電流と回路
国際バカロレア · IB Diploma · Physics · HL トピック 10 3:44 英語ナレーション・英語＋中文字幕 burning-in
章
Transcript
A lamp becomes dimmer when another is added in series.
当串联接入另一盏灯时，原有灯泡会变暗。
Current 电流 and energy transfer depend on the whole circuit, not only one lamp.
电流和能量传递取决于整个电路，而不仅仅是一盏灯。
Current is charge passing per time.
电流是单位时间内通过的电荷量。
Potential difference is energy transferred per charge.
电势差是每单位电荷所转移的能量。
Resistance is potential difference divided by current for a stated operating point.
电阻是在给定工作点下电势差与电流之比。
Current is the rate of flow of charge.
电流是电荷流动的速率。
Potential difference is energy transferred per unit charge.
电势差是单位电荷所转移的能量。
Current is the same through components in series.
串联电路中各元件的电流相同。
Potential differences add around the series path.
沿串联回路，电势差相加。
In parallel, branches share the same potential difference, while branch currents sum at a junction.
在并联电路中，各支路两端的电势差相同，而支路电流在节点处相加。
In the lab: Place an ammeter in series and a voltmeter in parallel.
在实验室中，将电流表串联接入电路，将电压表并联接入电路。
For an I-V investigation, change voltage in steps, reverse polarity when appropriate and limit current to reduce heating.
在进行I-V特性研究时，分步改变电压，必要时反转极性，并限制电流以减少发热。
Known: a resistor has 6.0 V across it and carries 0.30 A.
已知：一个电阻两端的电压为 6.0 V，流过的电流为 0.30 A。
Use R equals V divided by I.
使用公式 R = V/I。
R equals 6.0 divided by 0.30, which is 20 ohms.
R = 6.0 / 0.30 = 20 欧姆。
Power equals V times I.
功率 P = VI。
Power equals 6.0 times 0.30, which is 1.8 W.
功率 P = 6.0 × 0.30 = 1.8 W。
These quantities describe the same operating point.
这些物理量描述的是同一个工作点。
Your turn: A component has 12 V across it and carries 0.40 A.
轮到你了。 一个元件两端电压为12伏，通过的电流为0.40安培。
Find resistance.
求其电阻。
Pause the video and work it out.
请暂停视频，自己算一算。
Check: 12 volts divided by 0.40 amps is 30 ohms.
核对：12 伏 ÷ 0.40 安，电阻为 30 欧。
Current is not used up by a lamp.
电流不会被灯泡消耗。
A filament heats up, so its resistance need not remain constant as voltage changes.
灯丝发热，因此其电阻不一定随电压变化而保持恒定。
Not true: A lamp uses up electric current as charge passes through it.
这种说法不正确。 灯泡并不会在电荷流过时消耗电流。
Current is a flow of charge, and charge is conserved: the same current enters and leaves a lamp.
电流是电荷的流动，而电荷是守恒的。 流入和流出灯泡的电流大小相同。
The lamp transfers energy from the charges to heat and light; it does not consume the charges.
灯泡的作用是将电荷携带的能量转化为热能和光能，而不是消耗电荷本身。
Current is charge passing per time.
电流是单位时间内通过的电荷量。
Current is the same through components in series.
串联电路中各元件的电流相同。
Watch for: current is not used up by a lamp.
请注意：电流不会被灯泡消耗掉。
A component has 9 volts across it and carries 0.25 amperes.
某元件两端电压为9伏，通过的电流为0.25安培。
Find the resistance and the power.
求该元件的电阻和消耗的功率。
A lamp does not use up current.
灯泡不会消耗电流。
The result is a resistance of 36 ohms and a power of 2.25 watts.
结果是电阻为36欧姆，功率为2.25瓦。
The ammeter is in the main loop, so all the current passes through it.
电流表串联在主回路中，所以全部电流都流过它。
The voltmeter is connected across the resistor, so it measures the potential difference between its ends; R is V divided by I.
电压表并联在电阻两端，测量其两端的电势差；电阻等于电压除以电流。
Connect two 30 Ω resistors in parallel across 12 V.
将两个 30 Ω 电阻并联在 12 V 电源两端。
Find total resistance, total current and power; compare one resistor alone.
求总电阻、总电流和功率；并与单个电阻的情况进行比较。
The total resistance is 15 ohms, the current 0.80 A, and the power 9.6 W.
总电阻为 15 欧，总电流为 0.80 A，功率为 9.6 W。
Each branch draws 0.40 A and 4.8 W.
每条支路的电流为 0.40 A、功率为 4.8 W。
One resistor alone draws 0.40 A and 4.8 W; the same voltage does not imply the same total current.
单独一个电阻时电流为 0.40 A、功率为 4.8 W；电压相同并不意味着总电流相同。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。