Mistake Master
Student view — seeing the site as a student does
CED objectives

Electric Power

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletPower Lab · two bulbs, series or parallel, each bulb's power from its own current and voltage and once from a borrowed one

Electric power is the rate of energy conversion in a circuit element, $P = IV$, which for a resistor also reads $P = I^2R = V^2/R$. All three forms are equivalent, and all three require that the current, voltage and resistance in them belong to the same element. Because series elements share a current and parallel elements share a voltage, the larger resistance dissipates more in series while the smaller dissipates more in parallel. Energy follows from power by multiplying time, $E = Pt$, and the kilowatt-hour is that product, equal to $3.6$ million joules.

Three errors dominate. Feeding a power formula a value borrowed from elsewhere, classically $V_{\text{battery}}^2/R$ for one resistor in a series chain, which routinely returns more power than the whole battery delivers. Ranking bulb brightness by counting components, so a bulb added in parallel is expected to dim the others when with an ideal source it leaves them untouched and simply draws more from the battery. And interchanging power with energy, reporting watts where joules or kilowatt-hours belong, or dividing by the elapsed time instead of multiplying by it.

one current, two different shares of the voltage 12 V 4 Ω 8 Ω drops 4 V P = 4 W drops 8 V P = 8 W I = 1 A the same 1 A returns here 12² / 4 = 36 W borrows the battery's volts, and 36 W > the 12 W supplied series: current is shared, so use P = I²R and the BIGGER R wins
The struck line is the standard slip. Its own arithmetic exceeds the battery's entire output, which is the check that catches a borrowed voltage before anything else does.
identical bulbs, identical battery, opposite outcomes V/2 each each at V² / 4R a QUARTER of full, not half full V each at V² / R adding the second dims neither
Counting bulbs predicts the same answer on both sides and is wrong on one of them. What decides the outcome is whether the added bulb shares the voltage or gets its own copy of it.

The work

3 ways in · any order
Lesson
Electric Power

Derives the three power forms from P = IV, enforces that every symbol come from the element being asked about, and replaces bulb-counting with a recomputation of each bulb's own dissipated power.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: feeding a power formula the battery's voltage instead of the element's own, ranking brightness by how many components are in the circuit, importing the series rule into a parallel network, and interchanging watts with joules and kilowatt-hours. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
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