Electric Power
Introduced| English | Meaning |
|---|---|
| power/ˈpaʊə/ | The rate of energy transfer or work done. |
| watt/wɒt/ | The SI unit of power, equal to one joule per second. |
| dissipation/ˌdɪsɪˈpeɪʃn/ | dissipation |
Where does a battery's energy go — and how fast?
- A battery hands energy to charges; the circuit spends it.
- Power 功率 is how fast that energy is delivered or used.
- A bright bulb spends energy quickly; a dim one spends it slowly.
- Power ties current and voltage to heat, light, and motion.
Power = current × voltage
- Each charge drops energy $qV$ as it crosses a component.
- Charges arrive at a rate $I$, so the energy per second is $P = IV$.
- Its unit is the watt 瓦特 (1 W $= 1$ J/s).
- Power is simply how many joules leave the source every second.
Power in a resistor
Set the voltage and resistance and watch the current and power that result.
Electric power delivered to a component is:
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$P = IV$ — current times voltage, in watts.
A device draws $4\ \text{A}$ at $12\ \text{V}$. Find its power (in W).
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$P = IV = 4 \times 12 = 48\ \text{W}$.
The SI unit of power is the ____.
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1 watt $= 1$ J/s.
Three forms of the same law
- Combine with Ohm's law to get $P = I^2 R$ and $P = \dfrac{V^2}{R}$.
- Use whichever fits what you know: $I$ and $R$, or $V$ and $R$.
- All three give the same power for an ohmic resistor.
- $P = I^2 R$ shows why a big current heats a wire so strongly.
Which is not a correct form of electric power for a resistor?
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The valid forms are $IV$, $I^2R$, $V^2/R$. $IR^2$ is wrong.
Useful power and wasted heat
- In a resistor, all the power becomes heat — that is dissipation 耗散.
- In a motor, most becomes useful motion, with some lost to heat.
- Heaters and kettles are just resistors chosen to dissipate a lot.
- Thin wires with high current can overheat — a fire risk.
Select all true statements about electric power.
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P = IV, resistor power heats, energy = P×t. Power (rate) ≠ energy (total).
Paying for energy: the kilowatt-hour
- Your electricity bill charges for energy, not power.
- Energy $=$ power $\times$ time; one kilowatt-hour is $1\ \text{kW}$ for $1$ hour.
- A $2\ \text{kW}$ heater for $3$ hours uses $6\ \text{kWh}$.
- Power tells the rate; multiply by time to get the bill.
Your electricity bill charges for energy (power × time), not for power itself.
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Energy = power × time, billed in kilowatt-hours.
A bulb runs at $0.5\ \text{A}$ from a $230\ \text{V}$ supply. Find its power.
- $P = IV = 0.5 \times 230 = 115\ \text{W}$.
- It converts $115$ joules of energy every second.
Keep power (rate, in watts) separate from energy (total, in joules or kWh). Your bill is for energy $=$ power $\times$ time, not for power. A high-power device used briefly can cost less than a low-power one left on for days.
Power is energy per second: $P = IV$ (watts), also $P = I^2R = V^2/R$ for a resistor. In a resistor it all becomes heat (dissipation). Energy $=$ power $\times$ time — billed as kilowatt-hours.
Write from 30 watts and 600 joules
- A resistor carries 2 amperes at 15 volts for 20 seconds.
- Find the power and the energy transferred.
A resistor carries 2 amperes from a 15 volt supply for 20 seconds. What are the power and the energy transferred, and what happens to that energy?
2 times 15 is 30 watts. 30 watts times 20 seconds is 600 joules, dissipated as heat.
What are the power and the energy?
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IV is 30 watts, and power times 20 seconds is 600 joules of heat.