Electrical Power, Energy & Circuit Design — Print · Grade 12 Science
Chapter: Electricity, Magnetism & CircuitsElectrical Power, Energy & Circuit Design
Learning goals
- I can calculate component power and energy with correct circuit values.
- I can distinguish efficiency, kilowatt-hours, and source internal losses.
Readings for this unit
Before this unit
Current, voltage, resistance, series/parallel circuits, and time conversion.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why is the lowest wattage not enough to choose a useful design?
Illustrative evidence case. Two devices draw different power but must deliver the same agreed service.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why is the lowest wattage not enough to choose a useful design?
Your first model
Define service, operating time, source limits and comparison boundary.
Build the language from the idea
- kilowatt-hour
- Energy delivered by 1 kW over 1hour, equal to 3.60MJ.
- internal resistance
- A simplified source resistance causing voltage drop and heating.
- emf
- Source energy per charge under the modeled open-circuit condition.
Electrical power is energy-transfer rate. For a component with voltage V and current I, P=VI. An ohmic resistor also gives P=I²R=V²/R because V=IR. A device’s actual resistance can depend on temperature; not every electronic component follows a constant-R model.
Electrical energy transferred over a constant-power interval is E=Pt. The kilowatt-hour is an energy unit: 1 kWh=1000W×3600 s=3.60×10⁶ J. It is not a kilowatt per hour. A household bill can be modeled from energy amount times a supplied price per kWh, but real tariffs include other charges.
In a series circuit, current is the same through components and voltage drops add; in parallel, each branch has the same potential difference and branch currents add. Power calculations must use each component’s own voltage or current, not indiscriminately the supply total.
Batteries have internal resistance. In a simple discharging model, terminal voltage V=emf−Irinternal, and internal heating is I²rinternal. A short circuit can cause a large current and dangerous heating; a battery is not an unlimited constant-voltage source at all loads.
A practical design specifies required function, maximum current, operating duration, and energy capacity. Efficiency compares useful output with electrical input. Different lamps can use different powers for similar useful illumination; comparison should specify light output and lifetime rather than watts alone.
Build only low-voltage circuits with an appropriate educational kit and adult guidance; never use wall sockets, improvised mains wiring, or short a battery. Heating-appliance examples below are paper problems. Circuit calculations do not substitute for equipment ratings or electrical installation knowledge.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Connect power to electrical quantities
Electrical power P=VI uses the voltage across and current through the same component at the same operating point. For an ohmic resistor, substituting V=IR gives I²R and V²/R. These are equivalent forms under that model, not independent laws for arbitrary devices. A larger resistance lowers power at fixed V but increases power at fixed I; the held-constant condition determines the comparison.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
At fixed 12 V, what happens to power if resistance doubles from 24 to 48 Ω?
Visual reasoning — after your attempt
It halves from 6 to 3 W; current also halves.
For an appropriate steady circuit, electrical power transferred to a component is P=VI. Ohmic resistor forms I²R and V²/R follow by substitution. They are not interchangeable statements about a fixed real device under every source condition; identify what is held constant.
At fixed voltage, a larger ohmic resistance reduces power; at fixed current, a larger resistance increases it. The source condition changes the comparison.
Pause and explain
Why can resistance-power comparisons appear contradictory?
Check your reasoning after answering
They hold different variables constant. Use the actual source constraint before selecting P=V²/R or P=I²R.
Part 2 · Use energy units and duty cycle
A kilowatt-hour is an energy unit, not a power per hour:1 kWh=3.60×10⁶ J. Multiply power in kW by on-time in hours. A duty cycle is the fraction of a period during which a device operates in the specified mode, so a rated load running part-time uses less energy than continuous full-power operation. Include standby or multiple modes when they matter rather than multiplying a nameplate rating by every hour automatically.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What energy does a 2 kW load use at 25% duty over 8 h if off-state power is negligible?
Visual reasoning — after your attempt
2×0.25×8=4 kWh under the stated operating model.
Energy is power integrated over time; constant power gives E=PΔt. Kilowatt-hours measure energy, not power. A device’s maximum rated power is not necessarily its average use; duty cycle and operating behavior can change the total.
A 100 W device operating for 10 hours uses 1 kWh at constant power. A 50% on-off duty cycle over that interval changes the calculation.
Pause and explain
Why does a watt rating alone not determine daily energy use?
Check your reasoning after answering
Operating duration, duty cycle, and actual operating power are also needed.
Part 3 · Add loads within a stated source model
For ideal parallel loads sharing 12 V, source current is the sum of branch currents. Here 2+3+1=6 A and source output power is 72 W. A real source and conductors have operating limits and losses, so an ideal current sum is only part of a design comparison. Consider voltage regulation, continuous versus peak demand and the stated component ratings. Use supplied-data design here rather than modifying household wiring.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
If one 3 A branch is switched off, what ideal total remains?
Visual reasoning — after your attempt
3 A and 36 W at unchanged 12 V, assuming the source maintains that voltage.
Parallel loads can be designed to share a source voltage, while total current is the sum of branch currents. Source capacity and wiring constraints need separate evidence. A proposed circuit should specify required service, branch behavior, and component ratings rather than just a total watt sum.
Two 12 V loads at 12 W and 6 W draw 1.0 A and 0.5 A under the ideal stated conditions.
Pause and explain
Does a supply voltage label establish sufficient capacity?
Check your reasoning after answering
No. Available current, power, regulation, and other ratings must be compatible with the loads.
Part 4 · Separate service efficiency and design evidence
A fair device comparison holds the delivered service comparable, such as an agreed illumination pattern and duration. Lower input power can reduce operational energy if the service is actually met, but lifespan, production, repair and disposal need a wider life-cycle account. An attractive label is not a substitute for measured performance. State criteria and constraints before selecting an option, then monitor whether the promised service and savings occur in use.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What is the operational energy difference over 2000 h for these options?
Visual reasoning — after your attempt
(12−8) W×2000 h=8000 Wh=8 kWh, for the same stated service and duration.
A useful comparison defines the same service and operating interval. Lower electrical input can reflect greater efficiency, reduced service, or a different duty cycle. Evaluate measured output alongside input, and retain uncertainty before selecting a design.
A lamp using fewer watts is not automatically better for a task if it provides insufficient illumination or a shorter service life.
Pause and explain
What makes an energy-saving comparison meaningful?
Check your reasoning after answering
Equivalent useful service, comparable conditions, duration, and relevant performance measures are needed in addition to input energy.
Apply the whole unit · Account for two ideal low-voltage loads
Illustrative regulated 12.0 V source supplies two parallel loads rated 12.0 W and 6.00 W at that voltage. Both operate 4.00 hours daily; neglect source losses.
Your task
Find branch and total current, daily energy, and required source power. Explain why an actual source choice still requires ratings and why a lower-power alternative needs a service comparison.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
Branch currents=P/V: 1.00 A and 0.500 A; total 1.50 A.
Total load power=18.0 W; daily energy=18.0×4.00=72.0 Wh=0.0720 kWh.
Ideal source must supply the 18.0 W load; actual source losses and ratings need a separate account.
The calculation uses ideal stated load behavior at 12 V and does not select a real supply. A lower-power replacement can reduce energy only if it still provides the required service under comparable use. Actual measured voltage, current, duty cycle, and source efficiency can change the input-energy budget.
Scientific references and source-model foundations
Parent / teacher teaching notes
Ask for an energy-versus-power unit check and equivalent service criteria. Keep the activity to supplied diagrams or approved enclosed low-voltage learning equipment. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
The key idea
Compute component power with its own voltage/current and distinguish watts from energy.
Worked examples
Example 1
- P=12×.50=6W.
- Time=600 s; E=6×600.
Example 2 · Worked application
- 60 W=0.060 kW; energy=0.060×5=0.30 kWh.
- 5 h=18000 s; E=60×18000=1.08×10⁶ J.
- The two forms agree using 1 kWh=3.6×10⁶ J.
Kilowatt-hour is energy, not power per hour.
Example 3
- E=1.2×2=2.4 kWh.
- Cost=2.4×.20.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Kilowatt-hour is energy, not power per hour.
- A kWh is a unit of energy; watts and kilowatts measure power.
Kilowatt-hour is energy, not power per hour.
Example 5 · Reason from the visual model
- For ideal parallel loads sharing 12 V, source current is the sum of branch currents. Here 2+3+1=6 A and source output power is 72 W. A real source and conductors have operating limits and losses, so an ideal current sum is only part of a design comparison. Consider voltage regulation, continuous versus peak demand and the stated component ratings. Use supplied-data design here rather than modifying household wiring.
- 3 A and 36 W at unchanged 12 V, assuming the source maintains that voltage.
Calculated ideal branch demand; no household electrical work is part of this exercise.
Avoid these mistakes
- Watts measure power; watt-hours and joules measure energy.
Check your understanding
How would you compare two lamp designs fairly?
Discuss the answer
Compare equal useful light output, energy over the same use, lifetime, and relevant disposal or cost assumptions.
Return to the phenomenon
Return to the initial case. Why is the lowest wattage not enough to choose a useful design? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Power and time determine energy; a good design also tests performance, durability and impacts under transparent criteria.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why is the lowest wattage not enough to choose a useful design? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Power and time determine energy; a good design also tests performance, durability and impacts under transparent criteria.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A 15 W device is replaced by a 9 W device providing the same service for 2500 h/year. Find annual energy saved in kWh and joules, and name one uncounted life-cycle term.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year. Production, transport, replacement lifespan or disposal is outside this operational-only result.
Words to know
- kilowatt-hour
- Energy delivered by 1 kW over 1hour, equal to 3.60MJ.
- internal resistance
- A simplified source resistance causing voltage drop and heating.
- emf
- Source energy per charge under the modeled open-circuit condition.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
How would you compare two lamp designs fairly?
Check your explanation after answering
Compare equal useful light output, energy over the same use, lifetime, and relevant disposal or cost assumptions.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
1.8 W
P=VI.
Problem 2
Answer and explanation
Energy; it is power multiplied by time.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 3
Hints
List the given quantities, select a relationship, and check units before substituting.
Answer and explanation
0.008 kWh
.004 kW×2h=.008 kWh.
Problem 4
Answer and explanation
Very low external resistance permits large current; internal resistance then dissipates substantial I²r heat.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 5
Hints
Kilowatt-hour is energy, not power per hour.
Answer and explanation
1 kWh, equivalent to 3.6×10⁶ J.
Problem 6
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
The 10 W lamp uses less energy per equal operating time; broader impact also requires lifetime and other factors.
Problem 7
Answer and explanation
3 A and 36 W at unchanged 12 V, assuming the source maintains that voltage.
For ideal parallel loads sharing 12 V, source current is the sum of branch currents. Here 2+3+1=6 A and source output power is 72 W. A real source and conductors have operating limits and losses, so an ideal current sum is only part of a design comparison. Consider voltage regulation, continuous versus peak demand and the stated component ratings. Use supplied-data design here rather than modifying household wiring.
3 A and 36 W at unchanged 12 V, assuming the source maintains that voltage.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 8
Answer and explanation
Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year. Production, transport, replacement lifespan or disposal is outside this operational-only result.
Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year.
Production, transport, replacement lifespan or disposal is outside this operational-only result.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.