AP Physics C E&M- 11.4 Electric Power- Study Notes- New Syllabus
AP Physics C E&M- 11.4 Electric Power – Study Notes
AP Physics C E&M- 11.4 Electric Power – Study Notes – per latest Syllabus.
Key Concepts:
- Transfer of Energy in an Electric Circuit (Power)
Transfer of Energy in an Electric Circuit (Power)
Power in an electric circuit represents the rate at which energy is transferred, converted, or dissipated by a circuit element. It depends on both the current through the element and the potential difference across it.
Relevant Equation: 
\( \mathrm{P = I V} \)
- \(\mathrm{P}\) = power (watts, W)
- \(\mathrm{I}\) = current through the element (A)
- \(\mathrm{V}\) = potential difference across the element (V)
Derived Equations:
Using Ohm’s law (\( V = I R \)):
\( \mathrm{P = I V = I (IR) = I^2 R} \)
Or, equivalently:
\( \mathrm{P = \dfrac{V^2}{R}} \)
Interpretation:
- Power represents the rate of energy transfer in joules per second.
- Resistors and lightbulbs dissipate energy as heat or light.
- The brightness of a lightbulb is proportional to the power dissipated in it.
Energy Transfer in Time:
Over a time interval \( t \), the total electrical energy transferred by a circuit element is:
\( \mathrm{E = P t = (IV)t} \)
Key Idea: Power can be used both to calculate quantitative energy transfer and to predict qualitative effects (such as bulb brightness in series/parallel circuits).
Example
Two identical bulbs, each of resistance \(10 \, \Omega\), are connected in series across a \(20 \, V\) battery. Find the power dissipated in each bulb and compare their brightness to a single bulb connected across the same battery.
▶️ Answer/Explanation
Step 1: Total resistance: \( \mathrm{R_{total} = 10 + 10 = 20 \, \Omega} \).
Step 2: Current: \( \mathrm{I = \dfrac{V}{R} = \dfrac{20}{20} = 1.0 \, A} \).
Step 3: Voltage across each bulb: \( \mathrm{V = IR = (1.0)(10) = 10 \, V} \).
Step 4: Power per bulb: \( \mathrm{P = I V = (1.0)(10) = 10 \, W} \).
Step 5: Single bulb across \(20 \, V\): \( \mathrm{P = \dfrac{V^2}{R} = \dfrac{400}{10} = 40 \, W} \).
Final Answer: Each bulb in series dissipates \(10 \, W\) (dim), while a single bulb would dissipate \(40 \, W\) (brighter).
Example
The same two \(10 \, \Omega\) bulbs are connected in parallel across the same \(20 \, V\) battery. Find the power dissipated in each bulb and compare the brightness with the series case.
▶️ Answer/Explanation
Step 1: Voltage across each bulb = battery voltage = \(20 \, V\).
Step 2: Power per bulb: \( \mathrm{P = \dfrac{V^2}{R} = \dfrac{400}{10} = 40 \, W} \).
Step 3: Total power = \( 2 \times 40 = 80 \, W \).
Final Answer: Each bulb in parallel dissipates \(40 \, W\) (bright, same as a single bulb). Parallel bulbs are much brighter than in the series case.