Energy stored in a capacitor · 电容器储存的能量
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| energy stored/ˈenədʒi stɔːd/ | 储存的能量 | chǔ cún de néng liàng |
| defibrillator/dɪˈfɪbrɪleɪtə/ | 除颤器 | chú chàn qì |
Half the energy never reaches the capacitor
- Connect an ideal battery of e.m.f. $V$ to a capacitor through a wire. The battery pushes charge $Q = CV$ through a p.d. of $V$, so it gives out $QV$ of energy.
- The capacitor ends up holding $\tfrac12 QV$. Exactly half the energy has gone, as heat in the wire, and it makes no difference how good the wire is.
- The missing half is not an inefficiency to be engineered away. It is a consequence of the fact that the p.d. across the capacitor starts at zero and only reaches $V$ at the end.
- This lesson is the energy stored 储存的能量, where the factor of a half comes from, and what happens when charge is shared.
有一半能量根本没到电容器
- 把电动势为 $V$ 的理想电池经一根导线接到电容器上。电池把电荷 $Q = CV$ 推过 $V$ 的电压,所以它输出了 $QV$ 的能量。
- 电容器最终只存住 $\tfrac12 QV$。恰好一半的能量以导线中的热的形式没了,而且导线多好都一样。
- 少掉的那一半不是可以靠工程手段消除的低效。它是这样一个事实的结果:电容器上的电压从零开始,直到最后才达到 $V$。
- 这一课讲储存的能量(energy stored)、那个二分之一从哪里来,以及电荷被分享时会发生什么。
The three forms
- Charging a capacitor takes work, because each extra bit of charge must be pushed against the p.d. that is already there. Adding up all those bits gives:
- All three are the same expression with $Q = CV$ substituted differently. Choose the one that uses the two quantities the question gives you, and no rearranging is needed.
三种形式
- 给电容器充电需要做功,因为每一点新增的电荷都必须顶着已经存在的电压被推进去。把所有这些点加起来,得到:
- 三者是同一个表达式用 $Q = CV$ 作不同代换的结果。选用到题目所给两个量的那一个,就不需要任何移项。
The energy stored in a capacitor is: · 电容器储存的能量是:
Equivalent forms: $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$. · 等价的形式:$W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$。
A $2.0\ \mu\text{F}$ capacitor is charged to $1000\ \text{V}$. How much energy does it store? · 一个 $2.0\ \mu\text{F}$ 的电容器被充电到 $1000\ \text{V}$。它储存多少能量?
$W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 2.0 \times 10^{-6} \times (1000)^{2} = 1.0\ \text{J}$. · $W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 2.0 \times 10^{-6} \times (1000)^{2} = 1.0\ \text{J}$。
The area under the graph
- A graph of $V$ against $Q$ for a capacitor is a straight line through the origin, of gradient $1/C$.
- The energy stored is the area under that line up to the charge $Q$, which is a triangle of area $\tfrac12 QV$.
- That is where the half comes from: the p.d. grows from zero to $V$ as the capacitor charges, so the average p.d. during charging is $V/2$, not $V$.
A triangle, not a rectangle, and that is the whole reason for the half
图线下的面积
- 电容器的 $V$ 对 $Q$ 图是一条过原点的直线,斜率为 $1/C$。
- 所储存的能量是直到电荷 $Q$ 为止该线下的面积,那是一个面积为 $\tfrac12 QV$ 的三角形。
- 那个二分之一就从这里来:电压在充电过程中从零增长到 $V$,所以充电期间的平均电压是 $V/2$,不是 $V$。
是三角形不是矩形,而这就是那个二分之一的全部理由
Energy in a capacitor · 电容器中的能量
E = ½C·V²
Stored energy grows with the square of the voltage. · 储存的能量随电压的 平方 增长。
Why is there a factor of ½ in the stored-energy formula? · 为什么储存能量的公式里有一个 ½ 因子?
The p.d. rises from 0 to V as it charges, so the average is V/2 — hence the triangle area $\tfrac{1}{2}QV$. · 充电时电势差从 0 升到 V,所以平均是 V/2——因此三角形面积是 $\tfrac{1}{2}QV$。
The energy stored equals the ____ under the V–Q graph. · 储存的能量等于 V–Q 图线下的 ____。
The triangle under the line has area $\tfrac{1}{2}QV$ — the stored energy. · 线下的三角形面积是 $\tfrac{1}{2}QV$——即储存的能量。
Worked example: energy from a graph
- A graph of $Q$ against $V$ is a straight line through the origin passing through $(10\ \text{V},\ 1.2 \times 10^{-3}\ \text{C})$. Find the capacitance, the energy at $10\ \text{V}$, and the extra energy needed to reach $12\ \text{V}$.
- The gradient of $Q$ against $V$ is the capacitance: $C = 1.2 \times 10^{-3}/10 = 1.2 \times 10^{-4}\ \text{F} = 120\ \mu\text{F}$.
- Energy at $10\ \text{V}$: the area, $\tfrac12 QV = \tfrac12 (1.2 \times 10^{-3})(10) = 6.0 \times 10^{-3}\ \text{J}$.
- At $12\ \text{V}$: $\tfrac12 CV^2 = \tfrac12 (1.2 \times 10^{-4})(144) = 8.6 \times 10^{-3}\ \text{J}$, so the extra energy is $2.6 \times 10^{-3}\ \text{J}$.
- Note that a 20% rise in voltage costs a 44% rise in energy, because $W \propto V^2$.
例题:从图求能量
- $Q$ 对 $V$ 的图是一条过原点、经过 $(10\ \text{V},\ 1.2 \times 10^{-3}\ \text{C})$ 的直线。求电容、$10\ \text{V}$ 时的能量,以及升到 $12\ \text{V}$ 所需的额外能量。
- $Q$ 对 $V$ 的斜率就是电容:$C = 1.2 \times 10^{-3}/10 = 1.2 \times 10^{-4}\ \text{F} = 120\ \mu\text{F}$。
- $10\ \text{V}$ 时的能量:面积 $\tfrac12 QV = \tfrac12 (1.2 \times 10^{-3})(10) = 6.0 \times 10^{-3}\ \text{J}$。
- $12\ \text{V}$ 时:$\tfrac12 CV^2 = \tfrac12 (1.2 \times 10^{-4})(144) = 8.6 \times 10^{-3}\ \text{J}$,所以额外能量是 $2.6 \times 10^{-3}\ \text{J}$。
- 注意电压上升 20% 要付出能量上升 44% 的代价,因为 $W \propto V^2$。
A graph of charge against p.d. for a capacitor is a straight line through the origin. What does its gradient give? · 电容器的电荷对电压图是一条过原点的直线。它的斜率给出什么?
Q against V has gradient C, and the energy is the area under the V against Q line. Do not confuse the gradient with the area. · Q 对 V 的斜率是 C,而能量是 V 对 Q 线下的面积。不要把斜率和面积混为一谈。
Energy goes as the square
- Because $W = \tfrac12 CV^2$, doubling the p.d. quadruples the energy for the same capacitor.
- That is why a camera flash or a defibrillator 除颤器 charges its capacitor to hundreds or thousands of volts rather than using a larger capacitor at a low voltage: voltage is the cheaper way to buy energy.
- It is also why the same capacitor at half the voltage stores only a quarter of the energy, which is worth checking your answers against.
能量按平方增长
- 因为 $W = \tfrac12 CV^2$,对同一个电容器,电压加倍能量变四倍。
- 这就是相机闪光灯或除颤器(defibrillator)把电容器充到几百上千伏、而不是在低电压下用更大电容器的原因:用电压来买能量更便宜。
- 这也是同一个电容器在一半电压下只存四分之一能量的原因,值得拿来检验你的答案。
Charging a capacitor through a wire wastes about half the supplied energy as heat. · 通过一根导线给电容器充电,会以热的形式浪费约一半提供的能量。
The battery gives out $QV$ but the capacitor stores only $\tfrac{1}{2}QV$; the rest heats the wire, whatever its resistance. · 电池放出 $QV$,但电容器只储存 $\tfrac{1}{2}QV$;其余的把导线加热,无论它的电阻是多少。
A 120 uF capacitor is charged to 12 V. What is the energy stored, in mJ? · 一个 120 uF 的电容器被充到 12 V。储存的能量是多少 mJ?
W = C V^2 / 2 = 1.2e-4 x 144 / 2 = 8.6e-3 J. At 10 V it would store only 6.0 mJ, because energy goes as the square of the voltage. · W = C V^2 / 2 = 1.2e-4 x 144 / 2 = 8.6e-3 J。在 10 V 时只存 6.0 mJ,因为能量随电压的平方变化。
Doubling the p.d. across a capacitor doubles the energy it stores. · 把电容器两端的电压加倍,它储存的能量也加倍。
W is proportional to V squared, so doubling the p.d. gives four times the energy. That is why flash and defibrillator circuits use high voltages. · W 与 V 的平方成正比,所以电压加倍给出四倍能量。这就是闪光灯和除颤器电路使用高电压的原因。
Worked example: sharing charge
- A charged capacitor $C$ at p.d. $V$ is disconnected and then connected across an uncharged capacitor of capacitance $3C$. Find the final p.d. and the energy lost.
- Charge is conserved. The total charge is still $Q = CV$, now on a parallel pair of total capacitance $4C$, so the common p.d. is $V' = Q/4C = V/4$.
- Energy is not conserved. Before: $\tfrac12 CV^2$. After: $\tfrac12 (4C)(V/4)^2 = \tfrac18 CV^2$. Three-quarters of the energy has gone.
- It is dissipated as heat in the connecting wires, and as a spark or radiation, while the charge moves, however small the resistance. Saying where it went is the mark.
例题:分享电荷
- 一个充到电压 $V$、电容为 $C$ 的电容器被断开,然后接到一个未充电、电容为 $3C$ 的电容器两端。求最终电压和损失的能量。
- **电荷守恒。**总电荷仍是 $Q = CV$,现在处在总电容为 $4C$ 的并联对上,所以共同电压是 $V' = Q/4C = V/4$。
- **能量不守恒。**之前:$\tfrac12 CV^2$。之后:$\tfrac12 (4C)(V/4)^2 = \tfrac18 CV^2$。四分之三的能量没了。
- 它在电荷移动时以连接导线中的热的形式耗散,还有火花或辐射,而且无论电阻多小都如此。说出它去了哪里才是得分点。
A charged capacitor is connected across an uncharged one. Put the reasoning in order. · 一个充了电的电容器接到一个未充电的电容器两端。把推理按顺序排列。
Charge is conserved, energy is not. Naming where the lost energy goes is what the final mark is for. · 电荷守恒,能量不守恒。说出损失的能量去了哪里,正是最后一分的所在。
Energy during a discharge falls twice as fast
- During a discharge the charge and the p.d. both decay as $e^{-t/RC}$, but the energy goes as $V^2$, so it decays as $e^{-2t/RC}$.
- After one time constant the p.d. has fallen to $37\%$ of its start, but the energy has fallen to $e^{-2} = 14\%$.
- After two time constants the p.d. is $14\%$ and the energy only $1.8\%$.
- So the energy has an effective half-life of half that of the charge. A question that gives you a time constant and asks for the remaining energy is testing exactly this factor of two.
- The same square is why a capacitor charged to twice the p.d. stores four times the energy.
放电时能量下降快一倍
- 放电过程中电荷和电压都按 $e^{-t/RC}$ 衰减,但能量按 $V^2$ 变化,所以它按 $e^{-2t/RC}$ 衰减。
- 经过一个时间常数,电压降到起始值的 $37\%$,而能量已降到 $e^{-2} = 14\%$。
- 经过两个时间常数,电压是 $14\%$,而能量只剩 $1.8\%$。
- 所以能量的有效"半衰期"是电荷的一半。给出时间常数并问剩余能量的题,考的正是这个二倍因子。
- 同一个平方也解释了为什么电压加倍的电容器储存四倍的能量。
A capacitor discharges for exactly one time constant. What percentage of its stored energy remains? · 电容器放电恰好一个时间常数。储存的能量还剩百分之几?
The p.d. falls to 37%, but energy goes as V squared, so e^-2 = 13.5% remains. Quoting 37% for the energy is the trap this question is built on. · 电压降到 37%,但能量按 V 的平方变化,所以剩下 e^-2 = 13.5%。给能量报 37% 正是这道题设下的陷阱。
Marks that slip away
- The factor of $\tfrac12$ comes from the average p.d. being $V/2$ during charging, or equivalently from the area of a triangle. Say one of those, not "it is in the formula".
- $W \propto V^2$, so doubling the voltage gives four times the energy, not twice.
- In a charge-sharing question, charge is conserved but energy is not. Name where the lost energy goes.
- Half the battery's energy is lost as heat during charging whatever the wire's resistance. It is not a fault in the circuit.
容易丢掉的分
- 那个 $\tfrac12$ 来自充电期间平均电压是 $V/2$,或等价地来自三角形的面积。要说其中一条,而不是"公式里就有"。
- $W \propto V^2$,所以电压加倍给出四倍能量,不是两倍。
- 在电荷分享题里,电荷守恒而能量不守恒。要说出损失的能量去了哪里。
- 充电时无论导线电阻多大,电池能量的一半都会以热的形式损失。这不是电路的毛病。
You've got it
- energy stored $W = \tfrac12 QV = \tfrac12 CV^2 = \dfrac{Q^2}{2C}$; pick the form matching the quantities given
- it is the area under the $V$ against $Q$ line, a triangle, and the $\tfrac12$ is because the average p.d. while charging is $V/2$
- $W \propto V^2$: double the p.d., four times the energy
- when charge is shared, charge is conserved and energy is not: the difference is dissipated as heat in the wires, whatever their resistance
你掌握了
- 储存的能量 $W = \tfrac12 QV = \tfrac12 CV^2 = \dfrac{Q^2}{2C}$;选用与所给量匹配的那一种形式
- 它是 $V$ 对 $Q$ 线下的面积,一个三角形,而那个 $\tfrac12$ 是因为充电期间的平均电压是 $V/2$
- $W \propto V^2$:电压加倍,能量四倍
- 电荷被分享时,电荷守恒而能量不守恒:差额以导线中的热耗散,无论电阻多小