Power in a circuit tells you how fast electrical energy is being converted into other forms — usually heat (toaster, hair dryer), light (bulb), or motion (motor). When you pay your electricity bill, you’re literally paying for power × time = energy.
The MCAT tests circuit power with three equivalent formulas. The trick isn’t memorizing any one of them — it’s knowing which form to reach for depending on what the question gives you.
The Three Power Formulas
When to use each:
P=IV — when you know current and voltage directly.
P=I2R — when you know current and resistance. Common in series circuits, where current is the same through every resistor.
P=V2/R — when you know voltage and resistance. Common in parallel circuits, where voltage is the same across every resistor.
Brightness of Bulbs
The MCAT loves asking which bulb in a circuit is brightest. The rule is simple:
Brightness ∝ power dissipated by the bulb.
But which power formula to use depends on the configuration:
In a series circuit, all bulbs carry the same current. So P=I2R — the bulb with the highest resistance is brightest.
In a parallel circuit, all bulbs see the same voltage. So P=V2/R — the bulb with the lowest resistance is brightest.
Energy Dissipated
Power is energy per unit time. To find total energy dissipated by a resistor over a time t:
This is exactly how electricity bills work — you pay for energy in kilowatt-hours (kWh), not power. A 100 W bulb running 10 hours = 1 kWh ≈ 15¢.
RMS Values (AC Circuits)
The MCAT occasionally touches on alternating current (AC). In AC circuits, voltage and current oscillate sinusoidally — they’re constantly changing direction. To get a meaningful “average” power, we use root-mean-square (RMS) values.
The average power dissipated in an AC circuit uses RMS values: Pavg=IrmsVrms=Irms2R.
Two identical bulbs are connected in series to a battery. A third identical bulb is connected directly across the same battery (in parallel with the series pair). Which single bulb is brightest?
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The single bulb in parallel is brightest. The single bulb gets the full battery voltage. The two series bulbs split the voltage, so each gets half. Since P=V2/R and all bulbs have the same R, the one with the full voltage dissipates the most power.
An AC source has peak voltage 170 V. What's the RMS voltage? If connected to a 100 Ω resistor, what's the average power dissipated?
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Vrms=120 V; Pavg=144 W.Vrms=170/2≈120 V. P=Vrms2/R=14,400/100=144 W. (This is exactly a US wall-outlet circuit driving a 100 Ω resistor.)
A 1500 W microwave runs for 5 minutes. How much energy does it use, in joules and in kWh?
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450,000 J = 0.125 kWh.E=Pt=1500×300=450,000 J. In kWh: 1.5×(5/60)=0.125 kWh — about 2 cents of electricity.