AP Physics 2- 11.4 Electric Power- Study Notes- New Syllabus
AP Physics 2- 11.4 Electric Power – Study Notes
AP Physics 2- 11.4 Electric Power – Study Notes – per latest Syllabus.
Key Concepts:
- Energy Transfer and Power in Electric Circuits
Energy Transfer and Power in Electric Circuits
Power in Electric Circuits:
Power (\(\text{P}\)) describes the rate at which energy is transferred into, out of, or within an electric circuit.
\(\text{P} = \dfrac{\text{W}}{t} = \text{VI}\)
- \(\text{W}\): Work done (or energy transferred), in joules (J)
- \(\text{t}\): Time, in seconds (s)
- \(\text{V}\): Potential difference across the circuit element (volts, V)
- \(\text{I}\): Current through the circuit element (amperes, A)
Alternate Forms of Power:
Using Ohm’s law (\(\text{V} = \text{IR}\)), power can also be expressed as:
\(\text{P} = \text{I}^2 \text{R} \quad \text{or} \quad \text{P} = \dfrac{\text{V}^2}{\text{R}}\)
- \(\text{I}^2 \text{R}\): Describes the rate of thermal energy dissipation in resistors (Joule heating).
- \(\dfrac{\text{V}^2}{\text{R}}\): Useful when only voltage and resistance are known.
Energy Transferred in Circuits:
The total electrical energy transferred over time is given by:
\(\text{E} = \text{Pt} = \text{VIt}\)
Interpretations of Energy Transfer:
- Into the circuit: A battery or EMF source provides energy to charges.
- Within the circuit: Resistors convert energy into heat, lightbulbs convert energy into heat and light, capacitors store energy in electric fields.
- Out of the circuit: Useful work (like powering a motor or producing light).
Example :
A 60 W light bulb is connected to a 120 V supply. Find the current through the bulb.
▶️ Answer/Explanation
Step 1: Use the power formula: \(\text{P} = \text{VI}\).
Step 2: Rearrange: \(\text{I} = \dfrac{\text{P}}{\text{V}} = \dfrac{60}{120}\).
Step 3: Calculate: \(\text{I} = 0.50 \, \text{A}\).
Answer: Current = \(0.50 \, \text{A}\).
Example :
A 10 Ω resistor carries a current of 2 A. Find the rate of heat energy dissipated in the resistor.
▶️ Answer/Explanation
Step 1: Use the formula: \(\text{P} = \text{I}^2 \text{R}\).
Step 2: Substitute values: \(\text{P} = (2)^2 (10) = 40 \, \text{W}\).
Step 3: Interpretation: The resistor dissipates 40 J of energy as heat per second.
Answer: Power = \(40 \, \text{W}\).