---
title: "Electric Power in Circuits | AP Physics C: E&M"
description: "Review electric power in AP Physics C: E&M circuits, including P = IΔV, P = I²R, P = ΔV²/R, energy transfer, and bulb brightness."
canonical: "https://fiveable.me/ap-physics-c-e-m/unit-11/4-electric-power/study-guide/u2cRqQTlthIAJtwp"
type: "study-guide"
subject: "AP Physics C: E&M"
unit: "Unit 11 – Electric Circuits"
lastUpdated: "2026-06-09"
---

# Electric Power in Circuits | AP Physics C: E&M

## Summary

Review electric power in AP Physics C: E&M circuits, including P = IΔV, P = I²R, P = ΔV²/R, energy transfer, and bulb brightness.

## Guide

Electric power is the rate at which energy is transferred by a circuit element. The core relationship is $P = I \Delta V$.

Using [Ohm's law](/ap-physics-c-e-m/unit-11/3-resistance-resistivity-and-ohms-law/study-guide/TnRPkql9C75GQe0d "fv-autolink"), you can also write power as $$P = I^2R$$ or $$P = \frac{\Delta V^2}{R}$$. On [AP Physics C: E&M](/ap-physics-c-e-m "fv-autolink"), these equations help you compare energy transfer and lightbulb brightness in circuits.

## Why This Matters for the AP Physics C: E&M Exam

Electric power questions usually ask you to connect current, [potential difference](/ap-physics-c-e-m/key-terms/potential-difference "fv-autolink"), resistance, and energy transfer. You may calculate power directly, compare which bulb is brighter, or justify how changing a circuit changes the rate at which energy is delivered to a component.

## Energy Transfer in Circuits

### Power in Circuit Elements

Power measures the rate at which energy is transferred, converted, or dissipated by a circuit element. In electrical systems, power tells us how quickly energy flows through components like resistors, [capacitors](/ap-physics-c-e-m/unit-10/3-capacitors/study-guide/jGOEgdPDnNfNGUeR "fv-autolink"), or lightbulbs.

The fundamental equation for electric power is:
$$P = I \Delta V$$

Where:
- $$P$$ is power measured in watts (W)
- $$I$$ is current measured in amperes (A)
- $$\Delta V$$ is potential difference measured in volts (V)

This equation can be transformed using Ohm's Law ($$\Delta V = IR$$) to give us two additional useful forms:
- $$P = I^2 R$$ (useful when you know current and resistance)
- $$P = \frac{\Delta V^2}{R}$$ (useful when you know [voltage](/ap-physics-c-e-m/key-terms/voltage "fv-autolink") and resistance)

Higher power corresponds to a greater rate of energy transfer. For example, when comparing household appliances:
- A 1500W hair dryer consumes energy much faster than a 60W light bulb
- A 2000W microwave oven transfers energy more rapidly than a 100W television

### Lightbulb Brightness and Power

The brightness of a lightbulb directly correlates with its power consumption. This relationship provides a visible demonstration of power in action.

When more power is dissipated in a lightbulb:
- The filament gets hotter
- More energy is converted to light (and heat)
- The bulb appears brighter to our eyes

This relationship allows us to make qualitative predictions about lightbulb brightness in circuits:
- In a [series circuit](/ap-physics-c-e-m/key-terms/series-circuit "fv-autolink") with identical bulbs, they will have equal brightness
- In a parallel circuit with identical bulbs, they will also have equal brightness
- If bulbs have different resistances in the same circuit, their brightness will vary according to the power dissipated by each

To compare the brightness of different lightbulbs, calculate the power dissipated by each using any of the power equations, and the bulb with the highest power value will shine the brightest.

> **Boundary Statement**
>
> The AP Physics C: Electricity & Magnetism exam only covers the transfer of mechanical and electrical energy, but students should know that electrical energy can also be dissipated as thermal energy.

## Practice Problem 1: Power in a Resistor

> A 12Ω resistor is connected to a 9V battery. Calculate the power dissipated by the resistor.

**Solution**

To solve this problem, find the power dissipated by the [resistor](/ap-physics-c-e-m/key-terms/resistor "fv-autolink"). Since the voltage and resistance are given, use:
$$P = \frac{\Delta V^2}{R}$$

Substituting our values:
$$P = \frac{(9\text{ V})^2}{12\text{ Ω}} = \frac{81\text{ V}^2}{12\text{ Ω}} = 6.75\text{ W}$$

Therefore, the resistor dissipates 6.75 watts of power.

## Practice Problem 2: Comparing Lightbulb Brightness

> Three lightbulbs with resistances of 10Ω, 20Ω, and 30Ω are connected in parallel to a 120V source. Which bulb will be the brightest and which will be the dimmest?

**Solution**

To determine which bulb is brightest, we need to calculate the power dissipated by each bulb.

For bulbs in parallel, each experiences the full voltage of the source (120V). We can use $$P = \frac{\Delta V^2}{R}$$ to calculate the power for each bulb:

For the 10Ω bulb:
$$P_1 = \frac{(120\text{ V})^2}{10\text{ Ω}} = \frac{14,400\text{ V}^2}{10\text{ Ω}} = 1,440\text{ W}$$

For the 20Ω bulb:
$$P_2 = \frac{(120\text{ V})^2}{20\text{ Ω}} = \frac{14,400\text{ V}^2}{20\text{ Ω}} = 720\text{ W}$$

For the 30Ω bulb:
$$P_3 = \frac{(120\text{ V})^2}{30\text{ Ω}} = \frac{14,400\text{ V}^2}{30\text{ Ω}} = 480\text{ W}$$

Comparing the power values:
$$P_1 > P_2 > P_3$$

Therefore, the 10Ω bulb will be the brightest (1,440W), and the 30Ω bulb will be the dimmest (480W). This makes sense because when connected to the same voltage, the bulb with the lowest resistance will draw the most current and dissipate the most power.

## Vocabulary

- **circuit element**: A component in an electric circuit, such as a resistor or lightbulb, through which current flows and across which a potential difference exists.
- **current**: The flow of electric charge through a conductor, measured as the amount of charge passing through a cross-section per unit time.
- **electric potential difference**: The difference in electric potential energy per unit charge between two points in a circuit, measured in volts.
- **energy transfer**: The movement of energy from one location, system, or form to another within an electric circuit.
- **power**: The rate at which energy is transferred, converted, or dissipated by a circuit element, measured in watts.
- **resistance**: The opposition to the flow of electric current in a circuit, measured in ohms (Ω).
- **thermal energy**: Energy dissipated in the form of heat when electrical energy is converted within a circuit element.

## FAQs

### What determines the rate at which energy is delivered by a current?

Electric power determines the rate of energy transfer. In a circuit element, power is the product of current and potential difference: P = IΔV.

### What is the formula for electric power in a circuit?

The main formula is P = IΔV. For resistors, using Ohm's law gives P = I²R and P = ΔV²/R.

### When should I use P = I²R?

Use P = I²R when you know the current through a resistor and its resistance, or when current is easier to compare than voltage.

### When should I use P = ΔV²/R?

Use P = ΔV²/R when you know the potential difference across a resistor and its resistance, especially for bulbs or resistors in parallel.

### How does power relate to lightbulb brightness?

For AP circuit comparisons, the bulb dissipating more power is brighter. Calculate or compare power for each bulb using the quantities shared by the circuit connection.

### How is electric power tested on AP Physics C: E&M?

You may calculate power, compare energy transfer rates, justify bulb brightness, or explain how changing resistance, current, or potential difference affects power.

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