Amfani da dokar farko ta thermodynamics a cikin wasu hanyoyin thermodynamic (Isobaric Isothermal Isochoric)

30 Amfani da dokar thermodynamics ta farko a wasu hanyoyin thermodynamic (Isobaric Isothermal Isochoric) 1. Jadawalin da ke ƙasa yana nuna zagayowar thermodynamic da gas ke fuskanta. Aikin da gas ke yi akan tsarin ABCD shine … An sani: Matsi 1 (P1) = 2 x 105 Pa Matsi 2 (P2) = 4 x 105 … Kara karantawa

Dokar farko ta thermodynamics - matsaloli da mafita

30 Dokar farko ta thermodynamics - matsaloli da mafita 1. Ana ƙara 3000 J na zafi a cikin tsarin kuma tsarin yana yin 2500 J na aiki. Menene canjin kuzarin ciki na tsarin? An sani: Zafi (Q) = +3000 Joules Aiki (W) = +2500 Joules Ana so: canjin … Kara karantawa

Ka'idar kinetic na gas - matsaloli da mafita

1. Ina sarrafa iskar gas a cikin akwati mai rufewa da farko suna da girma V da matsin lamba P. Idan matsin lamba na ƙarshe shine 4P kuma girman ya kasance daidai, menene rabon kuzarin motsi na farko da ƙarfin motsi na ƙarshe.

An sani:

Matsi na farko (P1 ) = P

Matsi na ƙarshe (P2 ) = 4P

Ƙarar farko (V1 ) = V

Ƙarar ƙarshe (V 2 ) = V

Ana so: Rabon kuzarin motsi na farko zuwa makamashin motsi na ƙarshe (K E 1 : K E 2 )

Magani:

Alaƙa tsakanin matsin lamba (P), girma (V) da kuzarin motsi (KE) na iskar gas masu kyau :

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 18

Rabon kuzarin motsi na farko zuwa makamashin motsi na ƙarshe:

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 19

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2. Menene matsakaicin kuzarin motsi na translational na ƙwayoyin halitta a cikin iskar gas mai kyau a 57 oC.

An sani:

Zafin iskar gas (T) = 57 o C + 273 = 330 Kelvin

Boltzmann 's constant (k) = 1.38 x 10 -23 Joule/Kelvin

Ana so: Matsakaicin kuzarin motsi na fassara

Magani:

Alaƙa tsakanin kuzarin motsi (KE) da zafin iskar gas (T):

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 3

Matsakaicin kuzarin motsi na fassara:

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 4

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3. Iskar gas a 27 oC a cikin akwati mai rufewa . Idan kuzarin motsi na iskar gas ya karu sau biyu na kuzarin motsi na farko, don haka zafin ƙarshe na iskar gas ɗin shine …

An sani:

Zafin farko (T 1 ) = 27 o C + 273 = 300 K

Makamashin motsi na farko = KE

Ƙarfin motsi na ƙarshe = 4 K E

Ana so: Zafin jiki na ƙarshe ( T2 )

Magani:

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 5

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4. Iskar gas mai kyau tana cikin akwati mai rufewa, ana dumama ta yadda matsakaicin saurin barbashi na gas na ƙarshe ya ƙaru da sau 3 matsakaicin saurin farko. Idan zafin iskar gas na farko ya kai 27 oC , to zafin ƙarshe na iskar gas mai kyau shine…

An sani:

Zafin farko = 27 o C + 273 = 300 Kelvin

Saurin farko = v

Saurin ƙarshe = 2v

Ana so : Zafin iskar gas mai kyau na ƙarshe

Magani:

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 20

Matsakaicin gudu na ƙarshe = 2 x matsakaicin gudu na farko

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 7

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5. Moles uku na iskar gas suna cikin sararin juzu'i na lita 36. Kowace kwayar iskar gas tana da kuzarin motsi na Joules 5 x 10 -21 . Madaidaitan iskar gas na duniya = 8.315 J/mole.K da kuma madaidaitan Boltzmann = 1.38 x 10 -23 J/K. Menene matsin lamba na iskar gas a cikin akwati.

An sani:

Adadin moles (n) = moles 3

Ƙarar = lita 36 = 36 dm 3 = 36 x 10 -3 m 3

Madaidaitan Boltzmann (k) = 1.38 x 10 -23 J/K

Makamashin motsi (KE) = Joules 5 x 10 –21

Daidaiton iskar gas na duniya (R) = 8.315 J/mole.K

Ana nema : Matsi na iskar gas (P)

Magani:

Lissafa zafin jiki ta amfani da lissafin kuzarin motsi na iskar gas.

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 8

Lissafin matsin lamba na iskar gas ta amfani da lissafin th na dokar iskar gas mai kyau (a cikin adadin moles, n):

PV = n RT

P (36 x 10 -3 ) = (3)(8.315)(241.5)

P (36 x 10 -3 ) = 6024.22

Ka'idar motsa jiki ta iskar gas - matsaloli da mafita 9

Matsin iskar gas shine 1.67 x 10 5 Pascal ko kuma 1.67 yanayi.

Karin bayani

Dokar iskar gas mai kyau - matsaloli da mafita

1. Ina sayar da iskar gas a cikin akwati mai rufewa da farko suna da girma V da zafin jiki T. Zafin ƙarshe shine 5/4T kuma matsin lamba na ƙarshe shine 2P. Menene girman ƙarshe na iskar gas?

An sani:

Ƙarar farko (V1 ) = V

Zafin farko (T 1 ) = T

Zafin ƙarshe (T2 ) = 5/4 T

Matsi na farko (P1 ) = P

Matsi na ƙarshe (P2 ) = 2P

Ana so: Juzu'i na ƙarshe (V 2 )

Magani:

Dokar iskar gas mai kyau - matsaloli da mafita 1

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2. Kayyade girman 2.00 mol es na iskar gas (iskar da ta dace) a STP. STP = Matsakaicin Zafin Jiki da Matsi.

An sani:

Moles na iskar gas (n) = moles 2 na kankara

Matsakaicin zafin jiki (T) = 0 o C = 0 + 273 = 273 Kelvin

Matsakaicin matsin lamba (P) = 1 atm = 1.013 x 10 5 Pa

Ƙimar iskar gas ta duniya (R) = 8.315 Joule/mol e .Kelvin

Ana buƙata : Yawan iskar gas (V)

Magani:

Daidaito na Dokar Iskar Gas Mai Kyau (a cikin adadin moles, n)

Dokar iskar gas mai kyau - matsaloli da mafita 2

Yawan moles 2 na iskar gas shine lita 44.8.

Yawan mole 1 na iskar gas shine lita 45.4 / 2 = lita 22.4.

Girman mole 1 na kowace iskar gas shine lita 22.4.

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3. Lita 4 na iskar oxygen yana da zafin jiki na 27°C da matsin lamba na 2 atm (1 atm = 10 5 Pa ) a cikin akwati mai rufewa. Matsayin iskar gas na duniya (R) = 8.314 J.mol e −1 .K −1 da lambar Avogadro ( N = 6.02 x 10 23 kwayoyin halitta/mol e . Menene kwayoyin iskar oxygen a cikin akwati?

An sani:

Yawan iskar gas (V) = lita 4 = 4 dm 3 = 4 x 10 -3 m 3

Zafin iskar gas (T) = 27 o C = 27 + 273 = 300 Kelvin

Matsi na iskar gas (P) = 2 atm = 2 x 10 5 Pa

Daidaiton iskar gas na duniya (R) = 8.314 J.mol e −1 .K −1

Lambar Avogadro (N A ) = 6.02 x 10 23

Ana so : Menene ƙwayoyin iskar oxygen a cikin akwati (N)

Magani:

Dokar iskar gas mai kyau - matsaloli da mafita 3

A cikin mole 1 na iskar oxygen, akwai ƙwayoyin oxygen 1.93 x 10 23 .

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4. Akwati mai ɗauke da iskar neon (Ne, atom mass = 20 u) a yanayin zafi da matsin lamba na yau da kullun (STP ) yana da girman 2 m3 . Kayyade nauyin iskar neon!

An sani:

Nauyin atomic na neon = gram 20/mol e = 0,02 kg/mol e

Yanayin zafi na yau da kullun (T) = 0 o C = 273 Kelvin

Matsakaicin matsin lamba (P) = 1 atm = 1.013 x 10 5 Pascal

Ƙara (V) = 2 m 3

Ana nema : nauyin (m) na iskar neon

Magani:

A yanayin zafi da matsin lamba na yau da kullun (STP), 1 mole na kowace iskar gas, hada da neon gas, suna da girma 22.4 litas = Dubu dm3 = 0.0448 m3.

Dokar iskar gas mai kyau - matsaloli da mafita 4

A cikin girman mita 2 , akwai mole 44.6 na iskar neon.

Matsakaicin nauyin atomic na iskar neon shine gram 20/mole.

Wannan yana nufin cewa a cikin mole 1 akwai gram 20 ko 0.02 kg na iskar neon. Domin a cikin mole 1 akwai 0.02 kg na iskar neon sannan a cikin mole 44.6 akwai mole 44.6 x 0.02 kg/mole = 0.892 kg = gram 892 na iskar neon.

Karin bayani

Dokar Gay-Lussac (ƙari mai ɗorewa) - matsaloli da mafita

1. Ina hulɗa da iskar gas da farko suna da matsin lamba P da zafin jiki T. Iskar tana shiga tsarin isochoric don haka matsin lamba na ƙarshe ya ninka matsin lamba na farko sau 4. Menene zafin jiki na ƙarshe na iskar gas?

An sani:

Matsi na farko (P1 ) = P

Matsi na ƙarshe (P2 ) = 4P

Zafin farko (T 1 ) = T

Ana so: Zafin ƙarshe ( T2 )

Magani:

Tsarin dokar Gay-Lussac :

Dokar Gay-Lussac (juzu'i mai ɗorewa) - matsaloli da mafita 1

Zafin ƙarshe ya ninka zafin farko sau 4.

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2. A cikin akwati mai rufewa, iskar gas mai kyau da farko tana da zafin jiki na 27 oC . Idan matsin lamba na ƙarshe ya ninka matsin lamba na farko sau biyu, menene zafin jiki na ƙarshe?

An sani:

Matsi na farko (P1 ) = P

Matsi na ƙarshe (P2 ) = 2P

Zafin farko (T 1 ) = 27 o C + 273 = 300 K

Ana so: Zafin ƙarshe ( T2 )

Magani:

Dokar Gay-Lussac (juzu'i mai ɗorewa) - matsaloli da mafita 2

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3. Ana cika tayar zuwa matsin lamba na atm 2 a 27°C. Bayan tuƙi, zafin da ke cikin tayar ya tashi zuwa 47°C. Menene matsin lamba a cikin tayar yanzu?

An sani:

Matsin yanayi = 1 atm = 1 x 10 5 Pa

Matsin ma'aunin farko = 2 atm = 2 x 10 5 Pa

Matsi na farko (P1 ) = 1 atm + 2 atm = 3 atm = 3 x 10 5 Pa

Zafin farko (T 1 ) = 27 o C + 273 = 300 K

Zafin ƙarshe (T 1 ) = 47 o C + 273 = 320 K

Ana so: Zafin ma'aunin ƙarshe

Magani:

Dokar Gay-Lussac (juzu'i mai ɗorewa) - matsaloli da mafita 3

Matsi na ƙarshe na ma'auni = matsin lamba na ƙarshe - matsin lamba na yanayi

Matsin lamba na ƙarshe = 3.2 atm – 1 atm

Matsin lamba na ƙarshe = 2.2 atm

Karin bayani

Dokar Charles (matsin lamba akai-akai) - matsaloli da mafita

1. A cikin akwati mai rufewa, iskar gas ɗin tana faɗaɗa ta yadda ƙarar ƙarshe ta zama sau 3 na ƙarar farko (V = ƙarar farko, T = zafin farko ). Menene zafin ƙarshe?

An sani:

Ƙarar farko (V1 ) = V

Ƙarar ƙarshe (V2 ) = 3V

Zafin farko (T 1 ) = T

Ana so: Zafin ƙarshe ( T2 )

Magani:

Tsarin dokar Charles :

Dokar Charles (matsin lamba akai-akai) - matsaloli da mafita 1

Zafin gas na ƙarshe ya ninka zafin farko sau uku.

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2. Ina hulɗa da iskar gas da farko suna da girma V da zafin jiki T. Idan iskar gas ta shiga tsarin isobaric ta yadda zafin jiki ya ninka zafin farko sau biyu to girman gas ɗin ƙarshe shine…

An sani:

Ƙarar farko (V1 ) = V

Zafin farko (T 1 ) = T

Zafin ƙarshe (T2 ) = 2T

Ana so: juzu'i na ƙarshe (V 2 )

Magani:

Dokar Charles (matsin lamba akai-akai) - matsaloli da mafita 2

Ƙarar gas ta ƙarshe ta zama sau biyu na ƙarar farko.

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3. A cikin akwati mai rufewa, iskar gas mai kyau da farko tana da girman lita 2 kuma zafin jiki na digiri 27 Celsius . Idan girman gas na ƙarshe ya zama lita 3 to zafin ƙarshe shine…

An sani:

Ƙarar farko (V1 ) = lita 2 = 2 dm 3 = 2 x 10 -3 m 3

Ƙarar ƙarshe (V2 ) = lita 3 = 3 dm 3 = 3 x 10 -3 m 3

Zafin farko (T 1 ) = 27 o C + 273 = 300 K

Ana so: Zafin ƙarshe ( T2 )

Magani:

Dokar Charles (matsin lamba akai-akai) - matsaloli da mafita 3

Zafin ƙarshe shine 177 o C ko 177 + 273 = 450 Kelvin.

Karin bayani

Dokar Boyle (zafin jiki mai ɗorewa) - matsaloli da mafita

1. Wasu iskar gas masu kyau da farko suna da matsin lamba P da girma V. Idan iskar gas ta shiga tsarin isothermal ta yadda matsin lamba na ƙarshe ya ninka matsin lamba na farko sau 4, to girman gas na ƙarshe zai kasance…

An sani:

Matsi na farko (P1 ) = P

Matsi na ƙarshe (P2 ) = 4P

Ƙarar farko (V1 ) = V

Ana so: Juzu'i na ƙarshe (V 2 )

Magani:

Tsarin dokar Boyle :

PV = mai dorewa

P 1 V 1 = P 2 V 2

(P)(V) = (4P)( V2 )

V = 4 V2

V 2 = V / 4 = ¼ V

Ƙarar iskar gas ta ƙarshe ta ninka sau ¼ sau na farko.

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2. A cikin akwati mai rufewa, iskar gas ɗin tana faɗaɗa ta yadda ƙarar ƙarshe ta zama sau biyu na ƙarar farko (V = ƙarar farko, P = matsin lamba na farko). Matsi na ƙarshe na iskar gas shine…

An sani:

Matsi na farko (P1 ) = P

Ƙarar farko (V1 ) = V

Ƙarar ƙarshe (V2 ) = 2V

Ana so : Matsi na ƙarshe ( P2 )

Magani:

P 1 V 1 = P 2 V 2

PV = P2 ( 2V )

P = P 2 (2)

P 2 = P / 2 = ½ P

Matsin iskar gas ya ninka sau ½ na matsin lamba na farko.

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3. A cikin akwati mai rufewa, iskar gas mai matsin lamba na atm 2 da kuma girman lita 1. Idan matsin lamba na iskar gas ya zama atm 4 to girman iskar gas zai zama...

An sani:

Matsi na farko (P1 ) = 2 atm = 2 x 10 5 Pa

Matsi na ƙarshe (P2 ) = 4 atm = 4 x 10 5 Pa

Ƙarar farko (V1 ) = lita 1 = 1 dm 3 = 1 x 10 -3 m 3

Ana so : Juzu'i na ƙarshe (V 2 )

Magani:

P 1 V 1 = P 2 V 2

( 2 x 10 5 )(1 x 10 -3 ) = (4 x 10 5 ) V 2

( 1)(1 x 10 -3 ) = (2) V 2

1 x 10 -3 = (2) V 2

V 2 = ½ x 10 -3

V 2 = 0.5 x 10 -3 m 3 = 0.5 dm 3 = lita 0.5 s

Karin bayani

Capacitors a jere da layi ɗaya - matsaloli da mafita

1. An haɗa na'urori uku, C 1 = 2 μF, C 2 = 4 μF, C 3 = 4 μF, a jere da layi ɗaya . A ƙayyade ƙarfin na'urar capacitor guda ɗaya wanda zai yi tasiri iri ɗaya da haɗin.

Capacitors a jere da layi ɗaya - matsaloli da mafita 1An sani:

Capacitor C 1 = 2 μ F

Capacitor C 2 = 4 μ F

Capacitor C 3 = 4 μ F

Ana so: Daidaitaccen ƙarfin (C)

Magani:

Ana haɗa na'urorin haɗin C2 da C3 a layi ɗaya. Daidaitaccen ƙarfin:

C P = C 2 + C 3 = 4 + 4 = 8 μF

Ana haɗa na'urorin haɗin C 1 da C p a jere. Ƙarfin da ya dace:

1/C = 1/C 1 + 1/ C P = 1/2 + 1/8 = 4/8 + 1/8 = 5/8

C = 8/5 μ F

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2. An haɗa na'urori masu ƙarfin lantarki guda biyar, C 1 = 2 μF, C 2 = 4 μF, C 3 = 6 μF, C 4 = 5 μF, C 5 = 10 μF, a jere da layi ɗaya. A ƙayyade ƙarfin na'urar ƙarfin lantarki guda ɗaya wadda za ta yi tasiri iri ɗaya da haɗin.

An sani:

Capacitors a jere da layi ɗaya - matsaloli da mafita 2Capacitor C1 = 2 μF

Capacitor C 2 = 4 μ F

Capacitor C 3 = 6 μ F

Capacitor C 4 = 5 μ F

Capacitor C 5 = 10 μ F

Ana so: Daidaitaccen ƙarfin (C)

Magani:

Ana haɗa na'urorin haɗin C2 da C3 a layi ɗaya. Daidaitaccen ƙarfin:

CP = C2 + C3​

CP = 4 + 6

CP = 10 μ F

An haɗa na'urorin haɗin gwiwa C1 , C P , C4 da C5 a jere. Ƙarfin da ya dace:

1/C = 1/C 1 + 1/ C P + 1/ C 4 + 1/ C 5

1 / C = 1/2 + 1/10 + 1/5 + 1/10

1/C = 5/10 + 1/10 + 2/10 + 1/10

1/C = 9/10

C = 10/9 μ F

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3 C1 = 3 μF, C2 = 4 μF da kuma C3 = 3 μF, an haɗa su a jere da layi ɗaya. Ƙayyade makamashin lantarki a kan circuits.

An sani:

Capacitor C1 = 3 μFCapacitors a jere da layi ɗaya - matsaloli da mafita 3

Capacitor C 2 = 4 μ F

Capacitor C 3 = 3 μ F

Ana so: Daidaitaccen ƙarfin (C)

Magani:

Ana haɗa na'urorin haɗin C2 da C3 a layi ɗaya. Daidaitaccen ƙarfin:

CP = C2 + C3​

CP = 4 + 3

CP = 7 μ F

Ana haɗa na'urorin haɗin C 1 da C P a jere. Ƙarfin da ya dace:

1/C = 1/C 1 + 1/ C P

1/C = 1/3 + 1/7

1/C = 7/21 + 3/21

1/C = 10/21

C = 21/10

C = 2.1 μ F

C = 2.1 x 10 -6 F

Makamashin wutar lantarki a kan da'irori:

E = ½ CV 2

E = ½ ( 2.1 x 10 -6 )(12 2 )

E = ½ (2.1 x 10 -6 )(144)

E = (2.1 x 10 -6 )(72)

E = 151.2 x 10 -6 Joule

E = 1.5 x 10 -4 Joule

Karin bayani

Capacitors a cikin jerin - matsaloli da mafita

1. An haɗa na'urori huɗu, C 1 = 2 μF, C 2 = 1 μF, C 3 = 3 μF, C 4 = 4 μF, a jere . A ƙayyade ƙarfin na'urar capacitor guda ɗaya wanda zai yi tasiri iri ɗaya da haɗin.

An sani:

Capacitor C 1 = 2 μ F

Capacitor C 2 = 1 μ F

Capacitor C 3 = 3 μ F

Capacitor C 3 = 4 μ F

Ana so: Daidaitaccen ƙarfin

Magani:

Daidaitaccen ƙarfin:

1/C = 1/C 1 + 1/C 2 + 1/C 3 + 1/C 4

1/C = 1/2 + 1/1 + 1/3 + 1/4

1/C = 6/12 + 12/12 + 4/12 + 3/12

1/C = 25/12

C = 12/25

C = 0.48

Daidaiton ƙarfin haɗin gaba ɗaya shine 0.48 μF.

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2. Kayyade cajin da ke kan capacitor C 1 idan bambancin da ke tsakanin P da Q shine Volts 12 …

Capacitors a jere - matsaloli da mafita 1An sani:

Capacitor C 1 = 1 0 μ F = 1 0 x 10 -6 F

Capacitor C 2 = 2 0 μ F = 2 0 x 10 -6 F

Bambancin da zai iya faruwa (V) = Volts 12

Ana so: cajin da ke kan capacitor C 1 (Q 1 )

Magani:

Daidaitaccen ƙarfin:

1/C = 1/C 1 + 1/ C 2

1/C = 1/10 + 1/20 = 2/20 + 1/20 = 3/20

C = 20/3 μ F = (20/3) x 10 -6 F

Cajin wutar lantarki akan capacitor mai daidai:

Q = (C)(V) = (20/3)(12)(10 -6 ) = 80 x 10 -6 C

Q = 80 μC

Ana haɗa na'urorin haɗin kai a jere don haka cajin lantarki akan na'urorin haɗin kai daidai = cajin lantarki akan na'urar haɗin kai C 1 = cajin lantarki akan na'urar haɗin kai C 2.

Cajin wutar lantarki da ke kan capacitor C1 shine 80 μC.

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3. An haɗa na'urori biyu, C 1 = 2 μF da C 2 = 4 μF, a jere. Ana cajin na'urorin. Bambancin da ke tsakanin na'urorin C 1 shine Volts 2. Cajin wutar lantarki akan na'urorin C 2 shine…

An sani:

C apa c itor C 1 = 2 μF = 2 x 10 -6 F

C apa c itor C 2 = 4 μF = 4 x 10 -6 F

Bambancin da ke iya faruwa akan capacitor C 1 (V 1 ) = 2 Volts

Ana nema : Cajin wutar lantarki akan capacitor C 2.

Magani:

Cajin wutar lantarki akan capacitor C 1 :

Q1 = C 1 V 1 = (2 x 10 -6 ) (2) = 4 x 10 -6 C

Q1 = 4 μC​

Ana haɗa na'urorin haɗin kai a jere don haka cajin lantarki akan na'urar haɗin kai C 1 = cajin lantarki akan na'urar haɗin kai C 2.

Cajin da ke kan capacitor C2 shine 4 μC.

Karin bayani

Capacitors a layi daya - matsaloli da mafita

1. An haɗa na'urori huɗu , C 1 = 2 μF, C 2 = 1 μF, C 3 = 3 μF, C 4 = 4 μF, a layi ɗaya. Kayyade ƙarfin na'urar capacitor guda ɗaya wanda zai yi tasiri iri ɗaya da haɗin.

An sani:

Capacitor C 1 = 2 μ F

Capacitor C 2 = 1 μ F

Capacitor C 3 = 3 μ F

Capacitor C 3 = 4 μ F

Ana so: Daidaitaccen ƙarfin

Magani:

Daidaitaccen ƙarfin:

C = C 1 + C 2 + C 3

C = 4 μF + 2 μF + 3 μF = 9 μF

Daidaiton ƙarfin haɗin gaba ɗaya shine 9 μF.

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2. Kayyade cajin da ke kan capacitor C 2 idan bambancin da ke tsakanin maki A da B shine Volts 9…

Capacitors a layi daya - matsaloli da mafita 1

An sani:

C apa c itor C 1 = 20 μF = 20 x 10 -6 F

C apa c itor C 2 = 30 μF = 30 x 10 -6 F

Bambancin da ke tsakanin maki A da B (V AB ) = Volts 9

Ana so : cajin da ke kan capacitor C 2 (Q 2 )

Magani:

Bambancin da zai iya faruwa:

Ana haɗa na'urorin haɗin kai a layi ɗaya don haka bambancin yuwuwar tsakanin A da B (V AB ) = bambancin yuwuwar akan na'urar haɗin kai C 1 (V 1 ) = bambancin yuwuwar akan na'urar haɗin kai C 2 (V 2 ) = Volts 9.

Cajin wutar lantarki akan capacitor C 2 :

Q2 = C 2 V 2 = (30 x 10 -6 ) (9) = 270 x 10 -6 C

Q2 = 270 μC​

Cajin wutar lantarki akan capacitor C 2 shine 27 0 μC.

[irp]

3. An haɗa na'urori uku , C 1 = 4 μF, C 2 = 2 μF, C 3 = 3 μF, a layi ɗaya. Ana caji na'urorin. Bambancin yuwuwar akan na'urar capacitor C 2 shine Volts 4. Ƙayyade

(a) Cajin wutar lantarki akan capacitors C 1 , C 2 da C 3

(b) Cajin wutar lantarki akan daidai capacitor na duka haɗin

An sani:

C apa c itor C 1 = 4 μF = 4 x 10 -6 F

C apacitor C 2 = 2 μ F = 2 x 10 -6 F

C apacitor C 3 = 3 μ F = 3 x 10 -6 F

Bambancin da zai iya faruwa akan capacitor C 2 (V 2 ) = Volts 4

Ana nema : Cajin wutar lantarki akan capacitor C 3 (Q 3 )

Magani:

(a) Cajin wutar lantarki akan capacitor C 3

Bambancin da ke tsakanin capacitor C3 da capacitor :

Ana haɗa na'urorin haɗin kai a layi ɗaya don haka bambancin da ke tsakanin na'urorin haɗin kai da na'urorin haɗin kai yana da yawa. C3 (V3) = bambancin da ke tsakanin capacitor da capacitor C2 (V2) = bambancin da ke tsakanin capacitor da capacitor C1 (V1) = bambancin yuwuwar akan capacitor daidai gwargwado (V) = Volt 4

Cajin wutar lantarki akan capacitor C 1 :

Q1 = C 1 V 1 = (4 x 10 -6 )(4) = 16 x 10 -6 C

Q1 = 16 μC​

Cajin wutar lantarki akan capacitor C 2 :

Q2 = C 2 V 2 = (2 x 10 -6 )(4) = 8 x 10 -6 C

Q2 = 8 μC​

Cajin wutar lantarki akan capacitor C 3 :

Q3 = C 3 V 3 = (3 x 10 -6 )(4) = 12 x 10 -6 C

Q3 = 12 μC​

(b) Cajin wutar lantarki akan capacitor daidai gwargwado

Q = Q1 + Q2 + Q3

Q = 16 μ C + 8 μ C + 12 μ C = 36 μC

Madadin mafita:

Daidaitaccen ƙarfin:

C = C 1 + C 2 + C 3

C = 4 μF + 2 μF + 3 μF = 9 μF

C = 9 x 10 -6 F

Bambancin da ke tsakanin capacitor mai daidai da wannan:

V 1 = V 2 = V 3 = V = Volt 4

Cajin wutar lantarki akan capacitor mai daidai:

Q = CV = (9 x 10 -6 )(4) = 36 x 10 -6 C

Q = 36 μC

Karin bayani