11.4 Electric Power

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, you can also write power as $P = I^2R$ or $P = \frac{\Delta V^2}{R}$. On AP Physics C: E&M, these equations help you compare energy transfer and lightbulb brightness in circuits.

Electric power questions ask you to 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. Each task ties together current, potential difference, resistance, and energy transfer.

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, 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 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 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.

Practice Problem 1: Power in a Resistor

Solution

To solve this problem, find the power dissipated by the resistor. 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

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.