基本電學

抽象的電

講義

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電由於看不到感受不到,所以老師用水來比喻電的概念
因為水有一些特性可以用來比喻電
水會從高處往低處流動
1.水有高低,所以有水位差;電有高低,稱為電位差(電壓)
2.水會流動,所以有水流速;電會流動,所以有電流
3.水流經過水道設施,通暢的水道設施容易流動;
電流經過電阻,易導電的電阻則容易流動
4.水有水位變率,水位差越大,流速越快,水位變率越高;
電有電功率,電位差越大,電流越快,電功率越大(能量釋放)

電力輸送

(以水力輸送為比喻)

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1.水力輸送時,我們希望水道設施(通暢/阻塞)
2.水力輸送時,我們希望將水輸送至全國各地,所以需要(高水位差/低水位差)
3.水力輸送時,我們希望減少能量耗損,所以選擇(劇烈湍急/細水長流)的水流
小節:有效適切的電力輸送方案,會選擇
(高電阻/低電阻)、(高電壓/低電壓)、(高電流/低電流)

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電學單位

單位:電流(庫倫/秒)、電荷(庫倫)、電壓(焦耳/庫倫)

  • 電荷(量)=庫倫=6.25x1018個基本電荷(電子/質子)的電量=一坨電荷(量)
  • 電流=庫倫/秒=每秒流過多少庫倫=每秒流過多少坨正電荷(量)
  • 電壓=焦耳/庫倫=每庫倫電荷能量變化幾焦耳=每一坨電荷能量變化
  • 電阻=電壓/電流=電壓電流比 (歐姆發現固定導體通過的電流電壓比固定)

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電路

概述:通電時,電荷在電池得到能量,經過電器(電阻)時消耗能量

電壓(焦耳/庫倫):每庫倫電荷在電池得3焦耳能量,經過燈泡失3焦耳能量

(電壓為電荷的能量的變化,能量高低的變化也被稱為電位差)

電流(庫倫/秒):每秒流過3庫倫(電流的流速)

電阻=電壓/電流=1

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實際情形:只有電子(負電荷)在流動,電流(正電荷)只假想流

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電壓(焦耳/庫倫):每庫倫電荷在電池得3焦耳能量,經過燈泡失3焦耳能量

電流(庫倫/秒):每秒流過1.5庫倫(電流的流速)

電阻=電壓/電流=2

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電阻=電壓/電流,相同電壓,電阻越大,電流越小

EX:電阻超大(電阻窄長難導電)

*水道超塞

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EX:電阻超小(電阻寬短易導電)

*水道超寬,瀑布

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短路

(有一條水道超寬)

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串聯

電壓(焦耳/庫倫):每庫倫電荷在電池得3焦耳,經過兩個燈泡各耗1.5焦耳

電流(庫倫/秒):每秒流過1.5庫倫的電荷

電阻=電壓/電流=1

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電器串聯(同流):電流固定;電池電壓(電位差)=各電器電壓(電位差)和

串聯(同流)越多,總電流越小

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並聯

電壓(焦耳/庫倫):每庫倫在電池得3焦耳,在燈泡消耗3焦耳

電流(庫倫/秒):主電流=6庫倫/秒;分流=3庫倫/秒

電阻=電壓/電流=1

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電器並聯:電壓(電器)固定,電阻和電流成反比(電阻=電壓/電流)

支電流總和=主電流


電壓(焦耳/庫倫):每庫倫在電池得3焦耳,在某路消耗1焦耳再耗2焦耳

電流(庫倫/秒):主電流=2庫倫/秒;分流=1庫倫/秒

電阻=電壓/電流=1

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電器並聯:電壓(電器)固定,電阻和電流成反比(電阻=電壓/電流)

支電流總和=主電流

斷路:電路不接通

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小節:單位和公式

1.庫倫=6.25x1018個基本電荷(電子/質子)量

2.電壓=焦耳/庫倫=每庫倫電荷能量變化幾焦耳 (伏特)

3.電流=庫倫/秒=每秒流過幾庫倫的電荷 (安培)

4.電阻=電壓/電流=電壓電流比 (歐姆)

➡推導:電壓x電流=(庫倫/秒)x(焦耳/庫倫)=焦耳/秒=電功率

補充:電池串聯、並聯

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電的測量

安培計:測量電流、電阻極小

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伏特計:測量電壓、電阻極大

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檢流計:測量電流方向

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電阻測量:電阻=電壓/電流

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  • 失之毫釐,差之千里

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靜電現象

1.庫倫定律(靜電力)

a.庫倫=電荷量or帶電量
b.兩個帶電的物體之間的靜電力=9x109x帶電量(庫倫)乘積/距離平方

2.導體與絕緣體

a.絕緣體:電子不能原子之間自由移動(受到束縛)。 EX:塑膠、毛皮、玻璃
➔絕緣體可藉由摩擦帶電,微量電子發生轉移,稱為摩擦起電

b.導體:電子可在原子之間自由移動(自由電子)。EX:銀、銅、金
➔導體可藉由感應起電(和帶電棒感應)、接觸起電(和帶電棒接觸)的方式帶電

感應起電:和帶電棒感應

:KEY:正電棒:吸引電子、缺少電子;負電棒:排斥電子、電子太多

接觸起電:和帶電棒接觸

:KEY:正電棒:吸引電子、缺少電子;負電棒:排斥電子、電子太多

雷電現象


演練

1.庫倫(單位:C)表示的是(電荷的量/電流/電壓)

2.每分鐘有900庫倫通過導線截面,每秒流過___庫倫,為___安培(庫倫/秒)

3.導線的電流為0.1安培(庫倫/秒),1分鐘通過的電荷量為___庫倫

4.某燈泡每通過5庫倫的電荷釋放15J的能量,燈泡電壓為___伏特(焦耳/庫倫)

5.承上,若每秒有10庫倫的電荷流過,電功率=___(焦耳/秒)

6.電池提供3伏特(焦耳/庫倫)的電壓,接上燈泡後,

燈泡流過的電流為0.6A,則燈泡的電阻=____歐姆

7.電路如下,總電流為0.2安培,甲乙燈泡電壓分別為2伏特、1伏特

電池提供___伏特(焦耳/庫倫)的電壓

甲乙電流皆為0.2安培,甲的電阻為___歐姆;乙的電阻為___歐姆

甲的電功率=電流(庫倫/秒)x電壓(焦耳/庫倫)=___(焦耳/秒)

乙的電功率=電流(庫倫/秒x電壓(焦耳/庫倫)=___(焦耳/秒)

8.電路如下,總電流為2安培,並聯的甲乙燈泡的電阻皆為1歐姆

甲乙燈泡並聯,電壓相同,電流比=電阻比=___ : ___

甲燈電流+乙燈電流=2安培,甲燈電流=___安培;乙燈電流=___安培

甲燈電壓=乙燈電壓=電流x電阻=___伏特

9.電路如下,電池提供電壓2伏特(焦耳/庫倫),接者a、b、c三個電阻

ab電流大小相同,電壓比=電阻比= ___ : ___

a電壓+b電壓=2伏特,a電壓=___伏特;b電壓=___伏特

a電流=b電流=電壓/電阻=___安培

c電流=電壓/電阻=___安培,總電流=___安培

10.電路圖如下,甲電阻=1歐姆,乙電阻=2歐姆,且乙電壓=4伏特

甲乙電流大小相同,電阻電壓成正比,甲電壓:乙電壓= ___ : ___

甲電壓=___伏特,電路的總電壓=___伏特

甲電流=乙電流=電壓/電阻=___安培

11.有帶電物體A和B互相吸引以及帶電物體C、D互相吸引(如圖)

AB和CD之間的距離相同,靜電力比=電量乘積比= ___ : ___

CD之間的靜電力是AB之間靜電力的___倍

12.有帶電物體A和B互相吸引,將AB分開為原本距離的兩倍

A和B之間的靜電力和距離平方成正比

分開後距離平方變為___倍,AB之間的靜電力變為___倍

13.撕開免洗筷塑膠套時,手上的電子部分轉移到塑膠套上

塑膠帶微量的(負電/正電),手帶微量的(負電/正電),此時手和塑膠套

產生了(靜電力/萬有引力/摩擦力),塑膠套黏在手上,此現象稱為____

14.拿毛皮去摩擦塑膠尺,毛皮的電子部分轉移到尺上面

塑膠尺帶微量(負電/正電),毛皮帶微量的(負電/正電),此時毛皮和尺

產生了(靜電力/萬有引力/摩擦力),此現象稱為____

15.以下為感應起電的4個步驟

甲:一開始金屬球不帶電;乙:正電棒過來會(吸引/排斥)電子

丙:電子(由手往金屬球/金屬球往手);丁:金屬球帶(正電/負電)


參考影片:

https://www.youtube.com/watch?v=DqDtLFb27fw
https://www.youtube.com/watch?v=iXUIyq7wkqM


演練2

:KEY:電荷(量)的單位:庫倫=6.25x1018個基本電荷(電子/質子)

:KEY:電流的單位:安培=庫倫/秒 *每秒有幾庫倫電荷流過

:KEY:失去電子的東西帶正電;得到電子的東西帶負電

兩個帶電的東西相互吸引或排斥的力為靜電力

(靜電力與帶電量乘積成正比;距離平方成反比)

1.某電路中,電流為0.2安培(庫倫/秒),則每秒有多少庫倫的電荷流過?

(A)0.2庫倫 (B)0.4庫倫 (C.)0.6庫倫 (D)0.8庫倫

2.承上題,相當於每秒流過多少顆電子?

(一庫倫=6.25x1018個電子的電量)

(A)0.2x6.25x1018 (B)0.4x6x1018 (C.)6x1018 (D)0.2

3.某導電每分鐘通過180庫倫的電荷,電流大小為多少安培(庫倫/秒)?

(A)1.8 (B)3 (C.)18 (D)180

4.如圖,ABCD為帶電體,AB之間的距離=CD之間的距離,但由於AB的電量與CD不同,靜電力大小也不同,則AB之間的靜電力和CD之間的靜電力大小的比例為? (兩帶電物體之間的靜電力和帶電量乘積成正比)

(A)1:2 (B)2:1 (C.)3:2 (D)1:4

5.電流為正電荷的流動,由電池的正極流向負極,而實際情形是?

(A)電流是質子在流動 (B)實際是電子(負電荷)在流動,由負極往正極

(C.)實際是電子(負電荷)在流動,由正極往負極 (D.)是原子在流動

6.塑膠棒和毛皮摩擦後,塑膠棒帶負電,毛皮帶正電,過程中有何變化?

(A)塑膠棒的質子跑到毛皮 (B)毛皮的電子跑到塑膠棒

(C.)塑膠的電子跑到毛皮 (D)毛皮的質子跑到塑膠棒

7.甲乙燈泡並聯如圖,此電路總電流為4安培(庫倫/秒),甲燈泡處電流為

2安培(庫倫/秒),則乙燈泡處的電流大小為幾安培(庫倫/秒)?

(A)1 (B)2 (C.)3 (D)4

8.安培計是用來測量電流的,下列關於安培計的敘述何者錯誤?

(A)安培計的電阻很小,電流容易通過

(B)安培計與電阻串聯

(C.)安培計正極接電池正極、負極接電池負極

(D)燈泡的電阻越大,安培計測到的電流越大 (電阻=電壓/電流)

:KEY:電荷(量)的單位:庫倫=6.25x1018個基本電荷(電子/質子)

:KEY:電壓的單位:伏特=焦耳/庫倫 *每庫倫電荷能量變化幾焦耳

:KEY:電池提供電荷(庫倫)能量,電荷在電阻(燈泡)消耗能量後回到電池

:KEY:電壓為電荷的能量變化,又稱為電位差

8.有個電池電壓為3伏特(焦耳/庫倫),則10庫倫的電荷經過電池可得到幾焦耳的能量?

(A)3焦耳 (B)6焦耳 (C.)10焦耳 (D)30焦耳

9.每4庫倫的電荷流過燈泡,總共會釋放16焦耳的能量,則燈泡的電壓為多少伏特(焦耳/庫倫)?

(A)0.25焦耳 (B)2焦耳 (C.)4焦耳 (D)8焦耳

10.電荷通過兩4伏特(焦耳/庫倫)燈泡,每庫倫的電荷總共釋放幾焦耳能量?

(A)4焦耳 (B)8焦耳 (C.)16焦耳 (D)24焦耳

11.承上題,若測量AB兩點的電壓會是幾伏特(焦耳/庫倫)?

(A)4伏特 (B)8伏特 (C.)16伏特 (D)24伏特

12.如圖,電荷通過甲燈泡的電路,每庫倫電荷釋放4焦耳能量;通過乙燈泡的電路,每庫倫電荷會釋放4焦耳能量,若測AB兩點的電壓會是幾伏特?

(A)1 (B)2 (C.)3 (D)4

13.承上題,甲的電流大小為2安培(庫倫/秒),甲燈泡每秒能量變化幾焦耳?

(A)16焦耳/秒 (B)8焦耳/秒 (C.)4焦耳/秒 (D)2焦耳/秒

14.如圖,串聯兩個電壓2伏特(焦耳/庫倫)的電池,電荷經過甲電池(A到B),每庫倫電荷得到2焦耳能量;經過乙電池(C到D),每庫倫電荷得到2焦耳能量,若測量A到C點的電壓,會測到幾伏特(焦耳/庫倫)的電壓?

(A)4伏特 (B)3伏特 (C.)2伏特 (D)1伏特

15.如圖,電荷通過甲電池,每庫倫電荷釋放2焦耳能量;通過乙電池,每庫倫電荷會釋放2焦耳能量,若測AB兩點的電壓會是幾伏特(焦耳/庫倫)?

(A)4伏特 (B)3伏特 (C.)2伏特 (D)2伏特

16.伏特計是用來測電壓的儀器,關於伏特計的敘述何者錯誤?

(A)伏特計要與待測物並聯

(B)伏特計電阻要很大,不然電流會容易流到伏特計

(C.)伏特計正極接著電路的正極,負極接著電路的負極

(D)只有電池兩端測的到電壓(電荷的能量變化)

:KEY:電流:安培=庫倫/秒 *電荷的流量

:KEY:電壓:伏特=焦耳/庫倫 *電荷的能量變化

:KEY:電阻:歐姆=電壓(伏特)/電流(安培)

*意涵:電壓電流比、單位電流經過時電荷需要的能量變化

*發現:歐姆發現電流任一導體,電流大小和電壓比值固定,成正比

*形狀越窄長、越難導電的導線電阻會越大

17.大小為4安培的電流經過燈泡時,測得燈泡電壓為4伏特,則燈泡的電阻大小為多少?

(A)1Ω (B)2Ω (C.)3Ω (D)4Ω

18.如圖,電池電壓為4伏特,接上電阻為3歐姆的燈泡,則電路的電流為何?

(A)4安培 (B)3安培 (C.)2安培 (D)1安培

19.電路如下,電池電壓4伏特,有兩個電阻為1歐姆燈泡串聯,則左右兩個燈泡的電壓比為?

(A)4:1 (B)3:1 (C.)1:1 (D)1:3

20.承上題,左燈泡電壓+右燈泡電壓=4伏特,左右燈泡電壓各為多少伏特?

(A)1伏特、3伏特 (B)3伏特、1伏特 (C.)2伏特、2伏特 (D)1伏特、1伏特

21.電路如圖,甲乙燈泡的電流比為何?

(A)2:1 (B)1:2 (C.)1:1 (D)4:1

22.承上題,總電流=甲燈泡的電流+乙燈泡的電流=9安培,甲燈泡電流和乙燈泡的電流大小各為多少安培?

(A)3、6 (B)6、3 (C.)4、5 (D)5、4

23.承上,電荷經過甲燈泡時,電壓為幾伏特?

(A)6伏特 (B)9伏特 (C.)3伏特 (D)10伏特

24.電路如下,電池電壓6伏特,有甲乙兩個3歐姆的燈泡並聯,且總電流為4安培,則甲乙的的電流大小的比例為?

(A)1:2 (B)2:1 (C.)1:1 (D)4:1

25.承上題,總電流=甲燈泡電流+乙燈泡電流=幾安培?

(A)3安培 (B)2安培 (C.)4安培 (D)1安培

26.電阻越大,電流越難通過,下列何者不能將總電阻變大?

(A)不斷串聯燈泡(電阻)

(B)增加電阻的長度

(C.)讓電阻變窄(截面積減少)

(D)不斷並聯燈泡(電阻)

:KEY:導體(電子自由移動)可藉由和帶電棒感應或接觸的方式帶電

正電棒會吸引電子、缺少電子;負電棒會排斥電子、電子太多

:KEY:絕緣體可藉由摩擦起電(電子轉移)的方式帶電

27.圖為一帶電體使金屬球感應起電的五個步驟,下列何者錯誤?

(A)乙步驟時,正電棒吸引電子,質子不動

(B)丙步驟時,正電棒吸引手上的電子

(C.)最後因質子消失金屬球帶負電

(D)最後因電子較多金屬球帶負電

28.圖為一帶電體使金屬球接觸起電的四個步驟,下列何者錯誤?

(A)步驟2時,電子由金屬球流入正電棒(缺少電子)

(B)步驟3時,金屬球失去電子帶正電,和正電棒互相排斥

(C.)金屬球接觸起電最後會和正電棒相反

(D)過程中質子數量完全不變,只有電子轉移

29.關於摩擦起電的敘述何者錯誤?

(A)摩擦時發生電子轉移

(B)兩物體一個失去電子的帶正電、得到電子的帶負電

(C.)兩物體帶電量會相同(電子得到的量=失去的量)

(D)兩個不帶電的金屬球也能藉由摩擦起電

30.下列有關物質帶電方式,何者錯誤?

(A)摩擦起電本質是電子轉移,會帶相反電荷

(B)感應起電後,導體靠近帶電體的地方,兩者電性相反

(C.)接觸起電後,最後金屬球電性與原帶電體相同

(D)帶電體靠近不帶電金屬球,再把兩者分開,最後金屬球兩端帶異性電

補充

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