Design Energy Conversion Devices - Grade 11 Physics

1. How should the eight modules be connected?
   Answer: Connect all eight in series, giving $16 V$ and internal resistance $8.0 Ω$, which matches the load and delivers $8.0 W$ When a source with internal resistance $r$ drives a load $R_L$, maximum power transfer occurs when $r = R_L$. This is the key principle here: you need to arrange the modules so their combined internal resistance equals $8.0\ \Omega$. Each module is a $2.0\ \text{V}$ source with $1.0\ \Omega$ internal resistance. Connecting all eight in series adds both the voltages and the resistances: total EMF $= 8 \times 2.0 = 16\ \text{V}$, and total internal resistance $= 8 \times 1.0 = 8.0\ \Omega$. Since $r = R_L = 8.0\ \Omega$, the maximum power condition is satisfied. The delivered power is $P = \left(\frac{16}{8+8}\right)^2 \times 8 = (1.0)^2 \times 8 = 8.0\ \text{W}$. That confirms choice A is correct. Choice B (two in series, four strings in parallel) gives $4\ \text{V}$ and $r = \frac{2.0}{4} = 0.5\ \Omega$. The mismatch with $8.0\ \Omega$ is severe, so power drops to roughly $1.4\ \text{W}$. Choice C (four in series, two strings in parallel) gives $8\ \text{V}$ and $r = \frac{4.0}{2} = 2.0\ \Omega$. Closer, but still mismatched — power is only about $5.1\ \text{W}$. Choice D (all eight in parallel) gives $2\ \text{V}$ and $r = \frac{1.0}{8} = 0.125\ \Omega$, the worst mismatch possible, yielding under $0.5\ \text{W}$. The study tip: whenever a question asks for maximum power delivery, immediately calculate what internal resistance each arrangement produces and compare it to the load. The winning configuration is always the one where $r = R_L$, regardless of how the individual sources are split up.
2. Which design estimate and module connection are most appropriate?
   Answer: About $5.0 W$; connect six modules in series per string and six strings in parallel When a question combines energy calculations with circuit design, tackle the two parts separately: first find the power output, then determine the module arrangement needed to meet the voltage requirement. For the power calculation, start with the mechanical work per compression: $W = F \times d = 600 \times 0.005 = 3.0 \text{ J}$. Applying the 20% efficiency gives $0.20 \times 3.0 = 0.60 \text{ J}$ of electrical energy per compression. With 500 pedestrians per minute, that's $500 \times 0.60 = 300 \text{ J per minute}$, or $\frac{300}{60} = 5.0 \text{ W}$. So the power estimate is 5.0 W. For the module arrangement, the storage circuit needs 12 V input, and each module produces 2.0 V. To reach 12 V, you need $\frac{12}{2.0} = 6$ modules connected in series per string. The tile has 36 modules total, so you can create $\frac{36}{6} = 6$ parallel strings. This is exactly what B describes — making it the correct answer. Choice A gets the module arrangement right but miscalculates the power, likely forgetting to apply the 20% efficiency factor and arriving at an inflated ~25 W. Choice C correctly calculates 5.0 W but puts all 36 modules in a single series string, which would produce $36 \times 2.0 = 72 \text{ V}$ — far exceeding the required 12 V input. Choice D uses the wrong series count (12 modules gives 24 V, not 12 V) and undercalculates the power. A useful habit: always separate the energy/power calculation from the circuit design calculation, and verify that your series string voltage matches the specified requirement exactly.
3. What spring constant is required if the spring is initially compressed by $0.20 m$?
   Answer: About $726 N/m$, including both masses, the rise during spring expansion, and the conversion loss When a question involves a spring launcher with energy losses and multiple moving parts, your instinct should be to build a complete energy accounting from start to finish — every joule in must be traced carefully to every joule out. Here's the full picture for answer D. After separation, the payload ($0.20\text{ kg}$) rises $4.0\text{ m}$, so its kinetic energy at separation is $KE_{payload} = mgh = (0.20)(9.8)(4.0) = 7.84\text{ J}$. Since both masses have the same speed at separation, the platform's KE is half that (half the mass): $KE_{platform} = (0.10)(9.8)(4.0) = 3.92\text{ J}$. The system also rose $0.20\text{ m}$ during spring expansion, gaining gravitational PE: $\Delta PE_{rise} = (0.30)(9.8)(0.20) = 0.588\text{ J}$. Total useful energy at separation: $7.84 + 3.92 + 0.588 = 12.35\text{ J}$. Accounting for the 85% efficiency: $E_{spring} = 12.35 / 0.85 = 14.53\text{ J}$. Since $E_{spring} = \frac{1}{2}kx^2$, we get $k = \frac{2(14.53)}{(0.20)^2} \approx 726\text{ N/m}$, confirming D. Answer A fails because it ignores the platform's kinetic energy entirely — a significant chunk of energy is simply discarded. Answer B includes both masses and the initial rise correctly but skips the 85% efficiency factor, underestimating how much spring energy is actually needed. Answer C applies the efficiency factor but neglects the $0.20\text{ m}$ rise during spring expansion, losing the gravitational PE term. Your strategy: on spring-energy problems, sketch an "energy inventory" — list every form of energy the spring must supply, including gravitational PE during launch and any efficiency losses. Missing even one term shifts your answer by a meaningful margin.
4. What minimum usable energy capacity should the capacitor have for one typical stop?
   Answer: $72 kJ$, because it must store the generated energy the battery cannot accept When a problem involves energy conversion and storage systems, your job is to track where energy flows — and where the system has limits. Here, you need to find how much electrical energy is generated, how much the battery can absorb, and what's left over for the capacitor. Start by calculating the kinetic energy lost during braking: $\Delta KE = \frac{1}{2}(1500)(18^2 - 6^2) = \frac{1}{2}(1500)(324 - 36) = \frac{1}{2}(1500)(288) = 216 \text{ kJ}$. The motor-generator converts 75% of that into electrical energy: $0.75 \times 216 = 162 \text{ kJ}$. The battery accepts up to 90 kJ, so the capacitor must handle the remainder: $162 - 90 = 72 \text{ kJ}$. That confirms C — the capacitor needs a minimum usable capacity of 72 kJ. A is wrong because 54 kJ represents the energy lost due to imperfect conversion (the 25% that never becomes electrical), not the overflow the capacitor must handle. B is wrong because the capacitor doesn't need to match the battery's capacity — it only needs to cover what the battery cannot accept, which is a completely different quantity. D is wrong because 162 kJ is the total electrical energy generated; storing all of it in the capacitor would ignore the 90 kJ the battery already handles. A useful strategy: in multi-step energy problems, build a "flow chart" — total energy → converted energy → battery share → capacitor remainder. Each step has its own calculation, and confusing any two adjacent values is exactly how distractors are designed to catch you.
5. What minimum flywheel moment of inertia should the designer select?
   Answer: $15.3 kg·m^2$, accounting for efficiency, final speed, and the reserve Flywheel energy problems require you to work backwards from the load — start with what the device needs, then account for every efficiency loss and safety margin before sizing the flywheel. Here's the chain: the device demands $600\,W \times 100\,s = 60{,}000\,J$. But the generator is only 80% efficient, so the flywheel must supply $\frac{60{,}000}{0.80} = 75{,}000\,J$. Then add the 10% reserve: the flywheel must store $\frac{75{,}000}{0.90} = 83{,}333\,J$ of usable energy. Now apply the rotational energy formula: $\Delta E = \frac{1}{2}I(\omega_i^2 - \omega_f^2)$, so $83{,}333 = \frac{1}{2}I(120^2 - 60^2) = \frac{1}{2}I(10{,}800)$. Solving gives $I = \frac{83{,}333}{5{,}400} \approx 15.4\,kg{\cdot}m^2$, confirming C is correct. Choice A ignores both the efficiency loss and the reserve, and incorrectly assumes the flywheel spins all the way to zero — but the problem explicitly gives a final speed of $60\,rad/s$, so that unused energy stays locked in the flywheel. Choice B correctly applies the generator efficiency but forgets the 10% reserve requirement, yielding a flywheel too small to guarantee safe operation. Choice D misreads the generator's inefficiency as an extra output demand (essentially penalizing twice), which artificially inflates the required energy and gives an oversized, incorrect result. When you see multi-constraint energy problems, build a checklist: load energy → divide by efficiency → divide by usable fraction → solve for the unknown. Missing any one factor is exactly how distractors are designed.
6. Which gearbox design and operating conclusion are correct?
   Answer: Use a $25:1$ speed-increasing ratio; the required crank torque is about $4.3 N·m$ When a question involves a gearbox connecting a human-powered crank to a generator, your job is to match three things: the generator's ideal speed, the gear ratio needed to get there, and whether the human can actually supply enough power. Start with the gear ratio. The generator needs $100 \text{ rad/s}$ but the crank turns at $4.0 \text{ rad/s}$. To increase speed, you need a speed-increasing gearbox with ratio $\frac{100}{4.0} = 25:1$. This immediately eliminates B (which reduces speed) and confirms the ratio in D. Next, check power. The crank's available mechanical power is $P = \tau \cdot \omega = 5.0 \times 4.0 = 20 \text{ W}$. The generator converts 70% of that to electrical power: $0.70 \times 20 = 14 \text{ W}$. Since the radio needs only 12 W, you have enough — the radio can operate. Now verify the required crank torque in D. The generator needs enough mechanical input power: $\frac{12}{0.70} \approx 17.1 \text{ W}$. Since an ideal gearbox conserves power, the crank must supply $17.1 \text{ W}$. The required torque is $\tau = \frac{P}{\omega} = \frac{17.1}{4.0} \approx 4.3 \text{ N}\cdot\text{m}$, which is within the comfortable 5.0 N·m limit. D is correct. A is wrong because the ratio is 20:1 (not 25:1) and ignores generator efficiency entirely. B uses a speed-*reducing* ratio, which would make the generator spin slower, moving it away from its efficient point. C correctly identifies the 25:1 ratio but falsely claims gears multiply power demand — in an ideal gearbox, power is conserved, not amplified. Remember: gearboxes trade torque for speed (or vice versa), but never create or destroy power. Always apply efficiency after establishing power balance.
7. Which configuration is expected to produce the greatest electrical power?
   Answer: Flow rate $0.08 m^3/s$, head loss $1 m$, and combined efficiency $85$ When a question gives you multiple configurations with different flow rates, head losses, and efficiencies, your job is to calculate the output power for each one and compare — don't try to eyeball it. The formula is $P = \rho g Q h \eta$, where $h$ is the net head (gross head minus frictional loss), not the gross head alone. With $\rho g = 1000 \times 9.8 = 9800$, you can compute each case quickly: A: $h = 15 - 1 = 14 \text{ m}$, so $P = 9800 \times 0.08 \times 14 \times 0.85 \approx 9330 \text{ W}$ B: $h = 15 - 2 = 13 \text{ m}$, so $P = 9800 \times 0.10 \times 13 \times 0.72 \approx 9154 \text{ W}$ C: $h = 15 - 5 = 10 \text{ m}$, so $P = 9800 \times 0.12 \times 10 \times 0.76 \approx 8938 \text{ W}$ D: $h = 15 - 8 = 7 \text{ m}$, so $P = 9800 \times 0.14 \times 7 \times 0.75 \approx 7203 \text{ W}$ Option A produces the highest power at roughly 9330 W, confirming it's the correct answer. Options B and C are tempting because they have higher flow rates, but larger flow rates come with greater pipe friction and lower efficiency, eroding any gain. Option D is the worst performer — despite the highest flow rate, the massive 8 m head loss cuts the net head nearly in half. The key trap here is assuming more flow always means more power. Always account for all three factors — flow rate, net head, and efficiency — before drawing conclusions.
8. Which design best meets these conditions?
   Answer: Use $k≈62 N/m$ and moderate generator damping to preserve resonant motion When a question asks about optimal energy transfer in an oscillating system, think resonance: maximum power is extracted when the system's natural frequency matches the driving frequency, combined with moderate damping that allows sustained motion while still extracting energy. Start by calculating the spring constant needed to match the wave period of $T = 4.0\text{ s}$. Rearranging $T = 2\pi\sqrt{m/k}$ gives $k = \frac{4\pi^2 m}{T^2}$. Plugging in $m = 25\text{ kg}$ and $T = 4.0\text{ s}$: $k = \frac{4\pi^2 (25)}{(4.0)^2} = \frac{987}{16} \approx 62\text{ N/m}$ So the spring stiffness must be approximately $62\text{ N/m}$. That immediately narrows things down. Then, for sustained energy harvest, damping must be moderate — strong enough to extract energy, but not so strong it kills the oscillation. That's exactly what D describes: $k \approx 62\text{ N/m}$ with moderate damping to preserve resonant motion. A uses $k \approx 247\text{ N/m}$, which corresponds to a natural period of $T = 2.0\text{ s}$ — half the wave period, so the buoy is off-resonance and transfers far less energy. B gets the spring constant right but uses maximum damping, which would overdamp the system, suppressing oscillation and defeating the generator's purpose. C uses $k \approx 16\text{ N/m}$, giving a much longer natural period ($\approx 7.9\text{ s}$), again mismatched to the waves. Your study tip: always compute $k$ first using the given period, then evaluate the damping condition — questions like this often hide two separate criteria, and both must be satisfied.
9. What is the minimum number of standard panels required?
   Answer: Four panels, because three panels do not provide the required energy reserve When tackling solar power design problems, work through the energy chain step by step: generation → storage losses → usable output → reserve requirement. Start with the daily energy the station actually needs, including the 20% reserve: $30\,W \times 24\,h \times 1.20 = 864\,Wh$. Since all energy passes through the battery (80% efficient), the panels must deliver more than what the station consumes — divide by battery efficiency: $864 \div 0.80 = 1080\,Wh$ of raw panel output required per day. Now find what one panel produces. Each panel generates: $500\,W/m^2 \times 0.60\,m^2 \times 0.18 \times 6.0\,h = 324\,Wh/day$. Dividing the requirement by one panel's output: $1080 \div 324 \approx 3.33$ panels. Since you can't install a fraction of a panel, you must round up to 4 panels, making B correct. Choice A (two panels) ignores both the reserve requirement and storage losses — two panels yield only $648\,Wh$, far below the $1080\,Wh$ needed. Choice C (three panels) is the most tempting trap: three panels produce $972\,Wh$, which exceeds the station's raw consumption of $720\,Wh$, but fails to meet the full requirement once the 20% reserve and battery losses are both properly accounted for. Choice D (five panels) misapplies the losses by "doubling" penalties that should only appear once each, arriving at an inflated requirement. Your study tip: always apply efficiency factors as divisors on the requirement side (or multipliers on the supply side) — never apply the same loss twice, and always round panel counts up, never down.
10. What upper-reservoir volume allows the system to use all the available input energy and later recover the corresponding electrical output?
    Answer: About $8160 m^3$, storing the fraction of input energy delivered by the pumps Pumped-storage questions require you to carefully track where energy is lost at each stage. The key insight: the volume of water stored depends on how much energy actually reaches the reservoir — not how much electrical energy entered the system, and not how much comes out the other end. Here's the reasoning for C: The pumps are 80% efficient, so of the $5.0 \times 10^9 \text{ J}$ of input electrical energy, only $0.80 \times 5.0 \times 10^9 = 4.0 \times 10^9 \text{ J}$ is actually converted into gravitational potential energy. That stored energy equals $mgh = \rho V g h$, so solving for volume: $V = \frac{4.0 \times 10^9}{1000 \times 9.8 \times 50} \approx 8160 \text{ m}^3$. This is the correct reservoir volume — it reflects exactly the energy the pumps successfully deliver upward. A is wrong because it uses $6120 \text{ m}^3$, which corresponds to the recovered electrical output (applying both efficiencies: $0.80 \times 0.75 \times 5.0 \times 10^9$). That tells you how much electricity you get back, not how much water you stored. B incorrectly uses the full $5.0 \times 10^9 \text{ J}$ as if 100% of input energy becomes gravitational potential energy, ignoring pump inefficiency entirely and giving $10200 \text{ m}^3$. D applies both efficiencies to find the reservoir volume, which double-counts the turbine efficiency — turbine losses happen during recovery, not during filling. When solving energy-chain problems, always ask: "At which stage does each efficiency apply?" Apply only the relevant efficiency for the quantity you're solving for.

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