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CED objectives

Electric Power

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletPower Bench · each bulb's power three ways from its own readings, a fourth bar from the whole emf, and a clock that turns watts into joules

Electric power is $P = I\,\Delta V = I^2R = (\Delta V)^2/R$, and the three forms agree only when every symbol belongs to the same element, so a series resistor's power must be computed from its own current or its own potential difference rather than from the source voltage. Which form to reach for is decided by what the elements share: a shared current in series makes $P = I^2R$ the ranking rule and favours the larger resistance, while a shared potential difference in parallel makes $P = (\Delta V)^2/R$ the rule and favours the smaller one. Power is a rate, so energy is $P\,\Delta t$, and a kilowatt-hour is an energy equal to $3.6\times10^6$ J.

Three errors dominate. Using the source potential difference in a power formula for an element that receives only part of it, which is picking the formula to match the numbers on the page rather than the element in question. Ranking dissipated power by resistance alone without asking whether the elements share a current or a potential difference, which is why a $60$ W lamp outshines a $100$ W lamp when the two are wired in series. And reporting a power where an amount of energy was asked for, or dividing by the time instead of multiplying.

which element owns each symbol? 12 V 4 Ω 8 Ω I = 1 A, shared 12²/4 = 36 W the 12 V is across the PAIR I²R = 1² × 4 = 4 W the current IS the resistor's own cross-check with the element's own ΔV: IR = 4 V, so (ΔV)²/R = 16/4 = 4 W the three forms agree only when all three symbols name one element if a quantity belongs to the whole circuit, it does not belong in that element's formula
The 36 W route uses a number that was printed in the problem and belongs to something else. The 4 W route uses a number the resistor actually carries.
what do they share? that decides which form ranks them large R: BRIGHT small R SERIES: shared current P = I²R favours the LARGER R so a 60 W lamp outshines a 100 W lamp here: the higher rating belongs to the smaller resistance large R small R: BRIGHT PARALLEL: shared ΔV P = (ΔV)²/R favours the SMALLER R this is the case a bulb's printed rating assumes, since household fixtures are wired in parallel
The two bulbs are the same pair in both panels. Committing to one ranking rule gets one of these panels wrong every time.

The work

3 ways in · any order
Lesson
Electric Power ›

Matches every symbol in a power formula to one element, picks the ranking rule from what the elements share, and keeps a rate in watts distinct from an amount of energy in joules.

Skill check · 10 scenarios
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Diagnostic
10-item topic check ›

Ten items spanning the failure modes of this topic: using the source voltage in one element's power formula, ranking brightness by resistance without checking the connection, and reporting a rate where an energy was asked for. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
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Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions
Diagnostic first