Alternating currents · 交流电
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| peak value/piːk ˈvæljuː/ | 峰值 | fēng zhí |
| root-mean-square/ruːt miːn skweə/ | 均方根 | jūn fāng gēn |
| alternating current/ˈɔːltəneɪtɪŋ ˈkʌrənt/ | 交流电 | jiāo liú diàn |
| sinusoidal/ˌsaɪnəˈsɔɪdl/ | 正弦 | zhèng xián |
| angular frequency/ˈæŋɡjʊlə ˈfriːkwənsi/ | 角频率 | jiǎo pín lǜ |
| radians/ˈreɪdɪənz/ | 弧度 | hú dù |
| cathode-ray oscilloscope/ˈkæθəʊd reɪ ɒˈsɪləskəʊp/ | 示波器 | shì bō qì |
| time-base/taɪm beɪs/ | 时基 | shí jī |
The 230 volts that is never 230 volts
- A mains socket is labelled $230\ \text{V}$, yet an oscilloscope shows the voltage swinging up to about $325\ \text{V}$ and down to $-325\ \text{V}$, fifty times a second.
- Neither number is a mistake. The mains voltage really does reach $325\ \text{V}$, and it really does behave like a steady $230\ \text{V}$ battery when it heats a kettle.
- Reconciling those two facts is what this lesson is about.
- The tools are the peak value 峰值, the root-mean-square 均方根 value, and the fact that average a.c. power is half the peak power.
那个从来不是 230 伏的 230 伏
- 市电插座上标着 $230\ \text{V}$,可示波器显示电压每秒五十次地上摆到约 $325\ \text{V}$、下摆到 $-325\ \text{V}$。
- 两个数字都没错。市电电压确实达到 $325\ \text{V}$,而在烧水时它确实表现得像一节稳定的 $230\ \text{V}$ 电池。
- 把这两件事调和起来,就是这一课要做的。
- 用到的工具是峰值(peak value)、均方根(root-mean-square)值,以及交流平均功率是峰值功率一半这个事实。
Period, frequency and peak value
- An alternating current 交流电 keeps reversing direction; the mains is sinusoidal 正弦, so $I$ and $V$ follow sine waves in time.
- The period $T$ is the time for one complete cycle. The frequency $f$ is the number of complete cycles per unit time, so $f = 1/T$.
- The peak value is the maximum value of the current or voltage in a cycle. Those three sentences are the marked wordings.
- The mean value over a whole cycle is zero, since the wave spends as long negative as positive. That is precisely why an r.m.s. value is needed at all.
One line that never moves, one that never stops
周期、频率与峰值
- 交流电(alternating current)不断改变方向;市电是正弦(sinusoidal)的,所以 $I$ 和 $V$ 随时间按正弦变化。
- 周期 $T$ 是完成一个完整循环所用的时间。频率 $f$ 是单位时间内完整循环的个数,所以 $f = 1/T$。
- 峰值是一个循环中电流或电压的最大值。这三句话就是评分认可的措辞。
- 整个循环上的平均值是零,因为波形为负和为正的时间一样长。这正是为什么根本需要一个均方根值。
一条从不动的线,一条从不停的线
Alternating current · 交流电
I = a sin(bt)
AC is a sine · 正弦 wave — amplitude is the peak, b sets the frequency. · 交流是一条 正弦 波——振幅是峰值,b 决定频率。
An alternating current: · 交流电:
a.c. keeps swapping direction — mains does so 50 (or 60) times a second. · 交流电不断改变方向——市电每秒这样做 50(或 60)次。
Match each term to the definition the examiner marks. · 把每个术语与评分认可的定义配对。
Each is a fixed wording. The r.m.s. one is the only definition that mentions another current, because that is what it compares itself to. · 每一条都是固定措辞。均方根那条是唯一提到另一种电流的定义,因为它正是拿那个来比较的。
Reading the equation
- A supply is written $V = V_0 \sin(\omega t)$. The number in front is the peak value; the number multiplying $t$ is the angular frequency 角频率 $\omega = 2\pi f$.
- So $f = \omega / 2\pi$ and $T = 1/f$. The angle $\omega t$ is in radians 弧度, so put the calculator in radians before evaluating anything.
- A supply written with $\cos$ is the same wave started at its peak rather than at zero.
读懂那个方程
- 电源写成 $V = V_0 \sin(\omega t)$。前面的数是峰值;乘 $t$ 的数是角频率(angular frequency)$\omega = 2\pi f$。
- 所以 $f = \omega / 2\pi$,$T = 1/f$。角 $\omega t$ 以弧度(radians)计,所以算任何东西之前先把计算器切到弧度。
- 用 $\cos$ 写的电源是同一列波,只是从峰值而不是从零开始。
In $V = V_0\sin(\omega t)$ the coefficient of $t$ is the ____ frequency, equal to $2\pi f$. · 在 $V = V_0\sin(\omega t)$ 中,$t$ 的系数是____频率,等于 $2\pi f$。
omega is 2pif, so a supply written sin(100pit) has f = 50 Hz, not 100pi Hz. The angle is in radians, so a calculator left in degrees turns every such step into nonsense. · omega 是 2pif,所以写成 sin(100pit) 的电源 f = 50 Hz,而不是 100pi Hz。角以弧度计,所以计算器停在角度模式会把这类步骤全算错。
Worked example: unpacking a supply equation
- A supply has $V = 320\sin(100\pi t)$, with $V$ in volts and $t$ in seconds. Find the peak value, frequency, period and r.m.s. voltage, and the first time after $t = 0$ at which $V = 160\ \text{V}$.
- Peak: $V_0 = 320\ \text{V}$. Angular frequency: $\omega = 100\pi\ \text{rad/s}$, so $f = 100\pi / 2\pi = 50\ \text{Hz}$ and $T = 1/50 = 0.020\ \text{s}$.
- $V_{\text{r.m.s.}} = 320/\sqrt{2} = 226\ \text{V}$.
- For $V = 160\ \text{V}$: $\sin(100\pi t) = 0.5$, so $100\pi t = \pi/6$ and $t = 1/600 = 1.7 \times 10^{-3}\ \text{s}$. In degrees mode this last step gives nonsense.
例题:拆开一个电源方程
- 某电源 $V = 320\sin(100\pi t)$,$V$ 以伏计、$t$ 以秒计。求峰值、频率、周期和均方根电压,以及 $t = 0$ 之后第一次 $V = 160\ \text{V}$ 的时刻。
- 峰值:$V_0 = 320\ \text{V}$。角频率:$\omega = 100\pi\ \text{rad/s}$,所以 $f = 100\pi / 2\pi = 50\ \text{Hz}$,$T = 1/50 = 0.020\ \text{s}$。
- $V_{\text{r.m.s.}} = 320/\sqrt{2} = 226\ \text{V}$。
- 对 $V = 160\ \text{V}$:$\sin(100\pi t) = 0.5$,所以 $100\pi t = \pi/6$,$t = 1/600 = 1.7 \times 10^{-3}\ \text{s}$。在角度模式下最后这一步会给出荒唐的结果。
Mains supply is $50\ \text{Hz}$. What is its period? · 市电供电是 $50\ \text{Hz}$。它的周期是多少?
$T = \dfrac{1}{f} = \dfrac{1}{50} = 0.020\ \text{s}$. · $T = \dfrac{1}{f} = \dfrac{1}{50} = 0.020\ \text{s}$。
Reading a CRO trace
- A cathode-ray oscilloscope 示波器 is read exactly as in Topic 7.
- Horizontal: count the divisions for one full cycle and multiply by the time-base 时基 setting. That gives $T$, and $f = 1/T$.
- Vertical: count the divisions from the centre line to a peak and multiply by the $y$-gain. That gives $V_0$. Measuring peak to peak and halving is usually more accurate, because the centre line is easy to misjudge.
读示波器图
- 示波器(cathode-ray oscilloscope)的读法与主题 7 完全一样。
- 水平方向:数出一个完整循环占多少格,乘以时基(time-base)设置。这给出 $T$,而 $f = 1/T$。
- 竖直方向:数出从中线到峰的格数,乘以 $y$ 增益。这给出 $V_0$。量峰到峰再取一半通常更准,因为中线容易看偏。
On a CRO trace one complete cycle spans 4.0 divisions with the time-base at 5.0 ms per division. What is the frequency, in Hz? · 在示波器图上一个完整循环占 4.0 格,时基为每格 5.0 ms。频率是多少 Hz?
T = 4.0 x 5.0 ms = 20 ms, so f = 1/0.020 = 50 Hz. Count divisions for a WHOLE cycle, not a half. · T = 4.0 x 5.0 ms = 20 ms,所以 f = 1/0.020 = 50 Hz。要数一个完整循环的格数,不是半个。
Power in a resistor
- For a resistive load the instantaneous power is $P = I^2 R$, and with $I = I_0\sin(\omega t)$:
- This is always positive, since the current is squared, so a reversing current still heats the resistor. It pulses between zero and $I_0^2 R$ at twice the frequency of the current.
- The mean of $\sin^2$ over a cycle is $\tfrac{1}{2}$, so the average power is half the peak power.
Twice the frequency, and never negative
电阻中的功率
- 对电阻性负载,瞬时功率是 $P = I^2 R$,代入 $I = I_0\sin(\omega t)$:
- 由于电流被平方,这始终为正,所以来回反向的电流照样加热电阻。它在零与 $I_0^2 R$ 之间脉动,频率是电流的两倍。
- $\sin^2$ 在一个循环上的平均是 $\tfrac{1}{2}$,所以平均功率是峰值功率的一半。
两倍的频率,而且从不为负
The average power of a.c. in a resistor is ____ the peak power. · 交流在电阻中的平均功率是峰值功率的 ____。
The mean of $\sin^{2}$ over a cycle is $\tfrac{1}{2}$, so $\langle P\rangle = \tfrac{1}{2}P_{\text{peak}}$. · $\sin^{2}$ 在一个周期内的平均值是 $\tfrac{1}{2}$,所以 $\langle P\rangle = \tfrac{1}{2}P_{\text{peak}}$。
Worked example: showing it is half
- A supply of peak voltage $12\ \text{V}$ drives a $680\ \Omega$ resistor. Show by calculation that the mean power is half the peak power.
- Peak power, at the instant the voltage is at its peak: $P_0 = V_0^2/R = 12^2/680 = 0.212\ \text{W}$.
- Mean power, through the r.m.s. value: $V_{\text{r.m.s.}} = 12/\sqrt{2} = 8.49\ \text{V}$, so $\langle P\rangle = V_{\text{r.m.s.}}^2/R = 8.49^2/680 = 0.106\ \text{W}$.
- The two are shown separately, and the second is half the first. Writing "half of $0.212$" scores nothing, because that is the very thing being shown.
- The factor $\tfrac{1}{2}$ is just $(1/\sqrt{2})^2$: the r.m.s. factor, squared.
例题:把"一半"算出来
- 峰值 $12\ \text{V}$ 的电源驱动一个 $680\ \Omega$ 的电阻。用计算证明平均功率是峰值功率的一半。
- 峰值功率,取电压处于峰值的那一瞬:$P_0 = V_0^2/R = 12^2/680 = 0.212\ \text{W}$。
- 平均功率,经由均方根值:$V_{\text{r.m.s.}} = 12/\sqrt{2} = 8.49\ \text{V}$,所以 $\langle P\rangle = V_{\text{r.m.s.}}^2/R = 8.49^2/680 = 0.106\ \text{W}$。
- 两者是分别算出来的,而第二个是第一个的一半。写"$0.212$ 的一半"得不到分,因为那正是要证明的东西。
- 那个 $\tfrac{1}{2}$ 不过是 $(1/\sqrt{2})^2$:均方根因子的平方。
What r.m.s. actually means
- The r.m.s. value of an alternating current is the value of the direct (steady) current that would dissipate the same mean power in the same resistor. That sentence is the two-mark definition, and "by reference to the heating effect" is asking for exactly it.
- From $\langle P\rangle = I_{\text{r.m.s.}}^2 R = \tfrac{1}{2}I_0^2 R$ it follows that, for a sine wave:
- A definition that says only "$I_0/\sqrt{2}$" scores nothing, because that formula holds for a sine wave alone.
- Quoted a.c. values are r.m.s. values, which settles the opening puzzle: $230\ \text{V}$ mains has $V_0 = 230\sqrt{2} \approx 325\ \text{V}$, and components must be rated for the peak.
One dashed line that does the same heating
均方根到底是什么意思
- 交流电的均方根值,是在同一电阻上消耗同样平均功率的直流(稳恒)电流的值。这句话就是两分的定义,而"从热效应的角度"问的正是它。
- 由 $\langle P\rangle = I_{\text{r.m.s.}}^2 R = \tfrac{1}{2}I_0^2 R$ 得出,对正弦波:
- 只写"$I_0/\sqrt{2}$"的定义得零分,因为那个公式只对正弦波成立。
- 标称的交流值都是均方根值,这就解开了开头的谜题:$230\ \text{V}$ 市电的 $V_0 = 230\sqrt{2} \approx 325\ \text{V}$,元件必须按峰值来选耐压。
一条做同样多加热的虚线
An a.c. has a peak current of $2.0\ \text{A}$. What is the r.m.s. current? · 一个交流电的峰值电流是 $2.0\ \text{A}$。均方根电流是多少?
$I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}} = \dfrac{2.0}{\sqrt{2}} \approx 1.41\ \text{A}$. · $I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}} = \dfrac{2.0}{\sqrt{2}} \approx 1.41\ \text{A}$。
A 230 V (r.m.s.) mains supply has a peak voltage of about 325 V. · 一个 230 V(均方根)的市电供电的峰值电压约为 325 V。
$V_0 = V_{\text{rms}}\sqrt{2} = 230 \times \sqrt{2} \approx 325\ \text{V}$ — components must be rated for the peak. · $V_0 = V_{\text{rms}}\sqrt{2} = 230 \times \sqrt{2} \approx 325\ \text{V}$——元件必须按峰值来选额定值。
Which statements belong in the definition of an r.m.s. current? Select all · 所有 that apply. · 哪些说法属于均方根电流的定义?选出所有适用的。
The last one is a consequence for a sine wave, not the definition. A definition giving only I0/sqrt(2) scores nothing, because it is false for any other waveform. · 最后一条是正弦波的推论,不是定义。只给出 I0/sqrt(2) 的定义得零分,因为对任何其他波形它都不成立。
Worked example: a wave that is not a sine
- A current is $+2.0\ \text{A}$ for the first half of each cycle and $-1.0\ \text{A}$ for the second half. Find its r.m.s. value and the mean power in a $10\ \Omega$ resistor.
- Take the name literally. Square: $4.0\ \text{A}^2$ for half the time, $1.0\ \text{A}^2$ for the other half. Mean: $(4.0 + 1.0)/2 = 2.5\ \text{A}^2$. Root: $I_{\text{r.m.s.}} = \sqrt{2.5} = 1.6\ \text{A}$.
- Mean power $= I_{\text{r.m.s.}}^2 R = 2.5 \times 10 = 25\ \text{W}$.
- The sign of the current never matters, because it is squared away. And $I_0/\sqrt{2}$ would have given $1.4\ \text{A}$, which is wrong: that shortcut is for sine waves only.
例题:不是正弦的波
- 某电流在每个循环的前半是 $+2.0\ \text{A}$、后半是 $-1.0\ \text{A}$。求它的均方根值,以及在 $10\ \Omega$ 电阻上的平均功率。
- 照字面来做。平方:一半时间 $4.0\ \text{A}^2$,另一半 $1.0\ \text{A}^2$。取平均:$(4.0 + 1.0)/2 = 2.5\ \text{A}^2$。开方:$I_{\text{r.m.s.}} = \sqrt{2.5} = 1.6\ \text{A}$。
- 平均功率 $= I_{\text{r.m.s.}}^2 R = 2.5 \times 10 = 25\ \text{W}$。
- 电流的正负从来不要紧,因为它被平方掉了。而 $I_0/\sqrt{2}$ 会给出 $1.4\ \text{A}$,那是错的:那个捷径只对正弦波成立。
Put the steps for finding the r.m.s. value of a non-sinusoidal current in order. · 把求非正弦电流均方根值的步骤按顺序排列。
Root-mean-square, read backwards. For a square wave of +2.0 A and -1.0 A the mean square is 2.5 A^2 and the r.m.s. is 1.6 A, not the 1.4 A that I0/sqrt(2) would give. · 把"均方根"倒过来读就是步骤。对 +2.0 A 与 -1.0 A 的方波,均方是 2.5 A^2,均方根是 1.6 A,而不是 I0/sqrt(2) 给出的 1.4 A。
Marks that slip away
- The r.m.s. definition is about the same power in the same resistor, not about dividing by $\sqrt{2}$. Quote the sentence, not the formula.
- "Show that the mean power is half the peak" needs two separate calculations. Halving the first one is circular.
- Put the calculator in radians before evaluating $\sin(\omega t)$.
- $\omega$ is not the frequency. Divide by $2\pi$ first.
- The mean current over a cycle is zero; the mean power is not, because power depends on $I^2$.
容易丢掉的分
- 均方根的定义讲的是同一电阻上同样的功率,不是除以 $\sqrt{2}$。要背那句话,不是那个公式。
- "证明平均功率是峰值的一半"需要两个分开的计算。把第一个折半是循环论证。
- 算 $\sin(\omega t)$ 之前先把计算器切到弧度。
- $\omega$ 不是频率。先除以 $2\pi$。
- 一个循环上的平均电流是零;平均功率不是,因为功率取决于 $I^2$。
You've got it
- period is the time for one complete cycle, frequency the number of complete cycles per unit time, peak value the maximum in a cycle
- in $V = V_0\sin(\omega t)$ the coefficient of $t$ is $\omega = 2\pi f$, and the angle is in radians
- the power in a resistor pulses at twice the frequency and is never negative, so the mean power is half the peak power
- the r.m.s. value is the steady direct current dissipating the same mean power in the same resistor; only for a sine wave is it $I_0/\sqrt{2}$
你掌握了
- 周期是一个完整循环的时间,频率是单位时间内完整循环的个数,峰值是一个循环中的最大值
- 在 $V = V_0\sin(\omega t)$ 中 $t$ 的系数是 $\omega = 2\pi f$,而角以弧度计
- 电阻中的功率以两倍频率脉动且从不为负,所以平均功率是峰值功率的一半
- 均方根值是在同一电阻上消耗同样平均功率的稳恒直流;只有对正弦波它才是 $I_0/\sqrt{2}$