11.4 Electric Power

Electric power tells you how fast a circuit element transfers, converts, or dissipates energy, and you find it with $P = I \Delta V$. From that equation and Ohm's law you can also use $P = I^2 R$ and $P = \frac{(\Delta V)^2}{R}$, and because brightness rises with power, you can rank bulbs by comparing their power.

Why This Matters for the AP Physics 2 Exam

Power links current, voltage, and resistance to real energy transfer, so it shows up across circuit problems. You will use it to predict bulb brightness, explain where energy gets dissipated as heat, and back up claims with the right equation. Since the free-response section rewards clear justification, being able to say which power formula fits the values you know, and why, is a real advantage. Power reasoning also supports lab work where you collect current and voltage data and analyze how energy moves through a circuit.

more resources to help you study

practice multiple choiceFRQ practice & scoringcheatsheetsscore calculator

Key Takeaways

• Power is the rate of energy transfer, measured in watts (joules per second).
• The core relationship is $P = I \Delta V$, connecting current and potential difference.
• Using Ohm's law ($\Delta V = IR$), you can rewrite power as $P = I^2 R$ or $P = \frac{(\Delta V)^2}{R}$.
• Pick the power formula that matches the quantities you already know.
• Brightness increases with power, so power lets you compare bulbs qualitatively.
• Resistors convert electrical energy to thermal energy (Joule heating), so dissipated power shows up as heat.

Transfer of Energy in Electric Circuits

Energy constantly transfers into, out of, or within electric circuits. Electric power measures how quickly this energy moves or transforms.

The rate at which energy transfers, converts, or dissipates through a circuit element depends on two things: the current flowing through it and the electric potential difference (voltage) across it.

• Power is the rate of energy transfer, measured in watts (joules per second).
• A circuit element with high power is transferring energy rapidly.
• Depending on the device, that energy can leave as heat, light, sound, or mechanical motion.

The fundamental power equation relates current and voltage:

$P = I \Delta V$

Where:

• $P$ = power (watts)
• $I$ = current (amperes)
• $\Delta V$ = electric potential difference (volts)

From this equation, you can derive two more formulas using Ohm's law ($\Delta V = IR$):

$P = I^2 R = \frac{(\Delta V)^2}{R}$

Where:

• $R$ = resistance (ohms)

These three forms give you multiple ways to calculate power depending on which values you already have.

Predicting Bulb Brightness

Power helps explain many everyday electrical situations. The brightness of a light bulb increases with its power, so power lets you compare bulbs.

When comparing similar types of bulbs:

• A higher power bulb is brighter than a lower power bulb.
• A 100 W incandescent bulb produces much more light than a 40 W incandescent bulb.
• The difference comes from the higher power bulb transferring energy at a faster rate.

You can use power to qualitatively predict the relative brightness of bulbs in different circuit setups. This is a quick way to see how energy is distributed through a circuit.

How to Use This on the AP Physics 2 Exam

Problem Solving

• Start by listing what you know: current, voltage, or resistance. Then choose the matching power form.
• Use $P = I \Delta V$ when you have current and voltage.
• Use $P = I^2 R$ when you know current and resistance (common for resistors in series, where current is shared).
• Use $P = \frac{(\Delta V)^2}{R}$ when you know voltage and resistance (useful when voltage is fixed across a branch).
• Watch your units: amperes, volts, and ohms give you watts.

Free Response

• When you justify brightness, name the power formula you used and explain why it fits the situation.
• If resistance stays constant and voltage changes, state that power scales with the square of voltage.
• Tie dissipated power to energy conservation: electrical energy converts to thermal energy in resistors.

Common Trap

• The "brightest bulb" is not always the one with the most voltage or the most current. It is the one with the most power. Always compare power, not a single quantity.

Practice Problem 1: Power Calculation

Solution

Use $P = I^2 R$ since you know current and resistance.

Given:

• Resistance ($R$) = 20 Ω
• Current ($I$) = 3 A

Substitute: $P = I^2 R = (3 \text{ A})^2 \times 20 \text{ Ω} = 9 \times 20 = 180 \text{ W}$

The resistor dissipates 180 watts, which is converted to thermal energy (heat).

Practice Problem 2: Comparing Bulb Brightness

Solution

To compare brightness, compare the power of each bulb.

For the single bulb:

• It gets the full voltage: 120 V.
• If its resistance is $R$, its power is $P_{\text{single}} = \frac{(120 \text{ V})^2}{R} = \frac{14400}{R}$.

For the two bulbs in series:

• Each bulb gets half the voltage: 60 V (identical bulbs split voltage equally in series).
• Each series bulb has power $P_{\text{series}} = \frac{(60 \text{ V})^2}{R} = \frac{3600}{R}$.

Compare the powers: $\frac{P_{\text{series}}}{P_{\text{single}}} = \frac{3600/R}{14400/R} = \frac{3600}{14400} = \frac{1}{4}$

Each bulb in the series connection is only 1/4 as bright as the single bulb, because power is proportional to the square of voltage when resistance stays constant.

Common Misconceptions

• "Higher voltage always means brighter." Brightness tracks power, not voltage alone. A bulb can have higher voltage but lower power depending on current and resistance.
• "Current is used up as it powers a bulb." Current is the same on both sides of a single element; energy is transferred, but charge is conserved.
• "Only one power formula is correct." $P = I \Delta V$, $P = I^2 R$, and $P = \frac{(\Delta V)^2}{R}$ all give the same result. Pick the one that matches your known values.
• "Power and energy are the same thing." Power is the rate of energy transfer per second. Energy is power multiplied by time.
• "Resistors don't really lose energy." Resistors convert electrical energy into thermal energy, so dissipated power leaves the circuit as heat.

Related AP Physics 2 Guides

• 11.2 Simple Circuits
• 11.3 Resistance, Resistivity, and Ohm's Law
• 11.5 Compound Direct Current (DC) Circuits
• 11.8 Resistor-Capacitor (RC) Circuits
• 11.1 Electric Current
• 11.6 Kirchhoff's Loop Rule