Adeegsiga sharciga koowaad ee thermodynamics-ka qaar ka mid ah hababka thermodynamic-ka (Isobaric Isothermal Isochoric)

30 Adeegsiga sharciga koowaad ee thermodynamics-ka ee qaar ka mid ah hababka thermodynamics-ka (Isobaric Isothermal Isochoric) 1. Jaantuska hoose wuxuu muujinayaa wareegga thermodynamic-ka ee uu gaas la kulmo. Shaqada uu gaasku ka qabto habka ABCD waa … La yaqaan: Cadaadiska 1 (P1) = 2 x 105 Pa Cadaadiska 2 (P2) = 4 x 105 … Akhriso wax dheeraad ah

Sharciga ugu horreeya ee thermodynamics - dhibaatooyinka iyo xalalka

30 Sharciga ugu horreeya ee thermodynamics - dhibaatooyinka iyo xalalka 1. 3000 J oo kuleyl ah ayaa lagu daraa nidaam, 2500 J oo shaqo ahna waxaa sameeya nidaamku. Waa maxay isbeddelka tamarta gudaha ee nidaamka? La yaqaan: Kulaylka (Q) = +3000 Joules Shaqada (W) = +2500 Joules La Rabay: isbeddelka … Akhriso wax dheeraad ah

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka

1. Waxaan gaasaska ku shubaa weel xiran marka hore waxay leeyihiin mugga V iyo cadaadiska P. Haddii cadaadiska kama dambaysta ahi yahay 4P oo mugga uu yahay mid joogto ah, waa maxay saamiga tamarta kicinta bilowga ah iyo tamarta kicinta kama dambaysta ah.

La yaqaan:

Cadaadiska bilowga ah (P 1 ) = P

Cadaadiska kama dambaysta ah (P 2 ) = 4P

Mugga bilowga ah (V1 ) = V

Mugga kama dambaysta ah (V2 ) = V

La Doonayo: Saamiga tamarta dhaqdhaqaaqa bilowga ah ilaa tamarta dhaqdhaqaaqa ugu dambeysa (K E 1 : K E 2 )

Xalka:

Xiriirka ka dhexeeya cadaadiska (P), mugga (V) iyo tamarta dhaqdhaqaaqa (KE) ee gaasaska ku habboon :

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 18

Saamiga tamarta dhaqdhaqaaqa bilowga ah iyo tamarta dhaqdhaqaaqa ugu dambeysa:

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 19

[irp]

2. Waa maxay celceliska tamarta dhaqdhaqaaqa tarjumaadda ee molecules ee gaaska ku habboon 57 o C.

La yaqaan:

Heerkulka gaaska (T) = 57 o C + 273 = 330 Kelvin

Joogtada Boltzmann (k) = 1.38 x 10 -23 Joule/Kelvin

La Doonayo: Celceliska tamarta dhaqdhaqaaqa tarjumaadda

Xalka:

Xiriirka ka dhexeeya tamarta dhaqdhaqaaqa (KE) iyo heerkulka gaaska (T):

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 3

Tamarta dhaqdhaqaaqa tarjumaadda celceliska:

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 4

[irp]

3. Gaas ku jira 27 o C weel xiran . Haddii tamarta dhaqdhaqaaqa gaaska ay korodho 2 jibaar tamarta dhaqdhaqaaqa bilowga ah, sidaas darteed heerkulka ugu dambeeya ee gaaska waa…

La yaqaan:

Heerkulka bilowga ah (T 1 ) = 27 o C + 273 = 300 K

Tamarta dhaqdhaqaaqa bilowga ah = KE

Tamarta dhaqdhaqaaqa kama dambaysta ah = 4 K E

La Rabay: Heerkulka kama dambaysta ah ( T2 )

Xalka:

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 5

[irp]

4. Gaaska ugu habboon wuxuu ku jiraa weel xiran, waa la kululeeyaa si xawaaraha ugu dambeeya ee walxaha gaaska uu u kordho 3 jeer celceliska xawaaraha bilowga ah. Haddii heerkulka gaaska ee bilowga ah uu yahay 27 o C, markaa heerkulka ugu dambeeya ee gaaska ugu habboon waa…

La yaqaan:

Heerkulka bilowga ah = 27 o C + 273 = 300 Kelvin

Xawaaraha bilowga ah = v

Xawaaraha kama dambaysta ah = 2v

La Doonayo : Heerkulka ugu dambeeya ee gaaska ku habboon

Xalka:

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 20

Celceliska xawaaraha ugu dambeeya = 2 x xawaaraha celceliska bilowga ah

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 7

[irp]

5. Saddex mool oo gaas ah ayaa ku jira meel mug ah oo 36 litir ah. Molekul kasta oo gaas ah wuxuu leeyahay tamar firfircoon oo ah 5 x 10 -21 Joules. Joogtada gaaska guud = 8.315 J/mole.K iyo Joogtada Boltzmann = 1.38 x 10 -23 J/K. Waa maxay cadaadiska gaaska ee ku jira weelka.

La yaqaan:

Tirada boollada (n) = 3 boollada

Mugga = 36 litir = 36 dm 3 = 36 x 10 -3 m 3

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

Tamarta dhaqdhaqaaqa (KE) = 5 x 10 –21 Joules

Joogtada gaaska guud (R) = 8.315 J/mole.K

La Raadinayo : Cadaadiska gaaska (P)

Xalka:

Xisaabi heerkulka adoo isticmaalaya isle'egta tamarta dhaqdhaqaaqa ee gaaska.

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 8

Xisaabi cadaadiska gaaska adoo isticmaalaya isla'egta th ee sharciga gaaska ee ku habboon (tirada moles, n):

PV = n RT

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

P (36 x 10 -3 ) = 6024.22

Aragtida kinetic ee gaasaska - dhibaatooyinka iyo xalalka 9

Cadaadiska gaaska waa 1.67 x 10 5 Pascal ama 1.67 jawi.

Akhri wax dheeraad ah

Sharciga gaaska ee ugu habboon - dhibaatooyinka iyo xalalka

1. Waxaan gaasaska ku shubaa weel xiran marka hore waxay leeyihiin mugga V iyo heerkulka T. Heerkulka kama dambaysta ah waa 5/4T cadaadiska kama dambaysta ahna waa 2P. Waa maxay mugga kama dambaysta ah ee gaaska?

La yaqaan:

Mugga bilowga ah (V1 ) = V

Heerkulka bilowga ah (T 1 ) = T

Heerkulka kama dambaysta ah (T 2 ) = 5/4 T

Cadaadiska bilowga ah (P 1 ) = P

Cadaadiska kama dambaysta ah (P 2 ) = 2P

La Doonayo: Mugga Ugu Dambeeya (V 2 )

Xalka:

Sharciga gaaska ee ku habboon - dhibaatooyinka iyo xalalka 1

[irp]

2. Go'aami mugga 2.00 mol ee gaasaska (gaaska ku habboon) ee STP. STP = Heerkulka iyo Cadaadiska Caadiga ah.

La yaqaan:

Moles of gas (n) = 2 moles of ice

Heerkulka caadiga ah (T) = 0 o C = 0 + 273 = 273 Kelvin

Cadaadiska caadiga ah (P) = 1 atm = 1.013 x 10 5 Pa

Joogtada gaaska caalamiga ah (R) = 8.315 Joule/mol e .Kelvin

La Doonayo : Mugga gaasaska (V)

Xalka:

Isle'egta sharciga gaaska ee ugu habboon (tirada burooyinka, n)

Sharciga gaaska ee ku habboon - dhibaatooyinka iyo xalalka 2

Mugga 2 moles oo gaas ah waa 44.8 litir.

Mugga 1 mole oo gaas ah waa 45.4 litir / 2 = 22.4 litir.

Mugga 1 mole ee gaas kasta waa 22.4 litir.

[irp]

3. 4 litir oo gaas ogsijiin ah ayaa leh heerkul ah 27°C iyo cadaadis ah 2 atm (1 atm = 10 5 Pa) oo ku jira weel xiran. Joogtada gaaska guud (R) = 8.314 J.mol e −1 .K −1 iyo lambarka Avogadro ( N = A ) = 6.02 x 10 23 molecules/mol e . Waa maxay molecules-ka gaasaska ogsijiinta ee ku jira weelka?

La yaqaan:

Mugga gaasaska (V) = 4 litir = 4 dm 3 = 4 x 10 -3 m 3

Heerkulka gaasaska (T) = 27 o C = 27 + 273 = 300 Kelvin

Cadaadiska gaasaska (P) = 2 atm = 2 x 10 5 Pa

Joogtada gaaska guud (R) = 8.314 J.mol e −1 .K −1

Lambarka Avogadro (N A ) = 6.02 x 10 23

La Doonayo : Waa maxay molecules-ka gaasaska oksijiinta ee ku jira weelka (N)

Xalka:

Sharciga gaaska ee ku habboon - dhibaatooyinka iyo xalalka 3

1 mole oo gaasaska oksijiinta ah, waxaa ku jira 1.93 x 10 23 molikula oo oksijiin ah.

[irp]

4. Weel ay ku jirto gaas neon ah (Ne, cufnaanta atomiga = 20 u) heerkulka iyo cadaadiska caadiga ah (STP ) wuxuu leeyahay mug dhan 2 m 3. Go'aami cufnaanta gaaska neon!

La yaqaan:

Cufka atamka ee neon = 20 garaam/mol e = 0,02 kg/mol e

Heerkulka caadiga ah (T) = 0 o C = 273 Kelvin

Cadaadiska caadiga ah (P) = 1 atm = 1.013 x 10 5 Pascal

Mugga (V) = 2 m 3

La Doonayo : cufka (m) ee gaaska neon

Xalka:

Heerkulka iyo cadaadiska caadiga ah (STP), 1 mole gaas kasta, ku dar gaaska neon, mug leh 22.4 litirs = 22.4 dm3 = 0.0448 m3.

Sharciga gaaska ee ku habboon - dhibaatooyinka iyo xalalka 4

Mugga 2 m 3 , waxaa ku jira 44.6 moles oo gaas neon ah.

Cufka atomiga ee gaaska neon waa 20 garaam/mole.

Taas macnaheedu waa in 1 mole uu ku jiro 20 garaam ama 0.02 kg oo gaas neon ah. Sababtoo ah 1 mole waxaa ku jira 0.02 kg oo gaas neon ah marka 44.6 mole waxaa ku jira 44.6 mole x 0.02 kg/mole = 0.892 kg = 892 garaam oo gaas neon ah.

Akhri wax dheeraad ah

Sharciga Gay-Lussac (mugga joogtada ah) - dhibaatooyinka iyo xalalka

1. Waxaan marka hore gaasaska la macaamilaa waxay leeyihiin cadaadis P iyo heerkul T. Gaasku wuxuu maraa habka isochoric si cadaadiska ugu dambeeya uu u noqdo 4 jeer cadaadiska bilowga ah. Waa maxay heerkulka ugu dambeeya ee gaaska?

La yaqaan:

Cadaadiska bilowga ah (P 1 ) = P

Cadaadiska kama dambaysta ah (P 2 ) = 4P

Heerkulka bilowga ah (T 1 ) = T

La Rabay: Heerkulka kama dambaysta ah ( T2 )

Xalka:

Qaacidada sharciga Gay-Lussac :

Sharciga Gay-Lussac (mugga joogtada ah) - dhibaatooyinka iyo xalalka 1

Heerkulka kama dambaysta ah wuxuu noqonayaa 4 jeer heerkulka bilowga ah.

[irp]

2. Weel xiran, gaasaska ugu habboon waxay marka hore leeyihiin heerkul ah 27 o C. Haddii cadaadiska kama dambaysta ahi uu noqdo 2 jibaar cadaadiska bilowga ah, waa maxay heerkulka kama dambaysta ah?

La yaqaan:

Cadaadiska bilowga ah (P 1 ) = P

Cadaadiska kama dambaysta ah (P 2 ) = 2P

Heerkulka bilowga ah (T 1 ) = 27 o C + 273 = 300 K

La Rabay: Heerkulka kama dambaysta ah ( T2 )

Xalka:

Sharciga Gay-Lussac (mugga joogtada ah) - dhibaatooyinka iyo xalalka 2

[irp]

3. Taayir waxaa lagu shubaa cadaadis cabbirkiisu yahay 2 atm oo ah 27°C. Ka dib marka la wado, heerkulka taayirku wuxuu gaaraa 47°C. Waa maxay cadaadiska taayirku hadda ku jiro?

La yaqaan:

Cadaadiska jawiga = 1 atm = 1 x 10 5 Pa

Cadaadiska cabbirka bilowga ah = 2 atm = 2 x 10 5 Pa

Cadaadiska ugu horreeya ee dhammaystiran (P1 ) = 1 atm + 2 atm = 3 atm = 3 x 10 5 Pa

Heerkulka bilowga ah (T 1 ) = 27 o C + 273 = 300 K

Heerkulka kama dambaysta ah (T 1 ) = 47 o C + 273 = 320 K

La Doonayo: Heerkulka cabbirka kama dambaysta ah

Xalka:

Sharciga Gay-Lussac (mugga joogtada ah) - dhibaatooyinka iyo xalalka 3

Cadaadiska cabbirka kama dambaysta ah = cadaadiska kama dambaysta ah ee kama dambaysta ah - cadaadiska jawiga

Cadaadiska cabbirka kama dambaysta ah = 3.2 atm – 1 atm

Cadaadiska cabbirka kama dambaysta ah = 2.2 atm

Akhri wax dheeraad ah

Sharciga Charles (cadaadis joogto ah) - dhibaatooyinka iyo xalalka

1. Weel xiran, gaasku wuu ballaartaa si mugga ugu dambeeya uu u noqdo 3 jeer mugga bilowga ah (V = mugga bilowga ah, T = heerkulka bilowga ah ). Waa maxay heerkulka kama dambaysta ah?

La yaqaan:

Mugga bilowga ah (V1 ) = V

Mugga kama dambaysta ah (V2 ) = 3V

Heerkulka bilowga ah (T 1 ) = T

La Rabay: Heerkulka kama dambaysta ah ( T2 )

Xalka:

Qaacidada sharciga Charles :

Sharciga Charles (cadaadis joogto ah) - dhibaatooyinka iyo xalalka 1

Heerkulka ugu dambeeya ee gaasaska wuxuu noqdaa 3 jeer heerkulka bilowga ah.

[irp]

2. Waxaan marka hore gaasaska ka shaqeeyaa waxay leeyihiin mugga V iyo heerkulka T. Haddii gaasku maro habka isobaric si heerkulku u noqdo 2 jeer heerkulka bilowga ah markaa mugga ugu dambeeya ee gaasaska waa…

La yaqaan:

Mugga bilowga ah (V1 ) = V

Heerkulka bilowga ah (T 1 ) = T

Heerkulka kama dambaysta ah (T 2 ) = 2T

La rabay: mugga ugu dambeeya ( V2 )

Xalka:

Sharciga Charles (cadaadis joogto ah) - dhibaatooyinka iyo xalalka 2

Mugga ugu dambeeya ee gaasaska wuxuu noqonayaa 2 jeer mugga bilowga ah.

[irp]

3. Weel xiran, gaasaska ugu habboon waxay marka hore leeyihiin mug dhan 2 litir iyo heerkul dhan 27 o C. Haddii mugga ugu dambeeya ee gaasaska uu noqdo 3 litir markaas heerkulka ugu dambeeya waa…

La yaqaan:

Mugga bilowga ah (V1 ) = 2 litir = 2 dm 3 = 2 x 10 -3 m 3

Mugga kama dambaysta ah (V2 ) = 3 litir = 3 dm 3 = 3 x 10 -3 m 3

Heerkulka bilowga ah (T 1 ) = 27 o C + 273 = 300 K

La Rabay: Heerkulka kama dambaysta ah ( T2 )

Xalka:

Sharciga Charles (cadaadis joogto ah) - dhibaatooyinka iyo xalalka 3

Heerkulka kama dambaysta ah waa 177 o C ama 177 + 273 = 450 Kelvin.

Akhri wax dheeraad ah

Sharciga Boyle (heerkulka joogtada ah) - dhibaatooyinka iyo xalalka

1. Qaar ka mid ah gaasaska ugu habboon waxay marka hore leeyihiin cadaadis P iyo mugga V. Haddii gaasku maro habka isothermal-ka si cadaadiska kama dambaysta ahi uu u noqdo 4 jeer cadaadiska bilowga ah, markaa mugga ugu dambeeya ee gaaska waa…

La yaqaan:

Cadaadiska bilowga ah (P 1 ) = P

Cadaadiska kama dambaysta ah (P 2 ) = 4P

Mugga bilowga ah (V1 ) = V

La Doonayo: Mugga Ugu Dambeeya (V 2 )

Xalka:

Qaacidada sharciga Boyle :

PV = joogto ah

P 1 V 1 = P 2 V 2

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

V = 4 V 2

V 2 = V / 4 = ¼ V

Mugga ugu dambeeya ee gaasaska waa ¼ jeer mugga bilowga ah.

[irp]

2. Weel xiran, gaasku wuu ballaaraa si mugga ugu dambeeya uu u noqdo 2 jeer mugga bilowga ah (V = mugga bilowga ah, P = cadaadiska bilowga ah). Cadaadiska ugu dambeeya ee gaasaska waa…

La yaqaan:

Cadaadiska bilowga ah (P 1 ) = P

Mugga bilowga ah (V1 ) = V

Mugga kama dambaysta ah (V2 ) = 2V

La Rabay : Cadaadis kama dambays ah (P 2 )

Xalka:

P 1 V 1 = P 2 V 2

PV = P 2 (2V)

P = P 2 (2)

P 2 = B / 2 = ½ B

Cadaadiska gaastu wuxuu noqdaa ½ jeer cadaadiska bilowga ah.

[irp]

3. Weel xiran, gaasas leh cadaadis dhan 2 atm iyo mug dhan 1 litir. Haddii cadaadiska gaasku noqdo 4 atm markaas mugga gaasku wuxuu noqdaa...

La yaqaan:

Cadaadiska bilowga ah (P1 ) = 2 atm = 2 x 10 5 Pa

Cadaadiska kama dambaysta ah (P2 ) = 4 atm = 4 x 10 5 Pa

Mugga bilowga ah (V1 ) = 1 litir = 1 dm 3 = 1 x 10 -3 m 3

La Doonayo : Mugga kama dambaysta ah ( V2 )

Xalka:

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 = 0.5 litir s

Akhri wax dheeraad ah

Kondenser-yada taxanaha ah iyo kuwa is barbar socda - dhibaatooyinka iyo xalalka

1. Saddex kaabsoodh, C 1 = 2 μF, C 2 = 4 μF, C 3 = 4 μF, ayaa isku xiran taxane iyo is barbar socda . Go'aami awoodda kaabsoodh hal ah oo yeelan doonta saameyn la mid ah isku-darka.

Kondenser-yada taxanaha ah iyo kuwa is barbar socda - dhibaatooyinka iyo xalalka 1La yaqaan:

Kondenser C 1 = 2 μ F

Kondenser C 2 = 4 μ F

Kondenser C 3 = 4 μ F

La Doonayo: Awoodda u dhiganta (C)

Xalka:

Kondenser-yada C 2 iyo C 3 waxay ku xiran yihiin is barbar socda. Kondenser-yada u dhigma:

C P = C 2 + C 3 = 4 + 4 = 8 μF

Kondenser-yada C 1 iyo C p waxay ku xiran yihiin taxane. Awoodda u dhiganta:

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

C = 8/5 μ F

[irp]

2. Shan kaabsoodh, C 1 = 2 μF, C 2 = 4 μF, C 3 = 6 μF, C 4 = 5 μF, C 5 = 10 μF, ayaa isku xiran taxane iyo is barbar socda. Go'aami awoodda kaabsoodh hal ah oo yeelan doonta saameyn la mid ah isku-darka.

La yaqaan:

Kondenser-yada taxanaha ah iyo kuwa is barbar socda - dhibaatooyinka iyo xalalka 2Kondenser C1 = 2 μF

Kondenser C 2 = 4 μ F

Kondenser C 3 = 6 μ F

Kondenser C 4 = 5 μ F

Kondenser C 5 = 10 μ F

La Doonayo: Awoodda u dhiganta (C)

Xalka:

Kondenser-yada C 2 iyo C 3 waxay ku xiran yihiin is barbar socda. Kondenser-yada u dhigma:

CP = C 2 + C 3

CP = 4 + 6

CP = 10 μ F

Kondenser-yada C 1 , C P , C 4 iyo C 5 waxay ku xiran yihiin taxane. Awoodda u dhiganta:

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

[irp]

3 C1 = 3 μF, C2 = 4 μF iyo C3 = 3 μF, waxay ku xiran yihiin taxane iyo is barbar socda. Go'aami tamarta korantada ku saabsan wareegyada.

La yaqaan:

Kondenser C1 = 3 μFKondenser-yada taxanaha ah iyo kuwa is barbar socda - dhibaatooyinka iyo xalalka 3

Kondenser C 2 = 4 μ F

Kondenser C 3 = 3 μ F

La Doonayo: Awoodda u dhiganta (C)

Xalka:

Kondenser-yada C 2 iyo C 3 waxay ku xiran yihiin is barbar socda. Kondenser-yada u dhigma:

CP = C 2 + C 3

CP = 4 + 3

CP = 7 μ F

Kondenser-yada C 1 iyo C P waxay ku xiran yihiin taxane. Awoodda u dhiganta:

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

Tamarta korontada ee wareegyada:

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

Akhri wax dheeraad ah

Kondenser-yada taxanaha ah - dhibaatooyinka iyo xalalka

1. Afar kaabsoodh, C 1 = 2 μF, C 2 = 1 μF, C 3 = 3 μF, C 4 = 4 μF, ayaa isku xiran taxane ahaan . Go'aami awoodda kaabsoodh keliya oo yeelan doonta saameyn la mid ah isku-darka.

La yaqaan:

Kondenser C 1 = 2 μ F

Kondenser C 2 = 1 μ F

Kondenser C 3 = 3 μ F

Kondenser C 3 = 4 μ F

La Doonayo: Awoodda u dhiganta

Xalka:

Awoodda u dhiganta:

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

Awoodda u dhiganta ee isku-darka oo dhan waa 0.48 μF.

[irp]

2. Go'aami kharashka ku jira capacitor C 1 haddii farqiga u dhexeeya P iyo Q uu yahay 12 Volts…

Kondenser-yada taxanaha ah - dhibaatooyinka iyo xalalka 1La yaqaan:

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

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

Farqiga suurtagalka ah (V) = 12 Volts

La Rabay: dallacaadda ku jirta capacitor C 1 (Q 1 )

Xalka:

Awoodda u dhiganta:

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

Kharash koronto oo ku jira kalkulator u dhigma:

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

Q = 80 μ C

Kondenser-yadu waxay ku xiran yihiin taxane si dallacaadda korontada ee ku jirta kondenser-ka u dhigma = dallacaadda korontada ee ku jirta kondenser-ka C 1 = dallacaadda korontada ee ku jirta kondenser-ka C 2.

Dakhli koronto oo ku jira capacitor C 1 waa 80 μC.

[irp]

3. Laba kaabsade, C 1 = 2 μF iyo C 2 = 4 μF, ayaa isku xiran taxane ahaan. Kaabsadeyaashu waa la dallacayaa. Farqiga suurtagalka ah ee kaabsade C 1 waa 2 Volts. Dakhli koronto oo ku jira kaabsade C 2 waa…

La yaqaan:

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

Farqiga suurtagalka ah ee ku saabsan capacitor C 1 (V 1 ) = 2 Volts

La doonayo : Koronto ku shaqeeya capacitor C 2.

Xalka:

Kharashka korontada ee capacitor C 1 :

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

Q 1 = 4 μ C

Kondenser-yadu waxay ku xiran yihiin taxane si dallacaadda korantada ee kondenser-ka C 1 = dallacaadda korontada ee kondenser-ka C 2.

Kharashka ku jira capacitor C 2 waa 4 μC.

Akhri wax dheeraad ah

Kondenser-yada is barbar socda - dhibaatooyinka iyo xalalka

1. Afar kaabsoodh , C 1 = 2 μF, C 2 = 1 μF, C 3 = 3 μF, C 4 = 4 μF, ayaa isku xiran. Go'aami awoodda kaabsoodh hal ah oo yeelan doona saameyn la mid ah isku-darka.

La yaqaan:

Kondenser C 1 = 2 μ F

Kondenser C 2 = 1 μ F

Kondenser C 3 = 3 μ F

Kondenser C 3 = 4 μ F

La Doonayo: Awoodda u dhiganta

Xalka:

Awoodda u dhiganta:

C = C 1 + C 2 + C 3

C = 4 μF + 2 μF + 3 μF = 9 μF

Awoodda u dhiganta ee isku-darka oo dhan waa 9 μF.

[irp]

2. Go'aami kharashka ku jira capacitor C 2 haddii farqiga u dhexeeya dhibcaha A iyo B uu yahay 9 Volts…

Kondenser-yada is barbar socda - dhibaatooyinka iyo xalalka 1

La yaqaan:

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

Farqiga suurtagalka ah ee u dhexeeya dhibcaha A iyo B (V AB ) = 9 Volts

La Rabay : dallacaadda ku jirta capacitor C 2 (Q 2 )

Xalka:

Farqiga suurtagalka ah:

Kondenser-yadu waxay ku xiran yihiin is barbar socda si farqiga u dhexeeya A iyo B (V AB ) = farqiga suurtagalka ah ee kondenser-ka C 1 (V 1 ) = farqiga suurtagalka ah ee kondenser-ka C 2 (V 2 ) = 9 Volts.

Kharashka korontada ee capacitor C 2 :

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

Q 2 = 270 μ C

Dakhli koronto oo ku jira capacitor C 2 waa 27 0 μC.

[irp]

3. Saddex kaabsoodh, C 1 = 4 μF, C 2 = 2 μF, C 3 = 3 μF, ayaa isku xiran si is barbar socda. Kaabsoodhyadu waa la dallacayaa. Farqiga suurtagalka ah ee ku saabsan kaabsoodh C 2 waa 4 Volts. Go'aami

(a) Dalac koronto oo ku jirta kapaderada C 1 , C 2 iyo C 3

(b) Kharash koronto oo ku jira kalkuletar u dhigma isku-darka oo dhan

La yaqaan:

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

Farqiga suurtagalka ah ee ku saabsan capacitor C 2 (V 2 ) = 4 Volts

La Rabay : Dakhli koronto oo ku jira capacitor C 3 (Q 3 )

Xalka:

(a) Dakhli koronto oo ku jira capacitor-ka C 3

Farqiga suurtagalka ah ee capacitor C 3 :

Kondenser-ku waxay ku xiran yihiin is barbar socda si farqiga u dhexeeya capacitor-ka uu u dhexeeyo C3 (V3) = farqiga u dhexeeya capacitor ee capacitor C2 (V2) = farqiga u dhexeeya capacitor ee capacitor C1 (V1) = farqiga u dhexeeya capacitor-ka u dhigma (V) = 4 Volt

Kharashka korontada ee capacitor C 1 :

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

Q 1 = 16 μ C

Kharashka korontada ee capacitor C 2 :

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

Q 2 = 8 μ C

Kharashka korontada ee capacitor C 3 :

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

Q 3 = 12 μ C

(b) Kharash koronto oo ku jira kalkulator u dhigma

Q = Q 1 + Q 2 + Q 3

Q = 16 μ C + 8 μ C + 12 μ C = 36 μ C

Xal kale:

Awoodda u dhiganta:

C = C 1 + C 2 + C 3

C = 4 μF + 2 μF + 3 μF = 9 μF

C = 9 x 10 -6 F

Farqiga suurtagalka ah ee ku yimaada capacitor-ka u dhigma:

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

Kharash koronto oo ku jira kalkulatorka u dhigma:

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

Q = 36 μ C

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