Unit 11 · Topic 11.4

11.4 Electric Power

Electric power is the rate a circuit element transfers energy. This topic gives the three forms of the power equation for resistors and shows how to use power to rank the brightness of bulbs, one of the most common circuit questions.

Key terms

electric power
watt
energy dissipated in a resistor
bulb brightness

Power in a circuit

When charge q moves through a potential difference $\Delta V$ , its electric potential energy changes by $q\Delta V$ . Doing that for a current I, which moves I coulombs each second, transfers energy at the rate $P = I\Delta V$ . Power is measured in watts, and 1 W = 1 J/s = 1 A·V.

For a battery, $P = I\mathcal{E}$ is the rate it supplies energy to the circuit. For a resistor, P is the rate it turns electrical energy into thermal energy (and light, for a bulb). That conversion is called energy dissipated in a resistor.

Three forms for a resistor

Combine $P = I\Delta V$ with $\Delta V = IR$ to get two more forms: $P = I^2R$ and $P = \dfrac{(\Delta V)^2}{R}$ . All three are always true for a resistor; pick the one that uses what you know, or what stays the same.

Elements in series share the same current, so compare them with $P = I^2R$ : the bigger resistor gets more power. Elements in parallel share the same potential difference, so compare them with $P = \dfrac{(\Delta V)^2}{R}$ : the smaller resistor gets more power.

Total energy transferred is power times time when the power is steady, $E = P\Delta t$ . If the power changes, as it does in an RC circuit, integrate: $E = \displaystyle\int P\,dt$ . Utility companies bill in kilowatt-hours; 1 kW·h = 3.6 × 10⁶ J.

Ranking bulb brightness

A bulb's brightness depends on the power it uses. More power means a brighter bulb. For identical bulbs, the one with the most current (or the biggest potential difference across it) is brightest.

Treat a bulb as a resistor with a fixed resistance unless a question says otherwise. Then follow these steps:

Find the current through each bulb, or the potential difference across it, using series and parallel rules (11.5).
Use $P = I^2R$ for bulbs carrying the same current and $P = \dfrac{(\Delta V)^2}{R}$ for bulbs with the same potential difference.
Rank by power, and say which bulbs are equally bright.

Ratings on real devices

A bulb labeled 60 W, 120 V uses 60 W only when it has 120 V across it. Its rating tells you its resistance: $R = \dfrac{(\Delta V)^2}{P}$ . Put it in a different circuit and it uses a different power.

A high-wattage bulb has a smaller resistance than a low-wattage bulb rated for the same voltage. That's why the results can flip when the bulbs are wired in series.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A household bulb
A bulb is rated 60 W at 120 V. Find the current it draws at its rated voltage, its resistance, and the energy it uses in 2.0 hours.
Show the solutionHide the solution
1. Step 1: $I = \dfrac{P}{\Delta V} = \dfrac{60}{120} = 0.50\text{ A}$ .
2. Step 2: $R = \dfrac{(\Delta V)^2}{P} = \dfrac{120^2}{60} = 240\ \Omega$ . Check: $\Delta V = IR = (0.50)(240) = 120$ V.
3. Step 3: Energy: $E = P\Delta t = (60\text{ W})(7200\text{ s}) = 4.32 \times 10^5\text{ J}$ , which is 0.12 kW·h.
Answer: 0.50 A, 240 Ω and 4.3 × 10⁵ J (0.12 kW·h)
Example 2Calculator allowed
Rated bulbs in series (classic trap)
A 60 W bulb and a 100 W bulb, both rated for 120 V, are connected in series across a 120 V source. Assume each bulb's resistance stays constant. Which bulb is brighter?
Show the solutionHide the solution
1. Step 1: Find each resistance from its rating: $R_{60} = \dfrac{120^2}{60} = 240\ \Omega$ and $R_{100} = \dfrac{120^2}{100} = 144\ \Omega$ .
2. Step 2: In series they share one current: $I = \dfrac{120}{240 + 144} = 0.3125\text{ A}$ .
3. Step 3: With the same current, use $P = I^2R$ : $P_{60} = (0.3125)^2(240) \approx 23\text{ W}$ and $P_{100} = (0.3125)^2(144) \approx 14\text{ W}$ .
4. Step 4: The bulb labeled 60 W is brighter. The trap is trusting the label; the rating only applies at 120 V across each bulb on its own.
Answer: The 60 W bulb is brighter (about 23 W versus 14 W)
Example 3Calculator allowed
Ranking three identical bulbs
Bulb A is connected in series with a battery and with a parallel pair of bulbs, B and C. All three bulbs are identical. Rank their brightness.
Show the solutionHide the solution
1. Step 1: All the current from the battery passes through A. At the junction it splits equally between B and C, because they're identical, so each carries half of A's current.
2. Step 2: $P = I^2R$ with the same R: B and C each get $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ of A's power.
3. Step 3: So A is brightest, and B and C are equally bright.
Answer: A is brightest; B and C are equal and dimmer (each a quarter of A's power)

Common mistakes

Using $P = I^2R$ to argue that a bigger resistor is always brighter. That's true only when the current is the same, as in series. In parallel the smaller resistor is brighter.
Assuming a bulb's wattage label is what it uses in any circuit. Use the label to find R, then find the actual power.
Thinking current gets used up in a bulb. The same current enters and leaves; energy is what the bulb transfers.

On the exam

Brightness rankings with justification are a staple. Earn the points by naming the shared quantity (same current or same ΔV) and the equation you used to compare.
If a question asks for energy over time in a circuit where the current changes, set up $\displaystyle\int P\,dt$ rather than multiplying one value of power by the time.

Connected topics

Videos

Electric Power
Flipping PhysicsWatch on YouTube (opens in a new tab)
Topic 11.3 - Ohm's Law and Electrical Power
Lessons With LondotWatch on YouTube (opens in a new tab)
Electric power | Circuits | Physics | Khan Academy
Khan Academy PhysicsWatch on YouTube (opens in a new tab)
DC Circuits - Power Dissipated in Resistors
Physics NinjaWatch on YouTube (opens in a new tab)
Power dissipation in resistors in series versus in parallel
Khan AcademyWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 11.4 Electric Power. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A space heater uses 1500 W when connected to a 120 V outlet. What is the heater’s resistance?

Question 2 of 4Calculator allowed

A current of 3.0 A passes through an 8.0 Ω resistor for 2.0 minutes. How much electrical energy does the resistor convert to other forms?

Question 3 of 4Calculator allowed

The current in a resistor R decreases with time as $I(t) = I_0e^{-t/\tau}$ . How much energy does the resistor dissipate from t = 0 until the current is essentially zero?

Question 4 of 4Calculator allowed

Three identical bulbs are connected in series with an ideal battery. A wire with negligible resistance is then connected across the middle bulb. What happens to the bulbs?

0 of 4 answered