Time Constant | RC、RL 和 RLC 中的时间常数

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How To Find The Time Constant in RC and RL Circuits

June 8, 2024

💡 Key learnings:

关键学习点：

• Time Constant Definition: The time constant ($\tau$) is defined as the response time of a first-order linear time-invariant (LTI) system to a step input.
  时间常数定义：时间常数（$\tau$）定义为一阶线性时间不变（LTI）系统对阶跃输入的响应时间。

• RC Circuit Time Constant: In an RC circuit, the time constant is the product of resistance ® and capacitance ©.
  RC 电路时间常数：在 RC 电路中，时间常数是电阻（R）和电容（C）的乘积。

• Significance in RC Circuits: The time constant shows how long it takes for the current in a capacitor to drop to 36.7% of its initial value.
  在 RC 电路中的重要性： 时间常数显示电容器中的电流下降到其初始值的 36.7% 需要多长时间。

• RL Circuit Time Constant: The time constant of an RL circuit is defined as the ratio of inductance (L) to resistance ®.
  RL 电路时间常数：RL 电路的时间常数定义为电感（L）与电阻（R）的比率。

• Significance in RL Circuits: The time constant indicates how long it takes for the current in an inductor to reach 63.3% of its final value, highlighting the key concept of the “time constant of rl circuit”.
  在 RL 电路中的重要性：时间常数表示电感中的电流达到其最终值的 63.3% 所需的时间，突出了“RL 电路的时间常数”的关键概念。

What is the Time Constant?

什么是时间常数？

The time constant – usually denoted by the Greek letter $\tau$ (tau) – is used in physics and engineering to characterize the response to a step input of a first-order., linear time-invariant (LTI) control system.. The time constant is the main characteristic unit of a first-order LTI system.

时间常数（通常用希腊字母 $\tau$（tau）表示）在物理学和工程学中用于描述对一阶线性时间不变（LTI）控制系统的阶跃输入的响应。时间常数是一阶 LTI 系统的主要特征单位。

The time constant is commonly used to characterize the response of an RLC circuit.

时间常数通常用于表征 RLC 电路的响应。

Let’s derive the time constant for both RC and RL circuits to understand how they respond to changes.

让我们推导出 RC 和 RL 电路的时间常数，以了解它们如何响应变化。

Time Constant of an RC Circuit

RC 电路的时间常数

Let us take a simple RC circuit, as shown below.

让我们以一个简单的 RC 电路为例，如下所示。

Assume the capacitor is initially uncharged and the switch is closed at time t = 0. Once the switch is closed, electric current i(t) begins to flow through the circuit. Using Kirchhoff Voltage Law in that single mesh circuit, we get:

假设电容器最初未充电，并且开关在时间 $t=0$ 时闭合。一旦开关闭合，电流 $i(t)$ 开始流过电路。在单网孔电路中使用基尔霍夫电压定律，我们得到：

$$V=Ri(t)+\frac{1}{C}\int i(t)\mathrm{d}t$$

Differentiating both sides with respect to time $t$ , we get.

根据时间 $t$ 对两侧进行微分，我们得到：

$$\begin{array}{rl}R\frac{\mathrm{d}i(t)}{\mathrm{d}t}+\frac{i(t)}{C}& =0\\ \Rightarrow R\frac{\mathrm{d}i(t)}{\mathrm{d}t}& =\frac{-i(t)}{C}\end{array}$$

$$\Rightarrow -RC\frac{\mathrm{d}i(t)}{i(t)}=\mathrm{d}t$$

$$\Rightarrow \frac{\mathrm{d}i(t)}{i(t)}=-\frac{1}{RC}\mathrm{d}t$$

Integrating both sides we get,

对两边进行积分，我们得到：

$$\begin{array}{rl}\int \frac{di(t)}{i(t)}& =-\frac{1}{RC}dt\\ \Rightarrow {\log }_{e}i(t)+{\log }_{e}i(k)& =-\frac{t}{RC}\\ \Rightarrow {\log }_e(ki(t))& =-\frac{t}{RC}\\ \Rightarrow ki(t)& =e^{\frac{-t}{RC}}\cdots (i)\end{array}$$

Now, at $t=0$ , the capacitor behaves as a short circuit, so, just after closing the switch, the current through the circuit will be,

现在，在 $t=0$ 时，电容器表现为短路，因此，在关闭开关后，通过电路的电流将为：

$$I_0=\frac{V}{R}$$

Now, putting this value in equation $(i)$ , we get,

现在，将该值代入等式 $(i)$ 中，我们得到：

$$\begin{array}{rl}k\frac{V}{R}& =e^0\\ \Rightarrow k& =\frac{R}{V}\end{array}$$

Putting the value of $k$ at equation $(i)$ , we get,

将 $k$ 的值代入等式 $(i)$ 中，我们得到：

$$i(t)=\frac{V}{R}{e}^{-\frac{t}{RC}}$$

Now, if we put $t=RC$ in the final expression of circuit current $i(t)$ , we get,

现在，如果我们将 $t=RC$ 代入电路电流 $i(t)$ 的最终表达式中，我们得到：

$$\begin{array}{rl}{I}_{t=RC}& =\frac{V}{R}{e}^{-1}\\ & =0.367\frac{V}{R}=0.367I_0\text{ or }36.7\%\text{ of }{I}_0\end{array}$$

The equation shows that $RC$ is the time in seconds for the current in a charging capacitor to drop to $36.7\%$ of its initial value. The initial value is the when the capacitor starts charging.

该方程表明 $RC$ 是充电电容器中的电流下降到其初始值的 $36.7\%$ 所需的时间（以秒为单位）。初始值是电容器开始充电时的电流。

This term is quite significant in analyzing the behavior of capacitive as well as inductive circuits. This term is known as the time constant.

这个术语在分析电容电路和电感电路的行为时非常重要。这个术语被称为时间常数。

So time constant is the duration in seconds during which the current through a capacitive circuit becomes $36.7$ percent of its initial value. This is numerically equal to the product of resistance and capacitance value of the circuit. The time constant is normally denoted by $\tau$ (tau). So,

因此，时间常数是通过电容电路的电流变为其初始值的 $36.7\%$ 的持续时间（以秒为单位）。这在数值上等于电路的电阻和电容值的乘积。时间常数通常用 $\tau$ （tau）表示。所以：

$$\tau =RC$$

In a complex $RC$ circuit, the time constant will be the equivalent resistance and capacitance of the circuit.

在复杂的 $RC$ 电路中，时间常数将是电路的等效电阻和电容。

Let us discuss the significance of the time constant in more detail. To do this, let us first plot current $i(t)$ .

让我们更详细地讨论时间常数的重要性。为此，我们首先绘制电流 $i(t)$ 。

$Att=0$ , the current through the capacitor circuit is

在 $t=0$ 时，通过电容器电路的电流为：

$$I_0=\frac{V}{R}$$

At $t=RC$ , the current through the capacitor is

在 $t=RC$ 时，通过电容器的电流为：

$$0.3670I_0=0.367\frac{V}{R}$$

Let us consider another $RC$ circuit.

让我们考虑另一个 $RC$ 电路。

Circuit equations using KVL. of the above circuits are,

使用上述电路的基尔霍夫电压定律（KVL）的电路方程是：

$$\begin{array}{rl}V-R{i}^{\prime }(t)-2R\left[i^{\prime }(t)-i(t)\right]& =0\cdots \cdots (ii)\\ \Rightarrow V-3R{i}^{\prime }(t)+2Ri(t)& =0\\ \Rightarrow 2V-6R{i}^{\prime }(t)+4Ri(t)& =0\cdots \cdots (iii)\end{array}$$

and

以及：

$$\begin{array}{rl}2R\left[i(t)-i^{\prime }(t)\right]+\frac{1}{C}\int i(t)\mathrm{d}t& =0\cdots \cdots (iv)\\ \Rightarrow 2Ri(t)-2R{i}^{\prime }(t)+\frac{1}{C}\int i(t)\mathrm{d}t& =0\\ \Rightarrow 6Ri(t)-6R{i}^{\prime }(t)+\frac{3}{C}\int i(t)\mathrm{d}t& =0\cdots \cdots (v)\end{array}$$

From $(iii)$ and $(v)$

从 $(iii)$ 和 $(v)$ ：

$$\begin{array}{rl}\Rightarrow 2V+4R{i}^{\prime }(t)& =6Ri(t)+\frac{3}{C}\int i(t)\mathrm{d}t\\ \Rightarrow 2V& =2Ri(t)+\frac{3}{C}\int i(t)\mathrm{d}t\end{array}$$

Differentiating both sides with respect to time $t$ , we get,

根据时间 $t$ 对两侧进行微分，我们得到：

$$\begin{array}{rl}0& =2R\frac{\mathrm{d}i(t)}{\mathrm{d}t}+\frac{3}{C}i(t)\\ \Rightarrow \frac{3}{C}i(t)& =-2R\frac{\mathrm{d}i(t)}{\mathrm{d}t}\\ \Rightarrow \frac{\mathrm{d}i(t)}{i(t)}& =-\frac{3}{2RC}\mathrm{d}t\end{array}$$

Integrating both sides we get,

对两侧进行积分，我们得到：

$$\begin{array}{rl}{\log }_{e}i(t)-{\log }_{e}k& =-\frac{3}{2RC}t\\ \frac{i(t)}{k}& =e^{-\frac{3}{2RC}t}\\ i(t)& =K{e}^{-\frac{3}{2RC}t}\end{array}$$

$Att=0,$

在 $t=0$ 时：

$$\begin{array}{rl}{i}_0=\frac{V}{R}& =K{e}^{-0}\\ \Rightarrow K& =\frac{V}{R}\\ \therefore i(t)& =\frac{V}{R}{e}^{-\frac{3t}{2RC}}\end{array}$$

The time constant of this circuit would be $\frac{2RC}{3}$ sec. Now, the equivalent resistance of the circuit is,

该电路的时间常数为 $\frac{2RC}{3}$ 秒。现在，电路的等效电阻是：

$$resistance=\frac{2R}{3}$$

The time constant of the circuit has become.

电路的时间常数已经变为：

$$\tau =\frac{2R}{3}\cdot C$$

So, $timeconstant=equivalentresistance\times capacitance$

所以，$时间常数=等效电阻\times 电容$

Time Constant of an RL Circuit

RL 电路的时间常数

Let us consider an example of a series $RL$ circuit.

让我们考虑一个串联 $RL$ 电路的示例。

Applying Kirchhoff Voltage Law in the above circuit. We get,

在上述电路中应用基尔霍夫电压定律。我们得到：

$$V=Ri(t)+L\frac{\mathrm{d}i(t)}{\mathrm{d}t}$$

The equation can also be solved Laplace Transformation technique. For that, we have to take Laplace Transformation of the equation at both sides,

该方程也可以用拉普拉斯变换方法求解。为此，我们必须对两边的方程进行拉普拉斯变换：

$$\begin{array}{rl}\frac{V}{s}& =L[sI(s)-i(0^{+})]+RI(s)\\ \Rightarrow \frac{V}{s}& =LsI(s)+RI(s)\\ \Rightarrow I(s)& =\frac{V}{s(Ls+R)}\end{array}$$

Hence, in this equation.

因此，在这个方程中：

$$i(0^{+})=0$$

Since the current just after the switch is on, the current through the inductor will be zero. Now,

由于开关刚接通后的电流，因此通过电感的电流将为零。现在：

$$\Rightarrow I(s)=\frac{V_0}{Ls\left(s+\frac{R}{L}\right)}=\frac{V_0}{L}\frac{1}{s\left(s+\frac{R}{L}\right)}$$

Taking inverse Laplace of the above equation, we get,

取上述方程的逆拉普拉斯，我们得到：

$$i(t)=\frac{V_0}{R}\left[1-e^{-\frac{Rt}{L}}\right]$$

Now, if we put,

现在，如果我们把：

$$t=\frac{L}{R}$$

We get,

我们得到：

$$i\left(\frac{L}{R}\right)=\frac{V_0}{R}\left[1-e^{-1}\right]=0.633\frac{V_0}{R}$$

At the $RL$ circuit, at time $=\frac{L}{R}$ sec, the current becomes $63.3\%$ of its final steady state value. The $\frac{L}{R}$ is known as the time constant of an LR circuit, the current of the inductor circuit.

在 $RL$ 电路中，在时间 $=\frac{L}{R}$ 秒时，电流变为其最终稳态值的 $63.3\%$ 。 $\frac{L}{R}$ 称为 $LR$ 电路的时间常数，即电感电路的时间常数。

The time constant of an $LR$ circuit is the ratio of inductance to the resistance of the circuit. Let us take another.

$LR$ 电路的时间常数是电感与电路电阻的比值。让我们再举一个例子。

This circuit can be redrawn as,

该电路可以重绘为：

So, the time constant of the circuit would be

因此，该电路的时间常数为：

$$\tau =\frac{L_1+L_2}{R_1+R_2}\text{ sec}$$

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Time Constant τ “Tau” Formulas for RC, RL & RLC Circuits

Electrical Technology

Time constant also known as tau represented by the symbol of “τ” is a constant parameter of any capacitive or inductive circuit. It differs from circuit to circuit and also used in different equations. The time constant for some of these circuits are given below:
时间常数也称为 tau，由 “$\tau$” 符号表示，是任何电容或电感电路的常数参数。它因电路而异，也用于不同的方程式。其中一些电路的时间常数如下：

$\tau$ for RC Circuit:

$\tau$ 对于 RC 电路：

In this circuit, resistor having resistance “R” is connected in series with the capacitor having capacitance C, whose τ “time constant” is given by:

在该电路中，具有电阻 “R” 的电阻器与具有电容 C 的电容器串联，其 $\tau$ “时间常数” 由下式给出：

$\tau =RC$
$\tau =RC=\frac{1}{2\pi f_C}$

Where

• $\tau =RC=$ is the time constant in seconds
  $\tau =RC=$ 以秒为单位的时间常数
• R is the resistance in series in ohms (Ω)
  R 串联电阻，单位为欧姆（Ω）
• C is the capacitance of the capacitor in farads
  C 电容器电容，单位为法拉
• $f_C$ = cutoff frequency in hertz
  $f_C$ = 截止频率，单位为赫兹

$\tau$ for RL Circuit:

对于 RL 电路的 $\tau$：

Inductor of inductance “L” connected in series with resistance “R”, whose time constant “τ” in seconds is given by:
电感 “L” 的电感器与电阻 “R” 串联，其时间常数 “τ” 以秒为单位，由下式给出：

$\tau =L/R$

Where

• R is the resistance in series
  R 是串联电阻
• L is the Inductance of the Inductor
  L 是电感的电感

Universal time Constant “τ” Formula

通用时间常数 “$\tau$” 公式

$Change=Final-Start\left(1-\frac{1}{e^{t/\tau }}\right)$

Where:

• Final = Value of calculated variable after infinite time (Ultimate value)
  最终 = 无限时间后计算变量的值（最终值）
• Start = Initial value of calculated variable
  Start = 计算变量的初始值
• $e=$ Euler’s number (≈2.7182818)
  $e=$ 欧拉数（≈2.7182818）
• $t=Timeinsecondst=$ 时间（以秒为单位）
• $\tau =$ Time constant for circuits in seconds
  $\tau$ = 电路的时间常数，单位为秒

$\tau$ for RLC Circuit:

τ 对于 RLC 电路：

In RLC circuit, we have both RL and RC time constant combined, which makes a problem calculating the time constant. So we calculate what we call the Q-Factor (quality factor).

在 RLC 电路中，我们将 RL 和 RC 时间常数组合在一起，这使得计算时间常数成为问题。因此，我们计算出我们所谓的 Q 因子（品质因数）。

$\tau$ for Series RLC Circuit:

$\tau$ 代表 串联 RLC 电路：

$Qfactor=\frac{1}{R}\sqrt{\frac{L}{C}}$

$\tau$ for Parallel RLC Circuit:

$\tau$ 对于并联 RLC 电路：

$Qfactor=R\sqrt{\frac{C}{L}}$

Where

• R is the resistance in series
  R 串联电阻
• L is the Inductance of the Inductor
  L 电感器的电感
• C is the capacitance of the capacitor
  C 电容器的电容

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Why is the time constant 63.2% and not 50% or 70%?

为什么时间常数是 63.2% 而不是 50% 或 70%？

edited Sep 20, 2018 at 14:42 bariod
asked Sep 18, 2018 at 14:39 Bala Subramanian

I am studying about RC and RL circuits. Why is the time constant equal to 63.2% of the output voltage? Why is it defined as 63% and not any other value?
我正在研究 RC 和 RL 电路。为什么时间常数等于输出电压的 63.2%？为什么它被定义为 63% 而不是任何其他值？

Does a circuit start working at 63% of output voltage? Why not at 50%?
电路是否以 63% 的输出电压开始工作？为什么不在 50% 呢？

Answers 1

Other answers haven’t yet hit upon what makes e special: defining the time constant as the time required for something to drop by a factor of e means that at any moment of time, the rate of change will be such that–if that rate were continued–the time required to decay to nothing would be one time constant.

其他答案尚未指出 “e” 的特殊之处：将时间常数定义为某事物衰减至原来的 1/e 所需的时间，意味着在任何时刻，其变化率都具有这样的特性 —— 如果该变化率持续下去，衰减至零所需的时间将为一个时间常数。

For example, if one has a 1uF cap and a 1M resistor, the time constant will be one second. If the capacitor is charged to 10 volts, the voltage will fall at a rate of 10 volts/second. If it’s charged to 5 volts, the voltage will fall at a rate of 5 volts/second. The fact that the rate of change decreases as the voltage does means that the voltage won’t actually decay to nothing in one second, but the rate of decrease at any moment in time will be the current voltage divided by the time constant.

例如，如果有一个 1 微法的电容和一个 1 兆欧的电阻，时间常数将为 1 秒。如果电容充电至 10 伏，电压下降速率将为 10 伏 / 秒。如果充电至 5 伏，电压下降速率则为 5 伏 / 秒。电压下降时变化率也随之减小，这意味着电压实际上不会在 1 秒内衰减至零，但在任何时刻，电压的下降速率都等于当前电压除以时间常数。

If the time constant were defined as any other unit (e.g. half-life), then the rate of decay would no longer correspond so nicely with the time constant.

如果将时间常数定义为其他单位（例如半衰期），那么衰减速率就不再能与时间常数如此完美地对应。

Answers 2

It’s built into the mathematics of exponential decay associated with first-order systems. If the response starts at unity at $t=0$, then after one “unit of time”, the response is $e^{-1}=0.36788$. When you’re looking at a risetime, you subtract this from unity, giving $0.63212$ or $63.2\%$.

这源于与一阶系统相关的指数衰减数学原理。如果响应在 $t=0$ 时从 1 开始，那么经过一个 “时间单位” 后，响应为 $e^{-1}=0.36788$。在研究上升时间时，用 1 减去这个值，得到 $0.63212$，即 $63.2\%$。

The “unit of time” is referred to as the “time constant” of the system, and is usually denoted $\tau$ (tau). The full expression for the system response over time ($t$) is

这个 “时间单位” 被称为系统的 “时间常数”，通常用 $\tau$（希腊字母 tau）表示。系统响应随时间（$t$）变化的完整表达式为 $$V(t)=V_0{e}^{-\frac{t}{\tau }}$$

So the time constant is a useful quantity to know. If want to measure the time constant directly, you measure the time it takes to get to $63.2\%$ of its final value.

因此，时间常数是一个很有用的参数。如果要直接测量时间常数，可以测量达到最终值 $63.2\%$ 所需的时间。

In electronics, it works out that the time constant (in seconds) is equal to $R\times C$ in an R - C circuit or $L/R$ in an R - L circuit, when you use ohms, farads and henries as units for the component values. This means that if you know the time constant, you can derive one of the component values if you know the other.

在电子学中，当以欧姆、法拉和亨利作为元件值的单位时，时间常数（以秒为单位）等于 $R-C$ 电路中的 $R\times C$ 或 $R-L$ 电路中的 $L/R$。这意味着，如果知道时间常数，并且知道其中一个元件的值，就可以求出另一个元件的值。

Answers 3

The decay of an RC parallel circuit with capacitor charged to $V_o$

一个电容器充电至 $V_o$ 的 RC 并联电路的衰减情况

$v(t)=V_o(1e^{-t/\tau })$, where $\tau$ is the time constant $R\cdot C$.

$v(t)=V_o(1e^{-t/\tau })$，其中 $\tau$ 是时间常数，$\tau =R\cdot C$

So $v(\tau )/V_o$ is approximately 0.63212055882855767840447622983854$

所以 $v(\tau )/V_o$ 大约为 0.63212055882855767840447622983854$

In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs.
Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. and can be described by a first-order Ordinary Differential Equation (ODE).

换句话说，时间常数由 $RC$ 的乘积（或 $L/R$ 比值）定义，看似随意的电压是该定义以及指数衰减或充电发生方式的结果。

指数衰减在各种物理过程中很常见，比如放射性衰变、某些类型的冷却等等，并且可以用一阶常微分方程（ODE）来描述。

Suppose you want to know the time when the voltage is 0.5 of the initial voltage (or final voltage if charging from 0). It is (from the above) $t=-\ln (0.5)\tau$ or about 0.693RC

假设你想知道电压为初始电压的 0.5 时（如果是从 0 开始充电，则为最终电压的 0.5 时）的时间 。根据上述内容，时间 $t=-\ln (0.5)\tau$ ，或者大约为 0.693RC 。

Either way you do it, some irrational numbers pop up and dealing with $RC=\tau$ is the “natural” way.

无论你用哪种方式计算，都会出现一些无理数，而处理 $RC=\tau$ 是一种“自然”的方式。

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Why is $RC$ exactly 63.2% full charge, and how is $5RC$ considered 100%?

为什么$RC$对应的是恰好 63.2% 的满电荷量，以及为什么把 $5RC$ 视为 100% 的满电荷量呢？

edited Nov 29, 2016 at 12:10 Jonas Schäfer’s
asked Nov 29, 2016 at 11:14 Ryan Abbas’s

The voltage of a charging and discharging capacitor is given by the formulas:

充电和放电电容器的电压由以下公式给出：

$U_C=U{e}^{-t/(RC)}$

and

以及

$U_C=U\left(1-e^{-t/(RC)}\right)$

respectively, but I don’t understand why RC is equal to 63.2% of full charge.

分别是上述公式，但我不理解为什么 $RC$ 相当于满电荷量的 63.2%。

I also don’t understand how 5RC is found and why it is chosen to be considered a full charge.

我也不理解 $5RC$ 是如何得出的，以及为什么选择将其视为满电荷量。

Answer

Take a look at how a capacitor charges through a resistor:

看看电容器通过电阻充电的过程：

Capacitor charging follows the exponential relationship:

电容器充电遵循指数关系：

$$V_c=V_s(1-e^{-\frac{t}{RC}}),$$

Where $V_s$ is the target “supply” voltage and, if you let $t=CR$ you find that $V_c/V_s=0.632$ or $63\%$ for short: -

其中 $V_s$ 是目标“电源”电压，并且如果你令 $t=CR$，你会发现 $V_c/V_s=0.632$，简而言之就是 $63\%$：

$$\frac{V_C}{V_S}=(1e^{-1})=0.632120558$$

When you do the math of stacking 5 lots of “63%” on top of each other you get: -

当你进行将 5 个 “$63\%$” 依次累计的数学运算时，你会得到：

$V_c=0.99326\times V_s$

or, put another way, $5CR$ gets you to within $1\%$ of the full charging voltage.

或者换句话说，$5CR$ 能使你达到接近满充电电压的 $1\%$ 以内。

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The Time Constant

时间常数

注：这篇找不到出处了，仅记录，未校对。

1. Definition：When a capacitor discharges through a resistor, the time constant measures how long it takes for the capacitor to discharge. It is defined as the time taken for the charge, current, or voltage of a discharging capacitor to decrease to 37% of its original value. For a charging capacitor, it is the time taken for the charge or voltage to rise to 63% of its maximum value. 37% is 0.37 or $\frac{1}{e}$ (where $e$ is the exponential function, approximately equal to 2.718) multiplied by the original value ($I_0$, $Q_0$, or $V_0$). The time constant is represented by the Greek letter $\tau$ and measured in seconds (s).

   定义：电容器通过电阻放电时，时间常数用于衡量电容器放电所需的时长。其定义为：放电电容器的电荷、电流或电压减小到其原始值的 37% 所需的时间；对于充电电容器而言，则是电荷或电压上升到其最大值的 63% 所需的时间。37% 即 0.37，也可表示为 $\frac{1}{e}$（$e$ 为指数函数，约等于 2.718）乘以原始值（$I_0$、$Q_0$或$V_0$ ）。时间常数用希腊字母 $\tau$ 表示，单位为秒（$s$）。

2. Calculation Formula：The time constant is defined by the equation $\tau =RC$, where $R$ is the resistance of the resistor (in $\Omega$) and $C$ is the capacitance of the capacitor (in $F$).

   计算公式：时间常数由公式 $\tau =RC$ 定义 ，其中 $R$ 是电阻的阻值（单位：$\Omega$），$C$ 是电容器的电容（单位：$F$）。

3. Half - life：The half - life $t_{1/2}$ of a discharging capacitor is the time taken for the charge, current, or voltage to reach half of its initial value. It can be related to the time constant as $t_{1/2}=\ln (2)\tau \approx 0.69\tau =0.69RC$.

   半衰期：放电电容器的半衰期 $t_{1/2}$ ，是指电荷、电流或电压达到其初始值一半所需的时间，它与时间常数的关系为 $t_{1/2}=\ln (2)\tau \approx 0.69\tau =0.69RC$ 。

Charging and Discharging Equations

充电和放电方程

1. Discharging Equations：The exponential decay equation for current is $I=I_0{e}^{-\frac{t}{RC}}=I_0{e}^{-\frac{t}{\tau }}$, where $I$ is the current during discharge (in $A$), $I_0$ is the initial current before discharge (in $A$), $t$ is the time (in $s$), and $RC$ is the time constant $\tau$ (in $s$). This equation shows that the smaller the time constant $\tau$, the faster the exponential decay of the current during discharge. Also, the larger the initial current $I_0$, the longer it takes for the capacitor to discharge. Since the current $I$ is always decreasing during discharge, $I_0$ is always greater than $I$.

   放电方程：电流的指数衰减方程为 $I=I_0{e}^{-\frac{t}{RC}}=I_0{e}^{-\frac{t}{\tau }}$ ，式中 $I$ 为放电过程中的电流（$A$），$I_0$ 为放电前的初始电流（$A$），$t$ 为时间（$s$），$RC$ 即时间常数 $\tau$（$s$）。该方程表明，时间常数 $\tau$ 越小，放电时电流的指数衰减越快；初始电流 $I_0$ 越大，电容器放电所需时间越长。由于放电过程中电流 $I$ 始终在减小，所以 $I_0$ 总是大于 $I$ 。

   • Since the current is proportional to the voltage across the capacitor and the charge on the plates, the equation for charge as a function of time is $Q=Q_0{e}^{-\frac{t}{RC}}=Q_0{e}^{-\frac{t}{\tau }}$, where $Q$ is the charge on the capacitor plates (in $C$) and $Q_0$ is the initial charge on the capacitor plates (in $C$).

     由于电流与电容器两端的电压以及极板上的电荷成正比，因此电荷随时间变化的方程为 $Q=Q_0{e}^{-\frac{t}{RC}}=Q_0{e}^{-\frac{t}{\tau }}$ ，其中 $Q$ 是电容器极板上的电荷（$C$），$Q_0$ 是初始时刻电容器极板上的电荷（$C$）。

   • The equation for voltage as a function of time is $V=V_0{e}^{-\frac{t}{RC}}=V_0{e}^{-\frac{t}{\tau }}$, where $V$ is the voltage across the capacitor (in $V$) and $V_0$ is the initial voltage across the capacitor (in $V$).

     电压随时间变化的方程为 $V=V_0{e}^{-\frac{t}{RC}}=V_0{e}^{-\frac{t}{\tau }}$ ，其中 $V$ 是电容器两端的电压（$V$），$V_0$ 是初始时刻电容器两端的电压（$V$）。

2. Charging Equations：When charging, the increase of charge $Q$ and voltage $V$ of the capacitor also shows exponential characteristics. They increase over time but at a decreasing rate. The charging equation for charge is $Q=Q_0(1-e^{-\frac{t}{RC}})=Q_0(1-e^{-\frac{t}{\tau }})$, where $Q_0$ is the maximum charge stored when the capacitor is fully charged (in $C$).

   充电方程：充电时，电容器的电荷 $Q$ 和电压 $V$ 增加的过程也呈现指数特性，随时间增加但增速逐渐减慢。电荷的充电方程为 $Q=Q_0(1-e^{-\frac{t}{RC}})=Q_0(1-e^{-\frac{t}{\tau }})$ ，这里 $Q_0$ 是电容器充满电时存储的最大电荷（$C$）。

   • The charging equation for voltage is $V=V_0(1-e^{-\frac{t}{RC}})=V_0(1-e^{-\frac{t}{\tau }})$, where $V_0$ is the maximum voltage across the capacitor when fully charged (in $V$).

     电压的充电方程为 $V=V_0(1-e^{-\frac{t}{RC}})=V_0(1-e^{-\frac{t}{\tau }})$ ，$V_0$ 是电容器充满电时的最大电压（$V$）。

   • The charging equation for current is the same as the discharging equation, $I=I_0{e}^{-\frac{t}{RC}}=I_0{e}^{-\frac{t}{\tau }}$, but here $Q_0$ and $V_0$ are the final (or maximum) values, not the initial values.

     电流的充电方程与放电方程形式相同，为 $I=I_0{e}^{-\frac{t}{RC}}=I_0{e}^{-\frac{t}{\tau }}$ ，但此时 $Q_0$ 和 $V_0$ 是最终（或最大）值，而非初始值。

Tips

1. whether the capacitor is charging or discharging in the exam question, as the definition of the time constant varies depending on the state.

   电容器是处于充电还是放电状态，因为时间常数的定义会因状态不同而有所差异。

2. the exponential constant $e$ (approximately equal to 2.718) and its inverse function, the natural logarithm function $\ln (y)$. If $e^x=y$, then $x=\ln (y)$. The 0.37 in the definition of the time constant is derived from the exponential constant, and its accurate expression is: The time constant is the time taken for the charge of a capacitor to decrease to $\frac{1}{e}$ (approximately 0.3678) of its original value.

   指数常数 $e$（约等于2.718）及其反函数自然对数函数 $\ln (y)$ 的使用。若 $e^x=y$，则 $x=\ln (y)$ 。时间常数定义中的 $0.37$ 正是由指数常数得出，其准确表述为：时间常数是电容器电荷减小到其原始值的 $\frac{1}{e}$（约为0.3678）所需的时间。

3. the half - life equation $t_{1/2}=0.69RC$

   半衰期方程 $t_{1/2}=0.69RC$ 。

Worked Example

1 计算电阻值

A capacitor of 7 nF is discharged through a resistor of resistance $R$. The time constant of the discharge is $5.6\times 10^{-3}$ s. Calculate the value of $R$.

一个电容为 7 nF 的电容器通过电阻 $R$ 放电，放电的时间常数为 $5.6\times 10^{-3}$ s。计算电阻 $R$ 的值。

已知时间常数公式 $\tau =RC$，则 $R=\frac{\tau }{C}$。将 $\tau =5.6\times 10^{-3}$ s，$C=7\times 10^{-9}$ F 代入公式可得：

$R=\frac{5.6\times 10^{-3}}{7\times 10^{-9}}=8\times 10^5\Omega$

2 计算电流下降所需时间

The initial current through a circuit with a capacitor of 620 µF is 0.6 A. The capacitor is connected across the terminals of a 450 Ω resistor. Calculate the time taken for the current to fall to 0.4 A.
一个电容为 620 µF 的电容器所在电路的初始电流为 0.6 A，该电容器连接在一个 450 Ω 的电阻两端。计算电流下降到 0.4 A 所需的时间。

根据电流的放电方程 $I=I_0{e}^{-\frac{t}{RC}}$，可得 $t=-RC\ln (\frac{I}{I_0})$ 。

已知 $I_0=0.6$ A，$I=0.4$ A，$R=450$ Ω，$C=620\times 10^{-6}$ F ，将这些值代入公式：

$t=-450\times 620\times 10^{-6}\ln (\frac{0.4}{0.6})\approx 0.15s$

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via：

• Time Constant: What it is & How to Find it in an RLC Circuit | Electrical4U
  https://www.electrical4u.com/time-constant/

• Time Constant τ “Tau” Formulas for RC, RL & RLC Circuits
  https://www.electricaltechnology.org/2020/11/time-constant-τ-tau-formulas-equations-rc-rl-rlc-circuits.html

• circuit analysis - Why is the time constant 63.2% and not 50% or 70%? - Electrical Engineering Stack Exchange
  https://electronics.stackexchange.com/questions/396653/why-is-the-time-constant-63-2-and-not-50-or-70

• discharge - why is RC= exactly 63.2% full charge, and how is 5RC considered 100%? - Electrical Engineering Stack Exchange
  https://electronics.stackexchange.com/questions/272138/why-is-rc-exactly-63-2-full-charge-and-how-is-5rc-considered-100

  • How Long Does It Take to Discharge a Capacitor?
    https://www.learningaboutelectronics.com/Articles/How-long-does-it-take-to-discharge-a-capacitor
    —

• Charging and Discharging of Capacitor – Explanation and Important FAQs
  https://www.vedantu.com/jee-main/physics-charging-and-discharging-of-capacitor

• The Time Constant | AQA A Level Physics Revision Notes 2015
  https://www.savemyexams.com/a-level/physics/aqa/17/revision-notes/7-fields-and-their-consequences/7-7-capacitor-charge-and-discharge/7-7-2-the-time-constant/

• Capacitor Discharge Equations | Cambridge (CIE) A Level Physics Revision Notes 2023
  https://www.savemyexams.com/a-level/physics/cie/25/revision-notes/19-capacitance/19-3-discharging-a-capacitor/capacitor-discharge-equations/

• RC Network Category Page - Basic Electronics Tutorials
  https://www.electronics-tutorials.ws/category/rc

• Derivation for voltage across a charging and discharging capacitor
  https://mechatrofice.com/circuits/charging-capacitor-derivation

• Capacitor Discharge Time Calculator (with Examples)
  https://3roam.com/capacitor-discharge-time-calculator/