Potential dividers · 分压器
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| potential divider/pəˈtenʃl dɪˈvaɪdə/ | 分压器 | fēn yā qì |
| load/ləʊd/ | 负载 | fù zài |
| thermistor/ˈθɜːmɪstə/ | 热敏电阻 | rè mǐn diàn zǔ |
| threshold/ˈθreʃəʊld/ | 阈值 | yù zhí |
| potentiometer/pəˌtentɪˈɒmɪtə/ | 电位差计 | diàn wèi chà jì |
| null method/nʌl ˈmeθəd/ | 零点法 | líng diǎn fǎ |
| galvanometer/ˌɡælvəˈnɒmɪtə/ | 检流计 | jiǎn liú jì |
| balance length/ˈbæləns leŋθ/ | 平衡长度 | píng héng cháng dù |
The volume knob
- A volume or dimmer knob can set the output anywhere from off to full.
- Inside is a potential divider 分压器 — resistors sharing out the voltage.
- It turns a fixed supply into any voltage you want.
音量旋钮
- 音量或调光旋钮可以把输出设在从关到最大的任何位置。
- 里面是一个 分压器(potential divider)——把电压分配出去的电阻。
- 它把固定的电源变成你想要的任何电压。
Splitting the voltage
- Two resistors in series share the supply in proportion to their resistance.
- Tapped across $R_2$: $V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$.
分配电压
- 两个串联的电阻 按电阻的比例 分享电源电压。
- 从 $R_2$ 两端取出:$V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$。
Sharing voltage in series · 在串联中分配电压
In a series loop the same current flows everywhere and the cell's voltage splits across the components — that split is how a potential divider works. · 在一个串联回路里,处处电流相同,电池的电压分配到各元件上——这种分配正是分压器的原理。
A $12\ \text{V}$ supply is across $R_1 = 2.0\ \text{k}\Omega$ and $R_2 = 4.0\ \text{k}\Omega$ in series. What is the output across $R_2$? · $12\ \text{V}$ 的电源加在串联的 $R_1 = 2.0\ \text{k}\Omega$ 和 $R_2 = 4.0\ \text{k}\Omega$ 上。$R_2$ 两端的输出是多少?
$V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1+R_2} = 12 \times \dfrac{4.0}{2.0+4.0} = 8.0\ \text{V}$. · $V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1+R_2} = 12 \times \dfrac{4.0}{2.0+4.0} = 8.0\ \text{V}$。
In a potential divider, the p.d. across each resistor is proportional to its resistance. · 在分压器中,每个电阻两端的电势差与它的电阻成正比。
The same current flows through both, so $V = IR$ means the bigger resistor takes the bigger share. · 相同的电流流过两者,所以 $V = IR$ 意味着更大的电阻分得更大的份额。
Worked example: a divider, then a load
A $12\ \text{V}$ supply is connected across $R_1 = 3.0\ \text{k}\Omega$ and $R_2 = 1.0\ \text{k}\Omega$ in series. Find the output across $R_2$. A $1.0\ \text{k}\Omega$ load 负载 is then connected across the output.
- Unloaded: $V_{\text{out}} = 12 \times \dfrac{1.0}{3.0 + 1.0} = 3.0\ \text{V}$.
- With the load: it sits in parallel with $R_2$, giving $\dfrac{1.0 \times 1.0}{1.0 + 1.0} = 0.50\ \text{k}\Omega$ in that position.
- Loaded output: $V_{\text{out}} = 12 \times \dfrac{0.50}{3.0 + 0.50} = 1.7\ \text{V}$.
- Check: connecting the load lowered the output — the divider only delivers its design voltage to a load whose resistance is much larger than $R_2$.
例题:分压器,然后接负载
$12\ \text{V}$ 电源接在串联的 $R_1 = 3.0\ \text{k}\Omega$ 和 $R_2 = 1.0\ \text{k}\Omega$ 两端。求 $R_2$ 两端的输出。然后在输出端接上一个 $1.0\ \text{k}\Omega$ 的负载(load)。
- 未接负载: $V_{\text{out}} = 12 \times \dfrac{1.0}{3.0 + 1.0} = 3.0\ \text{V}$。
- 接上负载: 它与 $R_2$ 并联,该位置的电阻变为 $\dfrac{1.0 \times 1.0}{1.0 + 1.0} = 0.50\ \text{k}\Omega$。
- 带负载的输出: $V_{\text{out}} = 12 \times \dfrac{0.50}{3.0 + 0.50} = 1.7\ \text{V}$。
- 检查: 接上负载 降低 了输出——只有当负载电阻远大于 $R_2$ 时,分压器才能输出设计的电压。
A $9.0\ \text{V}$ supply is across $R_1 = 6.0\ \text{k}\Omega$ and $R_2 = 3.0\ \text{k}\Omega$ in series, output across $R_2$. A $3.0\ \text{k}\Omega$ load is connected across the output. What is the output voltage now, in V? · $9.0\ \text{V}$ 电源接在串联的 $R_1 = 6.0\ \text{k}\Omega$ 和 $R_2 = 3.0\ \text{k}\Omega$ 两端,输出取自 $R_2$。在输出端接上一个 $3.0\ \text{k}\Omega$ 的负载。现在的输出电压是多少(单位 V)?
The load in parallel with $R_2$ gives $1.5\ \text{k}\Omega$, so $V_{\text{out}} = 9.0 \times \dfrac{1.5}{6.0 + 1.5} = 1.8\ \text{V}$ — down from $3.0\ \text{V}$ unloaded. · 负载与 $R_2$ 并联得 $1.5\ \text{k}\Omega$,所以 $V_{\text{out}} = 9.0 \times \dfrac{1.5}{6.0 + 1.5} = 1.8\ \text{V}$——比未接负载时的 $3.0\ \text{V}$ 低。
Sensor circuits
- Swap a resistor for a thermistor 热敏电阻 → the output changes with temperature.
- Swap it for an LDR → the output changes with light.
A thermistor in a divider turns a temperature change into a changing output voltage
传感器电路
- 把一个电阻换成 热敏电阻(thermistor) → 输出随温度变化。
- 换成 LDR → 输出随光照变化。
分压器中的热敏电阻把温度变化变成变化的输出电压
Replacing one resistor in a divider with an LDR makes the output voltage depend on: · 把分压器中的一个电阻换成光敏电阻,使输出电压取决于:
An LDR's resistance changes with light, so its share of the voltage — the output — tracks the brightness. · 光敏电阻的电阻随光变化,所以它分得的电压——输出——随亮度变化。
Which way does the output go?
- The output across a resistor is that resistor's share of the supply, and the share grows when its resistance grows relative to the other.
- Thermistor as $R_1$, output across fixed $R_2$: temperature ↑ → thermistor resistance ↓ → $R_2$'s share ↑ → $V_{\text{out}}$ rises.
- LDR as $R_1$, light decreases: LDR resistance ↑ → $R_2$'s share ↓ → $V_{\text{out}}$ falls. Swap the two components to reverse the behaviour.
输出朝哪个方向变?
- 某个电阻两端的输出是它在电源电压中的 份额,当 它的 电阻相对于另一个增大时,份额就增大。
- 热敏电阻作 $R_1$,输出取自定值电阻 $R_2$:温度 ↑ → 热敏电阻的电阻 ↓ → $R_2$ 的份额 ↑ → $V_{\text{out}}$ 上升。
- LDR 作 $R_1$,光照 减弱:LDR 电阻 ↑ → $R_2$ 的份额 ↓ → $V_{\text{out}}$ 下降。把两个元件互换位置就能反转这种行为。
In each divider the output is taken across the fixed · 固定的 resistor. Match the change to what the output does. · 每个分压器的输出都取自 定值 电阻。把变化与输出的表现配对。
When the sensor's resistance falls (hotter thermistor, brighter LDR), the fixed resistor takes a larger share of the supply and the output across it rises. · 当传感器的电阻减小时(热敏电阻更热、LDR 更亮),定值电阻分得电源电压的更大份额,它两端的输出上升。
Switching at a threshold 阈值
- Feed the changing output to a transistor or comparator.
- It can switch a load on or off when the temperature or light passes a set threshold (e.g. a street lamp at dusk).
在阈值处切换
- 把变化的输出送入晶体管或比较器。
- 当温度或光照越过设定的 阈值(threshold) 时,它可以把负载 接通或断开(例如黄昏时的路灯)。
A sensor divider can switch a load on when the measured quantity crosses a set ____. · 当被测量越过一个设定的 ____ 时,传感器分压器可以接通一个负载。
Feed the output to a comparator/transistor; it flips the load when the voltage passes the threshold (e.g. a lamp at dusk). · 把输出送到比较器/晶体管;当电压越过阈值时它翻转负载(例如黄昏时的灯)。
The potentiometer 电位差计 (null method 零点法)
- A uniform wire taps off $V_x = V_{\text{full}}\dfrac{x}{L_0}$ along its length — because for a uniform wire $R \propto L$, and with the same current everywhere $V = IR \propto L$.
- To compare e.m.f.s, slide until the galvanometer 检流计 reads zero — at balance no current flows through the test cell, so its internal resistance does not matter: $\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$.
Slide the contact until the galvanometer reads zero — at balance the tapped length is proportional to the e.m.f.
电位差计(零点法)
- 一根均匀的导线沿其长度取出 $V_x = V_{\text{full}}\dfrac{x}{L_0}$——因为均匀导线 $R \propto L$,各处电流相同,所以 $V = IR \propto L$。
- 要比较电动势,滑动触点直到检流计(galvanometer)读数为 零——平衡时没有电流通过待测电池,所以它的内阻无关紧要:$\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$。
滑动触点直到检流计读数为零——平衡时取出的长度与电动势成正比
At the balance point of a potentiometer, the current through the cell being measured is: · 在电位差计的平衡点,流过被测电池的电流是:
At balance the wire's p.d. exactly opposes the cell's e.m.f., so no current flows through it (a null). · 在平衡时,导线的电势差恰好与电池的电动势相抵,所以没有电流流过它(零值)。
At balance, the measured cell's internal resistance does not affect the result. · 在平衡时,被测电池的内阻不影响结果。
With zero current through the cell, there is no $Ir$ drop — so internal resistance has no effect. That is the strength of a null method. · 电池中没有电流,就没有 $Ir$ 压降——所以内阻没有影响。这就是零值法的优势。
Worked example: a balance length
A driver cell puts $2.0\ \text{V}$ across a uniform wire of length $100\ \text{cm}$. A test cell balances at a balance length 平衡长度 of $64\ \text{cm}$.
- e.m.f. of the test cell: $\varepsilon = 2.0 \times \dfrac{64}{100} = 1.28\ \text{V}$.
- If the test cell has internal resistance: no change — at balance no current flows through it, so there are no lost volts.
- If the driver cell has internal resistance: the p.d. across the wire is less than $2.0\ \text{V}$, so a longer length is needed for the same $1.28\ \text{V}$: the balance point moves further along the wire.
- Check: the balance length can never exceed the wire, so a test cell with an e.m.f. above the p.d. across the wire cannot be balanced at all.
例题:一个平衡长度
驱动电池在长 $100\ \text{cm}$ 的均匀导线两端加 $2.0\ \text{V}$。待测电池在 平衡长度(balance length) $64\ \text{cm}$ 处平衡。
- 待测电池的电动势: $\varepsilon = 2.0 \times \dfrac{64}{100} = 1.28\ \text{V}$。
- 如果待测电池有内阻: 没有变化——平衡时没有电流通过它,所以没有内阻压降。
- 如果驱动电池有内阻: 导线两端的电势差 小于 $2.0\ \text{V}$,所以同样的 $1.28\ \text{V}$ 需要更长的长度:平衡点沿导线 向远处 移动。
- 检查: 平衡长度不可能超过导线长度,所以电动势高于导线两端电势差的待测电池根本无法平衡。
A driver cell puts $1.5\ \text{V}$ across a $100\ \text{cm}$ uniform potentiometer wire. A test cell balances at $80\ \text{cm}$. What is its e.m.f., in V? · 驱动电池在 $100\ \text{cm}$ 长的均匀电位差计导线两端加 $1.5\ \text{V}$。待测电池在 $80\ \text{cm}$ 处平衡。它的电动势是多少(单位 V)?
The p.d. along the wire is proportional to length: $\varepsilon = 1.5 \times \dfrac{80}{100} = 1.2\ \text{V}$. · 导线上的电势差与长度成正比:$\varepsilon = 1.5 \times \dfrac{80}{100} = 1.2\ \text{V}$。
A divider's output is only $V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$ with nothing connected — a load in parallel with $R_2$ lowers it. In the potentiometer, "no current at balance" is what makes it a null method: the test cell's internal resistance drops out, but the driver cell's does not. And "the balance length is proportional to the e.m.f." holds only because the wire is uniform.
分压器的输出只有在 什么都不接 时才是 $V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$——与 $R_2$ 并联的负载会把它拉低。在电位差计中,"平衡时无电流"正是它成为 零点法 的原因:待测电池的内阻被消去,但 驱动 电池的内阻不会。另外,"平衡长度与电动势成正比"只因为导线是 均匀 的才成立。
You've got it
- a divider shares voltage by resistance: $V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$; a load across the output lowers it
- a thermistor or LDR makes the output respond to heat or light — the output across the fixed resistor rises when the sensor's resistance falls
- the null method: balance for zero current, so the test cell's internal resistance has no effect; $\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$
你掌握了
- 分压器按电阻分配电压:$V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$;接在输出端的负载会把它拉低
- 热敏电阻 或 LDR 让输出对热或光作出响应——传感器电阻减小时,定值电阻两端的输出上升
- 零点法:调到零电流时平衡,所以待测电池的内阻没有影响;$\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$