Resistance, electrical energy and power
Введено| English | Значение |
|---|---|
| resistance/rɪˈzɪstəns/ | сопротивление |
| ohms/əʊmz/ | омы |
| filament lamp/ˈfɪləmənt læmp/ | лампа накаливания |
| diode/ˈdaɪəʊd/ | диод |
| fixed resistor/fɪkst rɪˈzɪstə/ | постоянный резистор |
| kilowatt-hour/ˈkɪləwɒt ˈaʊə/ | киловатт-час |
Slowing the flow
- Push current through a thin wire and it gets hot; through a thick one, less so.
- Resistance 电阻 measures how hard it is for current to flow.
- It controls the current, heats your toaster, and sets your electricity bill.
Resistance
- Resistance $R$ is measured in ohms 欧姆 ($\Omega$). A bigger resistance gives a smaller current:
- To measure it: an ammeter in series reads $I$, a voltmeter across the component reads $V$, then divide.
A pylon carries the cables that bring electricity at very high voltage
Сопротивление
V = R·I
Ohm's law: voltage is proportional to current — the gradient is the resistance R.
A component has $12\ \text{V}$ across it and a current of $3.0\ \text{A}$ through it. What is its resistance, in Ω?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
$R = \dfrac{V}{I} = \dfrac{12}{3.0} = 4.0\ \Omega$.
What changes a wire's resistance
- A longer wire has more resistance — resistance is proportional to length ($R \propto l$).
- A thicker wire has less resistance — resistance is inversely proportional to the cross-sectional area ($R \propto \tfrac{1}{A}$).
- So doubling the length doubles $R$; doubling the area halves it.
I-V graphs for a resistor, a filament lamp 灯丝灯泡 and a diode 二极管
I–V characteristics
A resistor gives a straight line; a filament lamp and a diode do not. Pick a component and read I and R off the curve.
A wire is made twice as long (same thickness). Its resistance:
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Resistance is proportional to length, so doubling the length doubles the resistance.
Match each electrical relationship to its formula.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Each item links the term to its correct meaning.
Current–voltage graphs
- A fixed resistor 定值电阻: a straight line through the origin — resistance is constant, so $I \propto V$.
- A filament lamp: an S-shape that gets flatter — the wire heats up, so its resistance rises.
- A diode: current one way only — very high resistance the "wrong" way.
The I–V graph of a fixed resistor (at constant temperature) is:
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A fixed resistor has constant resistance, so $I \propto V$ — a straight line through the origin.
Electrical power and energy
- Power is the energy transferred each second, in watts (W):
- The energy transferred in a time $t$, in joules (J), is:
A device runs at $12\ \text{V}$ and draws $2.0\ \text{A}$. What is its power, in W?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
$P = IV = 2.0 \times 12 = 24\ \text{W}$.
The kilowatt-hour 千瓦时
- Electricity bills use a bigger unit, the kilowatt-hour (kWh): the energy a $1\ \text{kW}$ appliance uses in $1$ hour.
- Example: a $2\ \text{kW}$ heater for $3$ hours uses $6\ \text{kWh}$.
A $2.0\ \text{kW}$ heater is on for $3.0$ hours. How much energy does it use, in kWh?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
energy = power × time = $2.0 \times 3.0 = 6.0\ \text{kWh}$.
If one kWh costs $20$ cents, what does $6.0\ \text{kWh}$ cost, in cents?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
cost = energy × price = $6.0 \times 20 = 120$ cents.
Worked examples
1. A kettle on a $230\ \text{V}$ supply draws $9.0\ \text{A}$. $P = IV = 9.0 \times 230 = 2070\ \text{W} \approx 2.1\ \text{kW}$, and its resistance is $R = \dfrac{V}{I} = \dfrac{230}{9.0} = 26\ \Omega$.
2. A $2.0\ \text{kW}$ heater runs for $15$ minutes, which is $0.25\ \text{h}$. Energy $= 2.0 \times 0.25 = 0.50\ \text{kWh}$; at ¥0.60 per kWh the cost is $0.50 \times 0.60 =$ ¥0.30.
A component has 18 V across it and carries 2.0 A. What are its resistance and power? How many kilowatt-hours does a 1.5 kW heater use in 2.0 hours?
18 divided by 2 is 9 ohms. Current times voltage is 36 watts. 1.5 kilowatts times 2.0 hours is 3.0 kilowatt-hours.
What is the resistance?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
18 divided by 2.0 is 9 ohms.
You've got it
- $R = \dfrac{V}{I}$ (ohms); longer wire → more $R$, thicker wire → less $R$
- I–V graphs: resistor = straight line; lamp = S-curve (heats up); diode = one way
- power $P = IV$ (watts); energy $E = IVt$ (joules)
- kWh = power (kW) × time (h); cost = kWh × price
Write from a resistance of 9 ohms
- A component has 18 V across it and a current of 2.0 A.
- Find the resistance and the power. A 1.5 kW heater runs for 2.0 hours.