Discharging a capacitor · 电容器放电
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| discharge/dɪsˈtʃɑːdʒ/ | 放电 | fàng diàn |
| time constant/taɪm ˈkɒnstənt/ | 时间常数 | shí jiān cháng shù |
| exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ | 指数衰减 | zhǐ shù shuāi jiǎn |
The decay that never quite finishes
- A discharging capacitor loses 63% of its charge in the first time constant, then 63% of what remains in the next, and so on for ever.
- After five time constants less than 1% is left, but the curve never reaches zero. It cannot, because the rate of loss is always proportional to how much is left.
- Every quantity that behaves that way, radioactive decay included, gives the same curve, so learning it once serves twice.
- This lesson is the discharge 放电 equations, the time constant 时间常数, and the log plot that turns the curve into a straight line.
永远走不完的衰减
- 放电中的电容器在第一个时间常数内失去 63% 的电荷,在下一个时间常数内又失去剩余部分的 63%,如此永远进行下去。
- 五个时间常数之后剩下不到 1%,但曲线永远到不了零。它不可能到,因为损失的速率始终与还剩多少成正比。
- 每一个这样行为的量——包括放射性衰变——都给出同一条曲线,所以学一次可以用两次。
- 这一课讲放电(discharge)方程、时间常数(time constant),以及把曲线变成直线的对数图。
Where the exponential comes from
- Close the switch and the capacitor's p.d. drives a current through the resistor. Kirchhoff's second law gives $V_C = V_R$, and with $V_C = Q/C$, $V_R = IR$ and $I = -dQ/dt$:
- The rate of loss is proportional to the charge remaining. That is the defining property of exponential decay 指数衰减, and it is why the curve flattens as it falls.
指数从哪里来
- 合上开关,电容器的电压驱动电流通过电阻。基尔霍夫第二定律给出 $V_C = V_R$,而 $V_C = Q/C$、$V_R = IR$、$I = -dQ/dt$:
- 损失的速率与剩余的电荷成正比。这正是指数衰减(exponential decay)的定义性质,也是曲线在下降时越来越平的原因。
During discharge through a resistor, the charge on a capacitor: · 在通过电阻放电期间,电容器上的电荷:
The bigger the charge, the bigger the current — giving an exponential decay. · 电荷越大,电流越大——给出指数衰减。
Why is a capacitor's discharge exponential? · 电容器的放电为什么是指数式的?
Q/C = -R dQ/dt says exactly that. Every quantity with this property, radioactive decay included, gives the same curve. · Q/C = -R dQ/dt 说的正是这个。每个具有这种性质的量——包括放射性衰变——都给出同一条曲线。
The three equations
- Charge, p.d. and current all decay together, with the same time constant:
- The current is largest at the instant the switch closes, $I_0 = V_0/R$, because the full initial p.d. is then across the resistor. It falls as the p.d. falls.
- The area under the current-time graph is the charge that has flowed. All three curves have the same shape, so measuring $\tau$ from any one of them gives $RC$.
Steep at first, then ever flatter, and never quite zero
三个方程
- 电荷、电压和电流一起衰减,时间常数相同:
- 电流在开关合上的一瞬间最大,$I_0 = V_0/R$,因为那时全部初始电压都加在电阻上。它随电压下降而下降。
- 电流—时间图下的面积是已流过的电荷。三条曲线形状相同,所以从其中任何一条测出 $\tau$ 都能得到 $RC$。
一开始很陡,然后越来越平,而且永远到不了零
Charge / discharge curve · 充电 / 放电曲线
The voltage rises (or decays) exponentially with time constant τ = RC. · 电压随时间常数 τ = RC 指数上升(或衰减)。
Discharging a capacitor · 电容器放电
Q = Q₀·bᵗ
Charge decays · 衰变 exponentially through the resistor. · 电荷通过电阻 指数衰减。
The charge during discharge is given by: · 放电期间的电荷由下式给出:
Discharge falls toward zero: $Q = Q_0 e^{-t/RC}$ (the $1 - e^{-t/RC}$ form is for charging · 充电). · 放电向零下降:$Q = Q_0 e^{-t/RC}$($1 - e^{-t/RC}$ 形式是用于充电的)。
The time constant
- The time constant is:
- Its unit is the second, since an ohm times a farad is a second.
- It is the time for the decaying quantity to fall to $\dfrac{1}{e} \approx 37\%$ of its starting value. After $2\tau$ about 13.5% remains, and after $5\tau$ less than 1%.
- A large $R$ or a large $C$ means a slow discharge: a big store emptying through a narrow pipe.
时间常数
- 时间常数是:
- 它的单位是秒,因为欧姆乘法拉就是秒。
- 它是衰减量降到起始值 $\dfrac{1}{e} \approx 37\%$ 所需的时间。$2\tau$ 后约剩 13.5%,$5\tau$ 后不到 1%。
- $R$ 大或 $C$ 大都意味着放电慢:一个大储存器通过一根细管排空。
A capacitor of $500\ \mu\text{F}$ discharges through a $2000\ \Omega$ resistor. What is the time constant? · 一个 $500\ \mu\text{F}$ 的电容器通过一个 $2000\ \Omega$ 的电阻放电。时间常数是多少?
$\tau = RC = 2000 \times 500 \times 10^{-6} = 1.0\ \text{s}$. · $\tau = RC = 2000 \times 500 \times 10^{-6} = 1.0\ \text{s}$。
After one time constant, the charge falls to about ____ % of its starting value. · 经过一个时间常数后,电荷降到它起始值的约 ____ %。
It falls to $\dfrac{1}{e} \approx 0.37$, i.e. about 37%. · 它降到 $\dfrac{1}{e} \approx 0.37$,即约 37%。
After 5 time constants, less than 1% of the charge remains. · 经过 5 个时间常数后,剩下的电荷不到 1%。
$e^{-5} \approx 0.0067$, so under 1% is left — effectively fully discharged. · $e^{-5} \approx 0.0067$,所以剩下不到 1%——实际上完全放电了。
A capacitor charged to 24 V discharges through a 5.6 kilo-ohm resistor. What is the initial current, in mA? · 一个充到 24 V 的电容器通过 5.6 千欧的电阻放电。初始电流是多少 mA?
At the instant the switch closes the whole p.d. is across the resistor, so I0 = V0/R = 24/5600 = 4.3 mA. It then decays with the same time constant as the charge. · 开关合上的一瞬间全部电压都加在电阻上,所以 I0 = V0/R = 24/5600 = 4.3 mA。此后它以与电荷相同的时间常数衰减。
Worked example: a full discharge question
- A $470\ \mu\text{F}$ capacitor charged to $24\ \text{V}$ discharges through a $5.6\ \text{k}\Omega$ resistor. Find the time constant, the initial current, the p.d. after $4.0\ \text{s}$, and the time for the p.d. to fall to $6.0\ \text{V}$.
- $\tau = RC = (5.6 \times 10^3)(470 \times 10^{-6}) = 2.6\ \text{s}$, and $I_0 = V_0/R = 24/5600 = 4.3\ \text{mA}$.
- After $4.0\ \text{s}$: $V = V_0 e^{-t/RC} = 24\,e^{-4.0/2.63} = 5.3\ \text{V}$.
- For the time, take logarithms: $t = -RC\ln(V/V_0) = -2.63\ln(6.0/24) = 3.6\ \text{s}$.
- Keep $\tau$ unrounded inside the calculation and round only at the end. And note that $\ln(V/V_0)$ is negative, so the minus sign makes $t$ positive: that rearrangement is the most often asked and the most often botched.
例题:一道完整的放电题
- 一个充到 $24\ \text{V}$ 的 $470\ \mu\text{F}$ 电容器通过 $5.6\ \text{k}\Omega$ 的电阻放电。求时间常数、初始电流、$4.0\ \text{s}$ 后的电压,以及电压降到 $6.0\ \text{V}$ 所需的时间。
- $\tau = RC = (5.6 \times 10^3)(470 \times 10^{-6}) = 2.6\ \text{s}$,而 $I_0 = V_0/R = 24/5600 = 4.3\ \text{mA}$。
- $4.0\ \text{s}$ 后:$V = V_0 e^{-t/RC} = 24\,e^{-4.0/2.63} = 5.3\ \text{V}$。
- 求时间要取对数:$t = -RC\ln(V/V_0) = -2.63\ln(6.0/24) = 3.6\ \text{s}$。
- 计算过程中 $\tau$ 要不取整,只在最后取整。另外注意 $\ln(V/V_0)$ 为负,所以那个负号让 $t$ 为正:这个移项被问得最多,也最常出错。
After one time constant, about 37% of the original charge remains on the capacitor. · 一个时间常数之后,电容器上约剩原来电荷的 37%。
It falls to 1/e of its starting value, about 37%. After two it is at about 13.5%, and after five below 1%, though it never quite reaches zero. · 它降到起始值的 1/e,约 37%。两个之后约 13.5%,五个之后不到 1%,但它永远到不了零。
Worked example: reading $\tau$ off a graph
- The p.d. across a $2200\ \mu\text{F}$ capacitor falls from $6.0\ \text{V}$ to $2.2\ \text{V}$ in $4.5\ \text{s}$. Find the resistance.
- Check the ratio first: $2.2/6.0 = 0.367$, which is $1/e$. So $4.5\ \text{s}$ is one time constant.
- Then $R = \tau/C = 4.5/(2200 \times 10^{-6}) = 2.0 \times 10^3\ \Omega$.
- If the fall is not to $1/e$, use $RC = -t/\ln(V/V_0)$ instead. The answer is the same either way, but spotting the $1/e$ saves a step.
例题:从图上读出 $\tau$
- 一个 $2200\ \mu\text{F}$ 电容器两端的电压在 $4.5\ \text{s}$ 内从 $6.0\ \text{V}$ 降到 $2.2\ \text{V}$。求电阻。
- 先看比值:$2.2/6.0 = 0.367$,正是 $1/e$。所以 $4.5\ \text{s}$ 就是一个时间常数。
- 于是 $R = \tau/C = 4.5/(2200 \times 10^{-6}) = 2.0 \times 10^3\ \Omega$。
- 如果不是降到 $1/e$,就改用 $RC = -t/\ln(V/V_0)$。两种做法答案相同,但认出 $1/e$ 能省一步。
A capacitor with time constant 2.63 s is charged to 24 V. How long, in seconds, until the p.d. falls to 6.0 V? · 一个时间常数为 2.63 s 的电容器被充到 24 V。电压降到 6.0 V 需要多少秒?
t = -RC ln(V/V0) = -2.63 ln(0.25) = 3.6 s. ln(0.25) is negative, so the minus sign makes t positive; that rearrangement is the most often botched. · t = -RC ln(V/V0) = -2.63 ln(0.25) = 3.6 s。ln(0.25) 为负,所以负号让 t 为正;这个移项最常出错。
The log plot
- Take logarithms of $V = V_0 e^{-t/RC}$:
- So a graph of $\ln V$ against $t$ is a straight line with gradient $-1/RC$ and intercept $\ln V_0$.
- This is the standard experimental method: it turns a curve into a line, so $RC$ comes from a gradient rather than from a single reading, and any departure from a straight line shows the decay is not exponential.
对数图
- 对 $V = V_0 e^{-t/RC}$ 取对数:
- 所以 $\ln V$ 对 $t$ 的图是一条直线,斜率为 $-1/RC$,截距为 $\ln V_0$。
- 这是标准的实验方法:它把曲线变成直线,于是 $RC$ 来自一个斜率而不是单次读数,而任何偏离直线的地方都说明这个衰减不是指数式的。
The p.d. across a 2200 uF capacitor falls from 6.0 V to 2.2 V in 4.5 s. Put the steps of finding the resistance in order. · 一个 2200 uF 电容器两端的电压在 4.5 s 内从 6.0 V 降到 2.2 V。把求电阻的步骤按顺序排列。
If the fall is not to 1/e, use RC = -t/ln(V/V0) instead. The answer is the same, but spotting the 1/e saves a step. · 如果不是降到 1/e,就改用 RC = -t/ln(V/V0)。答案相同,但认出 1/e 能省一步。
Marks that slip away
- $\tau = RC$ is a time, in seconds, not a fraction. After one $\tau$ the quantity is at 37%, not 63% gone and stopped.
- The current is largest at the start, when the whole p.d. is across the resistor, and it decays with the same time constant as the charge.
- To find a time, take logs: $t = -RC\ln(V/V_0)$. Keep the time constant unrounded through the working.
- On a $\ln V$ against $t$ plot the gradient is $-1/RC$, so $RC$ is minus the reciprocal of the gradient.
容易丢掉的分
- $\tau = RC$ 是一个时间,以秒计,不是一个比例。一个 $\tau$ 之后这个量在 37%,不是"掉了 63% 就停了"。
- 电流在开始时最大,那时全部电压加在电阻上,而且它以与电荷相同的时间常数衰减。
- 求时间要取对数:$t = -RC\ln(V/V_0)$。整个计算过程中时间常数不取整。
- 在 $\ln V$ 对 $t$ 的图上斜率是 $-1/RC$,所以 $RC$ 是斜率的负倒数。
A graph of ln V against t for a discharging capacitor is a straight line. What is its gradient? · 放电电容器的 ln V 对 t 图是一条直线。它的斜率是多少?
Taking logs of V = V0 exp(-t/RC) gives ln V = ln V0 - t/RC. So RC is minus the reciprocal of the gradient, and the straight line itself confirms the decay is exponential. · 对 V = V0 exp(-t/RC) 取对数得 ln V = ln V0 - t/RC。所以 RC 是斜率的负倒数,而这条直线本身也证实了衰减是指数式的。
You've got it
- the rate of loss is proportional to the charge left, so the decay is exponential: $Q = Q_0e^{-t/RC}$, and $V$ and $I$ decay identically
- the current starts at $I_0 = V_0/R$ and the area under the current-time graph is the charge that has flowed
- the time constant $\tau = RC$ is the time to fall to $1/e \approx 37\%$; after $5\tau$ less than 1% remains
- take logs to find a time, $t = -RC\ln(V/V_0)$, and a plot of $\ln V$ against $t$ is a straight line of gradient $-1/RC$
你掌握了
- 损失速率与剩余电荷成正比,所以衰减是指数式的:$Q = Q_0e^{-t/RC}$,而 $V$ 和 $I$ 以同样方式衰减
- 电流从 $I_0 = V_0/R$ 开始,电流—时间图下的面积是已流过的电荷
- 时间常数 $\tau = RC$ 是降到 $1/e \approx 37\%$ 所需的时间;$5\tau$ 后剩下不到 1%
- 求时间取对数,$t = -RC\ln(V/V_0)$,而 $\ln V$ 对 $t$ 的图是斜率为 $-1/RC$ 的直线