节: CRC和LC电路 | 4.1 RC和LC电路

章节大纲

Objectives:
::目标:

Conceptualize how circuits involving combinations of resistors, capacitors, and inductors will work.
::构思涉及抵抗器、电容器和感应器组合的电路将如何运作。

Gain experience in electronics by building real and virtual circuits involving all of these new elements.
::通过建立涉及所有这些新要素的实际和虚拟电路,积累电子学经验。

RC and LC Circuits
::CRC和LC电路

The three “ passive ” circuit elements (resistors, capacitors, and inductors) in combination with voltage sources (batteries, wall outlets, and so on) and switches (mechanical or electronic in the form of transistors), can be combined to make all sorts of interesting and useful electric circuits. Let’s review the basic set of passive elements:
::三种 " 被动 " 电路元件(抵抗器、电容器和感应器),加上电压源(电池、墙插口等)和开关(机械或电子形式的晶体管),可以结合起来,使各种电路变得有趣和有用。

t
::t 时

RC Circuits
::RC 电路

Consider the circuit represented by the diagram shown here. At first the switch is open and the capacitor is completely uncharged. When the switch is closed, current will begin flowing out of the voltage source, through the resistor, and into the capacitor. No current flows across the capacitor-it represents an open circuit.
::考虑此处显示的图表所代表的电路。起初开关是开着的, 电容器是完全无电的。当开关关闭时, 电流将开始通过电压源、阻力器和电容器流动。没有电容器的电流代表开通的电路。

At first, there is very little charge on the capacitor, so $\begin{aligned} Q = C V = 0 \end{aligned}$ , and so the voltage across the capacitor is zero. This means the entire voltage drop $\begin{aligned} V_{0} \end{aligned}$ occurs across the resistor, and the current through the resistor is $\begin{aligned} I = \frac{V_{0}}{R} \end{aligned}$ - the highest it will ever be!
::首先,电容器上的电荷很少, 所以 CV=0, 所以电容器的电压是零。这意味着电压下降的V0 整个电压下降的V0 都发生在电阻器上, 而电流通过电阻器的电流是I=V0R, 将达到最高值 !

As the capacitor begins to fill, it gains charge and the voltmeter begins to register a voltage. Since there is now a drop across the capacitor, there must be a smaller drop across the resistor. This means the current through the resistor must be smaller as well. The circuit is beginning to slow down! Eventually, the capacitor is fully charged to $\begin{aligned} Q = C V_{0} \end{aligned}$ and no current flows through the resistor. This takes a characteristic time given by the formula.
::当电容器开始填充时, 电荷增加充电, 电压计开始注册电压。由于电容器现在有下降, 电压计必须下降小一点。这意味着电流通过电阻器时也必须缩小。电路开始放慢! 电路最终会完全充电到 QCV0 , 电压计没有流过电压。这需要公式给出的特性时间。

$\begin{aligned} τ = R C \dots measured in seconds \end{aligned}$

::RC... 以秒计

The differential equation describing the current flowing through the circuit can be determined by analyzing all the voltage drops in the circuit:
::描述电流通过电路的电流的不同方程式可以通过分析电路中的所有电压滴数来确定:

$\begin{aligned} V_{0} - I R + \frac{Q}{C} = 0 \end{aligned}$

::V0-IRC=0

Taking one derivative of this entire equation gives you
::取取整个方程式的一个衍生物给你

$\begin{aligned} R \frac{d I}{d t} + \frac{1}{C} I = 0 \end{aligned}$

::Rdidt+1CI=0

Simplifying you see
::简化您看到

$\begin{aligned} \frac{d I}{d t} = - \frac{1}{R C} I \end{aligned}$

::didt 1RCI

It is worth taking the time to verify that the following is an adequate solution
::值得花时间核实以下是适当的解决办法:

$\begin{aligned} I (t) = \frac{V_{0}}{R} e^{\frac{- t}{R C}} \dots the current exponentially decays, as described \end{aligned}$

::I(t) = V0Re- tRC... 描述的当前指数衰变

LC Circuits
::LC 电路

Consider the circuit represented by the diagram shown here. At first both switches are open and the capacitor is completely uncharged. When the leftmost switch is closed, the capacitor will begin to charge in exactly the way described previously (it’s just an RC circuit). Once the capacitor is charged, open the leftmost switch and forget about that side of the circuit.
::考虑这里所显示的图表所代表的电路。起初,两个开关都打开,电容器完全没有充电。当最左边的开关关闭时,电容器将开始按前述(这只是RC路段 ) 的方式充电。一旦电容器充电,打开最左边的开关,忘记电路的那一侧。

Now you close the second switch. What happens? Charge immediately begins to flow out of the capacitor and through the wires that make up the inductor. But the rapid change in current makes the inductor resist the flowing current with a “ back-emf . ” It takes some time for the current to “ get up to speed”: flowing out of the capacitor due to this inductive behavior. The system begins to oscillate with a characteristic time given by the formula
::现在,您关闭第二个开关。发生什么了? 电荷立即开始从电容器中流出,然后通过电源的电线流出。但电流的快速变化使得电源用“ 回光”抵抗流水。电流需要一段时间才能“跟上速度 ” : 电流由于这种感应行为而从电容器中流出。系统开始以公式给定的特性时间进行冲动。

$\begin{aligned} τ = \sqrt{L C} \dots measured in seconds \end{aligned}$

::LC... 以秒量度

The differential equation describing the charge stored in the capacitor can be determined by analyzing all the voltage drops in the (right-most portion of the) circuit:
::描述电容器中储存的电荷的差别方程式可以通过分析电路(电路中最右部分)的所有电压滴数来确定:

$\begin{aligned} - L \frac{d I}{d t} = \frac{Q}{C} \end{aligned}$

::- LdidtC

but noting that $\begin{aligned} I = \frac{d Q}{d t} \end{aligned}$ ... that is, current is just the rate at which charge flows, you can see that
::但注意到I=dQdt... 也就是说,当前只是收费流量的利率,你可以看到

$\begin{aligned} - L \frac{d^{2} Q}{d t^{2}} = \frac{Q}{C} \end{aligned}$

::-Ld2Qdt2C -Ld2Qdt2C

or
::或

$\begin{aligned} \frac{d^{2} Q}{d t^{2}} = - \frac{1}{L C} Q \end{aligned}$

::d2Qdt21LCQ

Again, it is worth taking the time that the following is an adequate solution to this equation:
::同样,值得花时间考虑的是,以下是这一方程式的适当解决办法:

$\begin{aligned} \begin{aligned} Q (t) & = A \sin (\frac{2 π}{\sqrt{L C}} t) + B \cos (\frac{2 π}{\sqrt{L C}} t) \dots take a derivative of this to get current \\ I (t) & = \frac{2 π}{\sqrt{L C}} A \cos (\frac{2 π}{\sqrt{L C}} t) - \frac{2 π}{\sqrt{L C}} B \sin (\frac{2 π}{\sqrt{L C}} t) \dots the current oscillates with time \end{aligned} \end{aligned}$

::Q(t) = Asin ( 2LCt) +Bcos ( 2LCt)... 使用此值的衍生物来获取当前I( t) = 2LCAcos ( 2LCt) - 2LCBsin ( 2LCt)... 当前振荡时间

Figure out how to design an OR circuit on your own with the following logic table. Again, answers will be provided on the next page.

章节大纲

Objectives: ::目标:

RC and LC Circuits ::CRC和LC电路

RC Circuits ::RC 电路

LC Circuits ::LC 电路

Objectives:
::目标:

RC and LC Circuits
::CRC和LC电路

RC Circuits
::RC 电路

LC Circuits
::LC 电路