Transformers and power transmission · 变压器与电力传输
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| transformer/trænsˈfɔːmə/ | 变压器 | biàn yā qì |
| primary/ˈpraɪməri/ | 初级 | chū jí |
| secondary/ˈsekəndəri/ | 次级 | cì jí |
| step-up/step ʌp/ | 升压 | shēng yā |
| step-down/step daʊn/ | 降压 | jiàng yā |
| transmission/trænˈsmɪʃn/ | 输电 | shū diàn |
Sending power across the country
- Power stations are far from cities, yet very little energy is lost on the way.
- The secret is the transformer 变压器, which changes the size of an a.c. voltage.
- It lets us send power at very high voltage — and that is what keeps the loss small.
把电力送过全国
- 发电站远离城市,然而在路上几乎没有能量损失。
- 秘密是变压器(transformer),它改变一个交流电压的大小。
- 它让我们以非常高的电压送电力——而那就是使损失小的原因。
The transformer
- A transformer has two coils on a soft-iron core: a primary 初级 (input) and a secondary 次级 (output).
- The a.c. in the primary makes a changing field in the core, which induces an a.c. e.m.f. in the secondary.
- The voltages and turns are linked by:
- It only works on a.c. — a steady d.c. makes no changing field.
变压器
- 一个变压器在一个软铁芯上有两个线圈:一个初级(primary,输入)和一个次级(secondary,输出)。
- 初级中的交流在芯中产生一个变化的场,它在次级中感应一个交流电动势。
- 电压和匝数由以下联系:
- 它只对交流起作用——一个稳定的直流产生不出变化的场。
The transformer equation · 变压器方程
Vs/Vp = Ns/Np
The voltage ratio equals the turns ratio — more turns on the secondary steps the voltage up. · 电压比等于匝数比——次级上更多的匝数升高电压。
A transformer works with a steady direct current (d.c.). · 一个变压器在一个稳定的直流(d.c.)下工作。
A transformer needs a changing magnetic field, so it works on a.c. only. A steady d.c. makes no change and induces nothing. · 一个变压器需要一个变化的磁场,所以它只对交流起作用。一个稳定的直流不造成变化也感应不出任何东西。
A transformer has $1000$ turns on the primary and $50$ on the secondary. The primary voltage is $240\ \text{V}$. What is the secondary voltage, in V? · 一个变压器初级有 $1000$ 匝,次级有 $50$ 匝。初级电压是 $240\ \text{V}$。次级电压是多少,以 V 计?
$V_s = V_p \times \dfrac{N_s}{N_p} = 240 \times \dfrac{50}{1000} = 12\ \text{V}$ — a step-down transformer. · $V_s = V_p \times \dfrac{N_s}{N_p} = 240 \times \dfrac{50}{1000} = 12\ \text{V}$——一个降压变压器。
Step-up 升压 and step-down 降压
- More turns on the secondary → a step-up transformer (voltage rises).
- Fewer turns on the secondary → a step-down transformer (voltage falls).
- If 100% efficient, power in = power out:
- So a step-up transformer raises the voltage but lowers the current.
A real transformer changes the voltage of an a.c. supply
升压和降压
- 次级上更多匝数 → 一个升压(step-up)变压器(电压上升)。
- 次级上更少匝数 → 一个降压(step-down)变压器(电压下降)。
- 如果 100% 高效,输入功率 = 输出功率:
- 所以一个升压变压器提高电压但降低电流。
一个真实的变压器改变一个交流供电的电压
Transformer · 变压器
Add turns to the secondary to step the voltage up, fewer to step it down — Vs/Vp = Ns/Np, exactly how the grid moves power efficiently. · 给次级添加匝数以升高电压,更少以降低它——Vs/Vp = Ns/Np,正是电网高效移动电力的方式。
A step-up transformer has: · 一个升压变压器有:
More secondary turns gives a higher output voltage — a step-up transformer. · 更多次级匝数给出一个更高的输出电压——一个升压变压器。
A 100%-efficient step-up transformer raises the voltage and lowers the current. · 一个 100% 高效的升压变压器提高电压并降低电流。
Power is conserved ($I_p V_p = I_s V_s$), so if the voltage goes up the current must come down. · 功率被守恒($I_p V_p = I_s V_s$),所以如果电压上升电流必须下降。
Why high voltage for transmission 输电
- Power wasted as heat in the cables is $P = I^2 R$ ($R$ = cable resistance).
- A step-up transformer raises the voltage and so lowers the current for the same power ($P = IV$).
- A smaller current means much less wasted power in the cables.
- A step-down transformer then lowers the voltage to a safe value before it reaches homes.
A transformer
为什么传输用高电压
- 在电缆中作为热浪费的功率是 $P = I^2 R$($R$ = 电缆电阻)。
- 一个升压变压器提高电压,因而为相同的功率降低电流($P = IV$)。
- 一个更小的电流意味着电缆中少得多的浪费功率。
- 然后一个降压变压器在它到达家庭之前把电压降到一个安全值。
一个变压器
Why is electricity sent across the country at very high voltage? · 为什么电以非常高的电压送过全国?
For a given power, raising the voltage lowers the current. Since the heat loss is $I^2R$, a smaller current wastes far less power. · 对一个给定的功率,提高电压降低电流。由于热损失是 $I^2R$,一个更小的电流浪费少得多的功率。
A cable carries a current of $20\ \text{A}$ and has a resistance of $5.0\ \Omega$. How much power is wasted as heat, in W? · 一根电缆携带 $20\ \text{A}$ 的电流且有 $5.0\ \Omega$ 的电阻。多少功率作为热被浪费,以 W 计?
$P = I^2 R = 20^2 \times 5.0 = 400 \times 5.0 = 2000\ \text{W}$. Lowering the current would cut this sharply. · $P = I^2 R = 20^2 \times 5.0 = 400 \times 5.0 = 2000\ \text{W}$。降低电流会大幅削减这个。
You've got it
- transformer: $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$; works on a.c. only
- more secondary turns = step-up (V↑); fewer = step-down (V↓)
- 100% efficient: $I_p V_p = I_s V_s$ — step-up raises V but lowers I
- transmit at high voltage → low current → less $I^2R$ loss in the cables
你掌握了
- 变压器:$\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$;只对交流起作用
- 更多次级匝数 = 升压(V↑);更少 = 降压(V↓)
- 100% 高效:$I_p V_p = I_s V_s$——升压提高 V 但降低 I
- 以高电压传输 → 低电流 → 电缆中更少的 $I^2R$ 损失