Electric Circuits
AP Physics C: Electricity and Magnetism Topic 11 8:42 English narration · English + 中文 subtitles burned in
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Transcript
You flip a switch, and the lamp lights at once.
你按下开关，灯立刻亮了。
So the electrons must be racing down the wire — right?
那么电子一定是在导线里飞奔——对吗？
Not at all.
完全不是。
Inside that copper wire, the carriers crawl forward at less than a millimetre per second.
在那根铜导线里，载流子每秒前进不到一毫米，比蜗牛还慢。
Slower than a snail. So why is the light instant?
那么，为什么灯会立刻亮？
What really travels down the wire?
导线里真正传播的到底是什么？
That question is this whole unit.
这个问题就是整个单元。
This is Unit Eleven: electric circuits.
这是第十一单元：电路。
We follow the charge — from the slow drift inside a wire, to resistance and power, to the two rules that crack any network, and on to circuits that change with time.
我们跟着电荷走—— 从导线内部缓慢的漂移，到电阻与功率，到能解开任何电路网络的两条定则， 再到随时间变化的电路。
Current is the rate at which charge passes a cross-section of a wire: one ampere is one coulomb per second.
电流是电荷通过导线某一横截面的速率：一安培就是每秒一库仑。
Zoom into the metal: the current is the number of carriers per cubic metre, times the charge on each, times their drift velocity, times the area.
放大到金属内部，就得到微观图像——电流等于每立方米的载流子数， 乘以每个载流子的电荷，乘以它们的漂移速度，再乘以横截面积。
Conventional current points the way positive charge would move — in a metal the electrons drift the other way.
常规电流的方向是正电荷会移动的方向；在金属中，电子其实朝相反方向漂移。
One warning: current has a direction, but it is a scalar — no components, no vector addition.
一个提醒：电流有方向，但它是标量——不能分解成分量，也不按矢量相加。
So how slow is that drift?
那么这个漂移到底有多慢？
An ordinary copper wire carries one point seven amperes.
一根普通铜导线通过一点七安培的电流。
First rearrange the formula: the drift speed is the current divided by the carrier number, the charge and the area.
先把公式变形：漂移速度等于电流，除以载流子数、电荷量和横截面积。
Now put the numbers in.
现在代入数字。
The answer is about one point three times ten to the minus four metres per second — a tenth of a millimetre each second.
答案约为一点三乘以十的负四次方米每秒——每秒零点一毫米。
Yet the lamp still lights the instant you close the switch.
可是，你一合上开关，灯依然立刻就亮。
Harder questions need current density: the charge flow per unit area.
更难的题目需要电流密度：单位面积上的电荷流量。
Unlike current, it really is a vector.
与电流不同，它确实是矢量。
A potential difference across a conductor sets up a field inside it, equal to the resistivity times the current density.
导体两端的电位差在其内部建立电场， 这个电场等于电阻率乘以电流密度。
And when the flow is not the same everywhere, integrate the current density over the cross-section — adding thin rings from the centre outwards.
而当各处的流动并不相同时，就要把电流密度对横截面积分， 从中心向外一圈一圈地累加。
Now the circuit.
现在来看电路本身。
In a closed loop, charge can flow all the way round.
在闭合回路中，电荷可以一路流回原处。
Open the switch and nothing flows anywhere in that loop.
断开开关，这个回路里就哪儿都没有电流。
A short circuit is a path with no potential difference across it, so charge rushes through it and ignores the element beside it.
短路是一条两端没有电位差的通路，于是电荷从短路冲过去， 而无视旁边的元件。
Learn the symbols — every question arrives as a circuit diagram.
要记住电路符号——每道题都以电路图的形式出现。
And learn the two connections: in series, the same current; in parallel, the same potential difference.
还要记住两种连接方式：串联，电流相同；并联，电位差相同。
Resistance measures how strongly an object opposes the flow of charge.
电阻衡量一个物体对电荷流动的阻碍程度。
Geometry decides most of it: the resistivity times the length, divided by the cross-sectional area.
几何形状决定了其中大部分：电阻率乘以长度，再除以横截面积。
Long and thin resists more; short and fat resists less.
又长又细，电阻大；又短又粗，电阻小。
Resistivity belongs to the material, not the shape, and in a metal it rises with temperature.
电阻率属于材料本身，与形状无关；在金属中，它随温度升高而增大。
Now the classic trick: stretch a wire to twice its length.
再看那道经典的陷阱题：把一根导线拉长到原来的两倍。
The volume is fixed, so the area halves — and the resistance is four times bigger.
体积不会改变，所以横截面积减半——于是电阻变成四倍。
Ohm's law ties them together: the current is the potential difference divided by the resistance.
欧姆定律把它们联系起来：电流等于电位差除以电阻。
Plot current against voltage — an ohmic material gives a straight line through the origin, whose gradient is one over the resistance.
画出电流对电压的图像——欧姆性材料给出一条过原点的直线， 它的斜率就是电阻的倒数。
A lamp filament is not ohmic: it heats up, its resistance rises, and the line bends.
灯丝不是欧姆性的：它会发热，电阻升高， 图线于是弯曲。
Charge falling through a potential difference gives up energy, so an element transfers energy at a rate equal to the current times the potential difference — and, with Ohm's law, in two more forms.
电荷通过电位差时会释放能量，所以一个元件传递能量的速率 等于电流乘以电位差——再借助欧姆定律，还有另外两种形式。
Use the one that matches the quantities you know.
用你已知量对应的那一个。
Power also ranks brightness: in series every bulb carries the same current, so the bigger resistance is brighter; in parallel every bulb has the same voltage, so the smaller one is.
功率还能给灯泡的亮度排序： 串联时每个灯泡电流相同，所以电阻大的更亮； 并联时每个灯泡电压相同，所以电阻小的更亮。
In the series loop there is one path only, so the same current passes through both lamps.
在串联回路里只有一条通路，所以同一个电流流过两个灯泡。
In the parallel circuit the current reaches a junction and splits, while both branches feel the full battery voltage.
在并联电路里，电流到达节点后分开，而两条支路都得到电池的全部电压。
Same current, or same voltage: that one question shapes every network problem.
电流相同，还是电压相同：这一个问题决定了每一道电路网络题的解法。
To solve a network, collapse it to one equivalent resistance.
要解一个电路网络，就把它化简。
In series the resistances add. In parallel add the reciprocals — always smaller than the smallest branch, because you opened extra paths.
串联时电阻直接相加； 并联时把倒数相加——结果总小于最小的那条支路，因为你多开了通路。
Collapse to one resistor, find the battery current, then expand back out.
一步步化简到只剩一个电阻，求出电池中的电流，再一步步展开回去。
Twelve volts across four ohms and two ohms in series: six ohms in total, two amperes, eight volts and four volts, and sixteen watts of heat in the larger resistor.
十二伏加在串联的四欧姆和二欧姆上：总共六欧姆，电流两安培， 两端分别是八伏和四伏，较大的电阻上有十六瓦的发热。
Real batteries are not ideal.
真实电池并不理想。
Model one as an ideal source of electromotive force with a small internal resistance in series.
把它看作一个理想电动势源，与一个小小的内阻串联。
When current flows, some of that force is used up inside, so the terminal voltage — what a voltmeter really reads — drops below it.
有电流流过时，一部分电动势消耗在电池内部， 所以端电压——电压表真正读到的值——会低于它。
Twelve volts, half an ohm and two amperes gives eleven volts.
十二伏、零点五欧姆、两安培，得到十一伏。
And the meters: an ammeter goes in series, ideally with zero resistance; a voltmeter goes in parallel, ideally with infinite resistance.
再说仪表：电流表要串联，理想情况下电阻为零； 电压表要并联，理想情况下电阻无穷大。
Two rules solve every circuit — Kirchhoff's laws.
两条定则可以解开每一个电路。
Kirchhoff's loop rule: the potential differences round any closed loop add to zero — conservation of energy.
回路定则：沿任意闭合回路， 各元件上的电位差之和为零——这是能量守恒。
Signs are where the marks are won: crossing a battery from minus to plus adds the electromotive force; crossing a resistor along the current subtracts the current times the resistance; against it, it adds.
分数就在符号上：从负极跨到正极，加上电动势； 顺着电流跨过电阻，减去电流乘电阻；逆着电流跨过，则是加。
Draw potential against position and you see it — a climb at the battery, a drop at each resistor, back to the start.
把电位随位置画出来就能看到——在电池处上升，在每个电阻处下降， 最后回到出发点。
Kirchhoff's junction rule: current in equals current out — conservation of charge.
节点定则：流入等于流出——这是电荷守恒。
Now a compound direct current circuit — the one the exam loves.
这就是考试最爱的电路。
Twelve volts and one ohm on the left, nine volts and one ohm on the right, both feeding a shared two ohm resistor in the middle.
左边十二伏、一欧姆，右边九伏、一欧姆， 两边共同供给中间那个二欧姆的电阻。
Find all three currents.
求三个电流。
Pause here and try it.
先暂停，自己试一试。
Start with the junction rule: the middle current is the sum of the other two.
先用节点定则：中间的电流等于另外两个之和。
Now walk each loop and substitute that sum.
再沿每个回路走一圈，把这个和代进去。
Solve the pair: three point six amperes on the left, zero point six zero on the right, and four point two through the middle.
解这一对方程： 左边三点六安培，右边零点六零安培，中间四点二安培。
Now add a capacitor.
现在加入电容。
Capacitors combine into an equivalent capacitance with the rules swapped: in parallel the capacitances add, in series you add the reciprocals — and capacitors in series carry the same charge, because charge is conserved.
电容的组合规则正好互换：并联时电容相加， 串联时把倒数相加——而串联的电容带有相同的电荷量，因为电荷守恒。
Put a resistor and a capacitor in one loop with a battery, and the loop rule gives a differential equation.
把一个电阻和一个电容与电池接成一个回路，用回路定则， 就得到一个微分方程。
Its solutions are exponentials, and one number sets the speed: the time constant, resistance times capacitance.
它的解是指数函数，而快慢由一个数决定： 时间常数，等于电阻乘以电容。
Watch a capacitor charge.
看电容充电。
The voltage climbs fast at first, then flattens out.
电压先快速上升，然后逐渐变平。
The steepness of that curve is the current — biggest at the very first instant, and only falling from there.
这条曲线的陡峭程度就是电流——最初一瞬间最大，此后只会减小。
After one time constant the capacitor holds about sixty-three percent of its final charge.
经过一个时间常数，电容带上约百分之六十三的最终电荷。
That gives the two limits every exam uses: at first an uncharged capacitor behaves like a plain wire; at steady state, long after, like a break in the circuit.
这给出了考试常用的两个极限：最初一瞬间，未充电的电容相当于一根导线； 很久以后，它相当于电路中的一个断口。
Numbers.
来代数字。
A five thousand ohm resistor, a two hundred microfarad capacitor, and a ten volt battery.
一个五千欧姆的电阻，一个两百微法的电容，一个十伏的电池。
Pause here and try it.
先暂停，自己试一试。
First the time constant: resistance times capacitance is exactly one second.
先求时间常数：电阻乘以电容，正好是一秒。
Next the first current: ten volts over five thousand ohms is two milliamperes.
再求最初的电流：十伏除以五千欧姆，等于二毫安。
Finally the charge after one time constant: about one point three millicoulombs, sixty-three percent of the full amount.
最后求一个时间常数后的电荷量：约为一点三毫库仑， 也就是满电荷量的百分之六十三。
Four marks students throw away.
四个学生常丢的分。
First, signs, not slogans: crossing a resistor along the current is a drop, so it enters the loop equation as a minus.
第一，写符号，不要背口号： 顺着电流跨过电阻是电位下降，所以在回路方程里带负号。
Second, the capacitor limits: at first a wire, long after a break — cover that branch with a finger and solve what is left.
第二，电容的两个极限：最初是导线，很久以后是断口—— 用手指盖住那条支路，先解剩下的电路。
Third, resistors add in series, but capacitors do the opposite.
第三，电阻串联相加，而电容正好相反。
Fourth, once current flows the terminal voltage is no longer the electromotive force.
第四，一旦有电流，端电压就不再等于电动势。
Get those four, and this unit is yours.
把这四点做对，这个单元就是你的了。