AP® Physics 2: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics-2/units/11/11-4)
Unit 11 · Topic 11.4
11.4 Electric Power
Electric power is the rate at which an element transfers energy, P = IΔV = I²R = (ΔV)²/R, measured in watts. A bulb's brightness rises with its power, so comparing power lets you rank bulbs in any circuit. Choosing the right form of the power equation is the key skill.
Key terms
- electric power
- watt
- brightness
- energy dissipated
Power in a circuit element
Each coulomb that passes through an element with potential difference ΔV across it transfers ΔV joules. With I coulombs passing each second, the energy transferred per second is:
Power is measured in watts (1 W = 1 J/s). In a resistor or bulb, that energy becomes thermal energy and light. In a battery, chemical energy is turned into electrical energy. The total energy transferred over a time interval is E = PΔt.
Three forms of the same equation
Combining P = IΔV with Ohm's law gives two more forms:
All three give the same answer for a given element. Choose the one that uses what's the same across the elements you're comparing.
- Elements in series share the same current, so compare with P = I²R: the bigger resistance gets more power.
- Elements in parallel share the same potential difference, so compare with P = (ΔV)²/R: the smaller resistance gets more power.
Brightness
A bulb's brightness increases with the power it uses. To rank bulbs, rank their powers. For identical bulbs, the one with more current through it (or more potential difference across it) is brighter.
A bulb rated “60 W at 120 V” uses 60 W only when it has 120 V across it. Connect it to a lower potential difference and it uses less power and glows more dimly. With R fixed, power goes as (ΔV)², so half the potential difference gives a quarter of the power.
Energy conservation in circuits
In any circuit, the power supplied by the battery equals the total power used by all the other elements. That's energy conservation, and it's a useful check on your answers: add up I²R for every resistor and compare with I × ε for the battery.
A real battery with internal resistance r also loses some power inside itself, I²r, which is why batteries warm up when they deliver a large current. The power delivered to the rest of the circuit is the terminal voltage times the current.
Power companies bill for energy, not power, using the kilowatt-hour: 1 kWh is 1000 W for one hour, or 3.6 × 10⁶ J.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A household bulb
A bulb is rated at 60 W when connected to 120 V. Find the current through it, its resistance, and the energy it uses in 2.0 hours.
Show the solutionHide the solution
- Step 1: Current: I = P/ΔV = 60 W / 120 V = 0.50 A.
- Step 2: Resistance: R = ΔV/I = 120/0.50 = 240 Ω (or R = (ΔV)²/P = 14,400/60 = 240 Ω).
- Step 3: Energy: E = PΔt = (60 W)(2.0 × 3600 s) = 432,000 J.
Answer: 0.50 A, 240 Ω, about 4.3 × 10⁵ J
- Example 2
Series versus parallel bulbs
Two identical bulbs, each with resistance R, are connected to an ideal battery of potential difference ε. Compare each bulb's power when they're in series and when they're in parallel, relative to a single bulb on the same battery.
Show the solutionHide the solution
- Step 1: Single bulb: P₀ = ε²/R.
- Step 2: Series: the total resistance is 2R, so each bulb gets ε/2. Each bulb's power = (ε/2)²/R = ε²/(4R) = P₀/4.
- Step 3: Parallel: each bulb is connected straight across the battery, so each gets the full ε. Each bulb's power = ε²/R = P₀.
- Step 4: So bulbs in parallel each glow as brightly as a single bulb, while bulbs in series are much dimmer.
Answer: Series: each bulb uses P₀/4. Parallel: each uses P₀.
- Example 3Calculator allowed
Which resistor gets more power? (classic trap)
A 2.0 Ω and a 4.0 Ω resistor are connected to a 12 V battery. Which one uses more power if they're in series? Which if they're in parallel?
Show the solutionHide the solution
- Step 1: Series: the current is 12/(2.0 + 4.0) = 2.0 A in both. Using P = I²R: 2.0 Ω uses 8.0 W and 4.0 Ω uses 16 W. The bigger resistor wins.
- Step 2: Parallel: each has 12 V across it. Using P = (ΔV)²/R: 2.0 Ω uses 72 W and 4.0 Ω uses 36 W. The smaller resistor wins.
- Step 3: The trap is assuming one rule works for both. Which resistor uses more power depends on how they're connected.
Answer: Series: the 4.0 Ω resistor (16 W vs. 8.0 W). Parallel: the 2.0 Ω resistor (72 W vs. 36 W).
Common mistakes
- Assuming a bigger resistance always means a brighter bulb. That's true in series, but in parallel the smaller resistance is brighter.
- Using a bulb's rated power when it's connected to a different potential difference.
- Thinking bulbs use up current. They transfer energy; the current in and out of a bulb is the same.
On the exam
- Bulb-brightness ranking questions are very common. Justify each ranking with the current through or potential difference across each bulb and the right power equation.
- Expect questions about how brightness changes when a bulb is added, removed or shorted; compare the power before and after.
Connected topics
Videos
Check yourself
4 questions on 11.4 Electric Power. Pick an answer to see if you got it, and why.
A lamp connected to a 120 V outlet carries a current of 0.50 A. At what rate does it use electrical energy?
A bulb is rated at 60 W when connected to 120 V. Assuming its resistance stays the same, what power does it use when connected to 60 V?
A 1500 W space heater runs for 2.0 hours. How much electrical energy does it use?
For a single ohmic resistor, which describes a graph of the power used (vertical axis) against the current in it (horizontal axis)?
0 of 4 answered