = 840.3 W (sheet prints 840.4). The rating is a property of the device and is unaffected; the toaster just runs under-powered.
Q7 · Max voltage across two rated resistors in seriesratings · the weaker link

150 Ω/2 W in series with 100 Ω/1 W. Largest voltage across the pair without exceeding either rating?

HintSeries ⇒ one common current. Find the maximum current each resistor tolerates (I = √(P/R)); the smaller one is the limit for both.
Solution
  1. 150 Ω/2 W: I ≤ √(2/150) = 0.1155 A. 100 Ω/1 W: I ≤ √(1/100) = 0.100 A.
  2. Limit is 0.100 A; then V = 0.1·(150+100) = 25 V.

Trap: adding the two rated voltages (17.3 V + 10 V = 27.3 V) overloads the 100 Ω resistor — they can't both be at their limit at once.

Q8 · Diameter from two resistance readingswire · R = ρL/A

50 m of 2 mm² wire has 0.56 Ω; 100 m of the same material has 2 Ω. Diameter of the second wire?

HintSame material ⇒ same ρ. Get ρ from the first wire, then A from the second, then d = √(4A/π). Keep A in mm² throughout if you like — ρ cancels anyway.
Solution
  1. Ratio form: R ∝ L/A, so A₂/A₁ = (L₂/L₁)·(R₁/R₂) = 2 · 0.56/2 = 0.56.
  2. A₂ = 0.56 · 2 = 1.12 mm².
  3. d = √(4·1.12/π) = √1.426 = 1.19 mm.

Via ρ explicitly: ρ = RA/L = 0.56·2×10⁻⁶/50 = 2.24×10⁻⁸ Ω·m, then A = ρL/R = 1.12×10⁻⁶ m² — same thing.

Q9 · Average power from a square wavegraph → average · §1.2

±10 V square wave across 10 Ω. Average power? (a) 5 W (b) 10 W (c) 100 W (d) 1000 W

HintPower is v²/R, and v² doesn't care about the sign of v. What is v² at every instant?
Solution
  1. v = ±10 ⇒ v² = 100 at all times.
  2. p = 100/10 = 10 W constant, so the average is 10 W — (b).

Trap: the average voltage is 0, but average power is not Vavg²/R. Average the power, not the voltage.

Q10 · Turns on a wire-wound resistorwire · length from R

0.2 mm constantan wire (ρ = 49×10⁻⁸ Ω·m) on a 1 cm former. Turns for 50 Ω?

HintTwo lengths: the wire length L = RA/ρ that gives 50 Ω, and the length of one turn, πD. Divide.
Solution
  1. Wire area A = π(0.1×10⁻³)² = 3.142×10⁻⁸ m².
  2. L = 50 · 3.142×10⁻⁸ / 49×10⁻⁸ = 3.206 m.
  3. One turn = π·0.01 = 0.03142 m; N = 3.206/0.03142 = ≈ 102 turns.
Q11 · Average of a triangle wavegraph → average

Current bounces linearly between 5 A and 10 A. Average value? (a) 2.5 (b) 5 (c) 7.5 (d) 10 A

HintAverage over one period = area under one period ÷ period. Each half-period is a trapezoid between 5 and 10.
Solution
  1. Over 0 to π the graph is a trapezoid of parallel sides 5 and 10 and width π: area π(5+10)/2.
  2. Average = π·7.5/π = 7.5 A — (c). (For any straight-line rise or fall the average is the midpoint.)
Q12 · Household load: current and kWhratings · energy

8 × 100 W lamps, 3 × 80 W fans, one ½ hp refrigerator, one 1000 W heater on 230 V. (a) Total current. (b) Energy per day if a quarter of the load runs all the time.

Hint1 hp = 746 W. Add the watts, divide by 230 V. For (b), energy = average power × hours; 1 kWh = 1000 W for 1 h.
Solution
  1. Load = 800 + 240 + 373 + 1000 = 2413 W.
  2. (a) I = 2413/230 = 10.5 A.
  3. (b) ¼ · 2413 W · 24 h = 14 478 Wh = 14.48 kWh.
Q13 · Energy absorbed by a real inductorgraph + R + L · §1.7

Current ramps 0 → 6 A over 2 s then holds; the coil has R = 1 Ω and L = 2 H. Energy absorbed in 4 s? (a) 144 J (b) 98 J (c) 132 J (d) 168 J

HintA "coil with resistance" is a resistor and an inductor. Resistor energy: ∫i²R dt piece by piece. Inductor energy: only ½Li² at the end matters (it absorbs nothing while i is constant).
Solution
  1. 0–2 s: i = 3t. Resistor: ∫₀² 9t² dt = 9·8/3 = 24 J.
  2. 2–4 s: i = 6. Resistor: 36·1·2 = 72 J. Resistor total 96 J.
  3. Inductor: ½·2·6² = 36 J stored by t = 2 s, unchanged after.
  4. Total 96 + 36 = 132 J — (c).

Trap: 144 J (option a) is what you get by pretending 6 A flowed for the whole 4 s in the resistor — the ramp matters.

Q14 · Exponential current, v = 5 di/dtcalculus signals · §1.7

i(t) = 3e−2t A enters the + terminal; v = 5 di/dt. (a) Charge delivered 0–2 s. (b) Power absorbed. (c) Energy absorbed in 3 s.

Hint(a) integrate i; (b) differentiate i for v, then p = vi — expect a negative sign; (c) integrate p. Note d/dt(e−2t) = −2e−2t.
Solution
  1. (a) q = ∫₀² 3e−2t dt = 1.5(1 − e−4) = 1.5 × 0.9817 = 1.4725 C.
  2. (b) v = 5·(−6e−2t) = −30e−2t; p = vi = −90e−4t W. Negative: the device is delivering power (an inductor discharging).
  3. (c) w = ∫₀³ −90e−4t dt = −22.5(1 − e−12) ≈ −22.5 J.
Q15 · Sinusoidal charge and voltagecalculus signals · units

q = 10 sin 4πt mC, v = 2 cos 4πt V. (a) Power at t = 0.3 s. (b) Energy delivered 0–0.6 s.

Hintq is in millicoulombs, so i comes out in mA. Differentiate for i, multiply for p; for (b) use cos²x = (1 + cos 2x)/2.
NoteThe sheet prints (a) as 164.5 W. With q in mC the current is 40π cos 4πt mA, so the power is 164.5 mW; the printed (b) 78.34 mJ is consistent with mW, not W. Worth a one-line check with the tutor.
Solution
  1. i = dq/dt = 40π cos 4πt mA = 0.04π cos 4πt A.
  2. p = vi = 0.08π cos² 4πt W. At 0.3 s: 4π·0.3 = 1.2π rad = 216°, cos = −0.809, cos² = 0.6545, p = 0.2513·0.6545 = 0.1645 W = 164.5 mW.
  3. (b) w = 0.08π ∫₀0.6 ½(1 + cos 8πt) dt = 0.04π[t + sin 8πt/(8π)]₀0.6. sin 4.8π = sin 0.8π = 0.5878; bracket = 0.6 + 0.0234 = 0.6234; w = 0.1257·0.6234 = 78.3 mJ.

Tutorial 2 · KVL and KCL

Pages: Module 1 lesson §4 (the two laws) and §6 (dependent sources); Reading a circuit diagram for node-spotting. Each circuit below is redrawn from the sheet; letters and arrows in blue are ones I added to fix the working.

The game on this sheet

Every question is one of three moves: KCL at one node (sum of currents leaving = 0), KVL round one loop (walk it, add rises, subtract drops, arrive back at 0), or a voltage walk between two nodes through whatever path is fully known. Dependent sources add one line: write what they depend on in terms of your unknown before you use them.

Q1 · v₀ and i₀ with a dependent current sourceKCL · one node pair

Find v₀ and i₀ using KCL.

HintEverything is in parallel: one top node, one bottom node, one voltage v₀ across all four branches. Write i₀ in terms of v₀ first, then KCL at the top node.
Solution
  1. i₀ = v₀/2 (2 Ω sees v₀).
  2. KCL, top node, currents leaving downward = 9 A entering: v₀/2 + i₀/4 + v₀/8 = 9.
  3. v₀(½ + ⅛ + ⅛) = ¾v₀ = 9 ⇒ v₀ = 12 V, i₀ = 6 A. Check: 6 + 1.5 + 1.5 = 9 ✓.
Q2 · Current through a dependent voltage sourceKCL + KVL · dependent source

Obtain I, the current through the 3VR source, using KVL.

HintName the current down the middle branch im (so VR = 2im). The left branch is a bare 2 A whatever the 3 Ω does. KCL at the top node links I and im; KVL round the middle–right loop gives the second equation.
Solution
  1. KCL at top node A: I arrives from the right branch; 2 A leaves left, im leaves down: I = 2 + im.
  2. Voltage A→B via the middle: 4 + 2im. Via the right (B up to A: rise 3VR, then drop 5I across 5 Ω): 3VR − 5I = 6im − 5I.
  3. Equate: 4 + 2im = 6im − 5I ⇒ 4im − 5I = 4. Put im = I − 2: 4I − 8 − 5I = 4 ⇒ I = −12 A.
  4. Check: im = −14, VR = −28; middle gives −24 V, right gives −84 + 60 = −24 V ✓.
Q3 · Two halves joined only by a dependent sourceKVL · read the gap

Obtain i using KVL. (a) 0.31 A (b) 1.25 A (c) 1.75 A (d) 2.5 A

NoteLook at the sheet's figure: nodes a and b are not joined by a wire — Vab is the open-circuit voltage across a gap. Read it as a wire and you get 0.5 A (not an option). Only the gap reading gives the key's 1.25 A.
HintWith the gap, the left half is a plain divider: Va follows at once. The right half is one loop whose only "source" is 4Vab, and Vb is set by the mesh current through the 1 Ω. Write Vab in terms of i, then KVL.
Solution
  1. Left: no current leaves node a to the right, so Va = 5·1/(1+1) = 2.5 V.
  2. Right loop, clockwise i: up through the 1 Ω at b, right through 3 Ω, down through the source (+ to −). Drops: i + 3i + 4Vab = 0 ⇒ i = −Vab.
  3. i flows up through the 1 Ω from ground to b, so Vb = −i and Vab = 2.5 + i.
  4. i = −2.5 − i ⇒ i = −1.25 A: 1.25 A in magnitude, flowing anticlockwise — (b).
Q4 · Voltage walk past unknown elementsKVL · voltage walk

Find (a) vR2, (b) v₂, given vR1 = 1 V. The blobs are unknown elements whose voltages are labelled.

HintKVL doesn't care what an element is, only its labelled voltage. Start at c = 0 V and walk: up the 8 V, across the 12 V — that's vR2. Keep walking to a; note b and c are the same node (bottom wire). Then walk a→b down the right side.
Solution
  1. From c (0 V) up through 8 V (+ at bottom): drop 8 → −8 V. Through 12 V left→right (− to +): rise 12 → +4 V. That node is the top of R₂: vR2 = 4 V.
  2. Continue: 7 V (+ left): drop 7 → −3 V. Blob "− 9 V +": rise 9 → Va = 6 V.
  3. a→b down the right: blob "− v₂ +" (rise v₂), 3 V source (− top, + bottom: rise 3), R₁ (+ top: drop 1). Vb = 6 + v₂ + 3 − 1 = 0 ⇒ v₂ = −8 V.
Q5 · KCL with negative labelsKCL · one node

1 A leaves the voltage source's + terminal into R₁. Find i₂.

HintOne KCL at the top node. A branch labelled "−3 A down" is a fact, not a puzzle: treat it exactly like "+(−3) A down".
Solution
  1. Into the top node: 1 A. Out (downward): i₂ + (−3) + 7.
  2. 1 = i₂ + 4 ⇒ i₂ = −3 A. The −2 V label never enters — resistor values weren't needed either.
Q6 · Five parallel branchesKCL · one node

Determine v.

HintAll five branches share the same v. Net source current into the top node = current the two 5 Ω resistors carry down.
Solution
  1. Into top node: 1 + 2 − 5 = −2 A.
  2. Out through resistors: v/5 + v/5 = 2v/5.
  3. 2v/5 = −2 ⇒ v = −5 V.
Q7 · Current-controlled current sourceKCL · two junctions

Find ix.

HintOnly two nodes (top and bottom). Call the top voltage v. Both resistors run from top to bottom round the outside. KCL at the right top junction (where i₁ arrives) links v and i₁; KCL at the left top junction gives the second equation.
Solution
  1. Right junction: in i₁; out 3i₁ down and v/20k through the 20 kΩ. i₁ = 3i₁ + v/20k ⇒ v = −40k·i₁.
  2. Left junction: in from the 5 kΩ side −v/5k (i.e. v/5k leaves leftward); out 4 mA and i₁: −v/5k = 4 mA + i₁ ⇒ 8i₁ = 4 mA + i₁.
  3. i₁ = 4/7 mA ≈ 0.571 mA. Bottom-right junction: in 3i₁ + v/20k = 3i₁ − 2i₁ = i₁, all leaving as ix: ix = i₁ = 4/7 mA.
Q8 · Four branch voltages from KVLKVL · four loops

Find V₁ to V₄.

HintFour small loops, one unknown each once you go in the right order: bottom-right loop has only V₄ unknown. Then bottom-left gives V₃, top-left gives V₁, top-right gives V₂. Walk each loop clockwise and honour every printed polarity.
Solution
  1. Bottom-right (clockwise from its bottom-left corner): up V₄ (rise), across 2 V left→right (drop), down 5 V (+ top: drop): V₄ − 2 − 5 = 0 ⇒ V₄ = 7 V.
  2. Bottom-left: up 4 V (+ at bottom: drop 4), across V₃ (+ left: drop), down V₄ (drop): −4 − V₃ − 7 = 0 ⇒ V₃ = −11 V.
  3. Top-left: up 3 V (rise), down V₁ (drop), leftward across V₃ from − to + (rise): 3 − V₁ + V₃ = 0 ⇒ V₁ = −8 V.
  4. Top-right: up V₁ (rise), down V₂ (− top: rise), leftward across 2 V from − to + (rise): V₁ + V₂ + 2 = 0 ⇒ V₂ = 6 V.
Q9 · Voltage walk along a chainKVL + KCL · chain

500 mA flows through the 7.3 Ω. Determine vx.

NoteWith + at the top of the source as drawn, the node voltage works out to −5.05 V. The sheet prints 5.05 V — the magnitude is right; ask which polarity the tutor intended, and either way this is the "fix the polarity before you write anything" habit in action.
HintWalk left to right: node 1's voltage is 2.3 minus the drop across 7.3 Ω. KCL at node 1 tells you what goes on through the series 2 Ω; another drop gives node 2. vx is node 2's voltage (bottom rail = 0).
Solution
  1. V₁ = 2.3 − 0.5·7.3 = −1.35 V.
  2. Through the 1 Ω: −1.35 A downward. KCL at node 1: on to the 2 Ω, 0.5 − (−1.35) = 1.85 A.
  3. V₂ = −1.35 − 2·1.85 = −5.05 V, so vx = −5.05 V (5.05 V in magnitude).
Q10 · KCL with a series current sourceKCL · one node

Current Ix in mA?

HintA resistor in series with a current source carries the source's current, full stop — the right-hand 100 Ω is a red herring. Then KCL at the middle node with unknown V.
Solution
  1. The right branch delivers exactly 10 mA into the middle node.
  2. KCL: (V − 1)/100 + V/100 = 0.01 ⇒ 2V − 1 = 1 ⇒ V = 1 V.
  3. Ix = 1/100 = 10 mA. (The 1 V source then carries no current at all.)

What the two sheets tell you

Nine families cover all 25 questions. The right-hand column is where each is taught — and which the practice drill randomises.

FamilyQuestionsWhere it's taughtDrill
Graph → area (charge, energy)T1 Q1, Q13Extras §1✓
Graph → slope (current from charge)T1 Q5Extras §1✓
Average value / average power of a waveformT1 Q9, Q11Extras §2✓
Nameplate ratings: R = V²/P, limits, kWhT1 Q2, Q3, Q6, Q7, Q12Extras §3✓
Wire: R = ρL/AT1 Q8, Q10Extras §4✓
Calculus-defined signals (p = vi, w = ∫p)T1 Q4, Q14, Q15Module 1 lesson §3, §5— (MATH U101 daily drill builds the calculus)
KCL at one node, incl. dependent sourcesT2 Q1, Q5, Q6, Q7, Q10Module 1 lesson §4, §6✓
KVL voltage walk / loop with unknown elementsT2 Q4, Q8Module 1 lesson §4✓
Chain / dependent-source loop (KCL + KVL together)T2 Q2, Q3, Q9Module 2 lesson (nodal)✓ chain
Classic traps on these sheets

1) Averaging voltage instead of power (T1 Q9). 2) Treating a nameplate power as fixed when the voltage changes — the resistance is fixed (T1 Q3, Q6). 3) Reading a gap as a wire (T2 Q3). 4) Forgetting the resistor part of a real coil (T1 Q13). 5) Unit prefixes — mC → mA, μs·A → μC (T1 Q1, Q15). 6) Writing an equation before fixing a polarity or current arrow (T2 Q9).