Quantities, units and measurement
引入| English | 释义 | 拼音 |
|---|---|---|
| charge/tʃɑːdʒ/ | 电荷 | diàn hè |
| current/ˈkʌrənt/ | 电流 | diàn liú |
| voltage/ˈvəʊltɪdʒ/ | 电压 | diàn yā |
| energy/ˈenədʒi/ | 能量 | néng liàng |
| power/ˈpaʊə/ | 功率 | gōng lǜ |
| kilowatt-hour/ˈkɪləwɒt ˈaʊə/ | 千瓦时 | qiān wǎ shí |
| passive sign convention/ˈpæsɪv saɪn kənˈvenʃn/ | 无源符号约定 | wú yuán fú hào yuē dìng |
| ammeter/ˈæmiːtə/ | 安培表 | ān péi biǎo |
| voltmeter/ˈvəʊltmiːtə/ | 电压表 | diàn yā biǎo |
| multimeter/ˌmʌltɪˈmiːtə/ | 万用表 | wàn yòng biǎo |
20 W against 2000 W
- A phone charger is labelled 20 W and a kettle 2000 W. Both move electric charge, yet one is a hundred times more powerful.
- To say why, you need six quantities and one sign rule. They are the language of every circuit question in this course.
Six quantities form one chain
- Charge 电荷 moves; its rate is current 电流 in amperes ($1\,\text{A} = 1\,\dfrac{\text{C}}{\text{s}}$). Voltage 电压 is energy per unit charge between two points, in volts.
- Power 功率 is energy per unit time, in watts: $p = vi$. Energy 能量 is power × time, in joules. One kilowatt-hour 千瓦时 is 3.6 MJ.
A 12 V battery drives a 0.25 A lamp: $p = 12 \times 0.25 = 3\,\text{W}$. In 2 minutes the lamp takes $w = pt = 3 \times 120 = 360\,\text{J}$.
Match each quantity to its unit.
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They form one chain: $1\,\text{A} = 1\,\dfrac{\text{C}}{\text{s}}$, $1\,\text{V} = 1\,\dfrac{\text{J}}{\text{C}}$ and $1\,\text{W} = 1\,\dfrac{\text{J}}{\text{s}}$.
The 12 V, 0.25 A lamp stays on for 5 minutes. How much energy does it take, in joules?
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$p = 12 \times 0.25 = 3\,\text{W}$, and $w = pt = 3 \times 300 = 900\,\text{J}$. Convert the minutes to seconds first.
The sign convention decides supply and demand
- Draw each current arrow before calculating. With the passive sign convention 无源符号约定, if the arrow enters an element's $+$ terminal, $p = vi$ is the power it absorbs.
- A negative absorbed power means the element supplies. The battery feeding the lamp absorbs −3 W: it supplies exactly what the lamp takes.
An element has its arrow entering the $+$ terminal with $v = 5\,\text{V}$ and $i = -2\,\text{A}$: $p = -10\,\text{W}$, so it supplies 10 W. No new physics, only bookkeeping.
An element's current arrow enters its + terminal, and measurements give v = 5 V, i = −2 A. What is true?
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p = vi = 5 × (−2) = −10 W absorbed, so it supplies 10 W. A negative current simply flows against the chosen arrow.
Prefixes carry the arithmetic
- Convert to SI units before computing: $3000\,\text{mA}\cdot\text{h} = 3\,\text{A} \times 3600\,\text{s} = 10\,800\,\text{C}$. A 3.7 V, 3 A·h phone battery stores $3.7 \times 3 = 11.1\,\text{W}\cdot\text{h} \approx 4.0 \times 10^{4}\,\text{J}$.
- Check answers by size: a phone battery should store tens of kilojoules, not megajoules.
An 18 W charger refills 11.1 W·h, ideally, in $\dfrac{11.1}{18} \approx 0.62\,\text{h}$: about 37 minutes, a sanity check against the packaging claim.
A 3.7 V battery stores 3 A·h of charge. Which energy value is correct?
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3 A·h at 3.7 V gives 11.1 W·h; 11.1 × 3600 ≈ 4.0 × 10⁴ J. The megajoule option forgets the hours-to-seconds step.
Meters have personalities
- An ammeter 安培表 goes in series and must be nearly a short circuit; a voltmeter 电压表 goes across the element and must be nearly an open circuit. A multimeter 万用表 is both, plus ohms.
- A meter left on its amps range and placed across a 9 V supply is a near short: with 0.5 Ω inside the supply and 0.1 Ω in the meter, $\dfrac{9}{0.6} = 15\,\text{A}$ flows, enough to blow the meter's fuse.
Why must a voltmeter have a very large resistance?
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A voltmeter sits in parallel with the element; a large resistance keeps its current negligible so the measured voltage is undisturbed. An ammeter is the opposite — nearly a short, in series.
When the meter changes the reading
- A voltmeter is a resistor too. Across a large resistance it carries a real share of the current, and the voltage it reports drops.
- Rule of thumb: the meter's resistance should be at least a hundred times the resistance it is connected across.
Two 10 MΩ resistors in series across 12 V share it equally: 6 V each. A 10 MΩ voltmeter across the lower one makes that part $10 \parallel 10 = 5\,\text{M}\Omega$, so it reads $12 \times \dfrac{5}{10 + 5} = 4\,\text{V}$. The check itself changed the circuit.
Two 1 MΩ resistors in series sit across 10 V. A voltmeter with 1 MΩ resistance is connected across the lower resistor. What does it read, in volts? Give two decimal places.
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The lower part becomes $1 \parallel 1 = 0.5\,\text{M}\Omega$, so the meter reads $10 \times \dfrac{0.5}{1 + 0.5} \approx 3.33\,\text{V}$ instead of 5 V. A 100 MΩ meter would read about 4.98 V.
Three common slips
- "The voltage at A" with no second point. Voltage is always between two points; name them, or say that the second one is ground.
- Mixed units. Minutes with watts, or mA with kΩ. Convert first; then the unit of the answer comes out by itself ($\dfrac{\text{V}}{\Omega} = \text{A}$).
- Trusting a meter blindly. Ask what resistance it adds and where.
"Node A is at 5 V" has a meaning only once a reference point, such as ground, has been chosen.
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Voltage is always between two points. "5 V at a node" means 5 V between that node and the chosen reference node.
Recap
- Charge → current → voltage → power → energy: $p = vi$, $w = pt$.
- Passive sign convention: arrow into $+$ gives absorbed power; negative means supplied.
- Ammeter in series (near short), voltmeter across (near open), and both disturb the circuit a little.