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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

AP Physics 2-Concise Summary Notes- All Topics

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 = \(40 \, \text{W}\).

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