Su'aalaha tusaalaha kondenser - wareegyada taxanaha iyo kuwa is barbar socda

11 Tusaalooyin dhibaatooyinka capacitor-ka - wareegyada taxanaha iyo kuwa is barbar socda

1. Saddex capacitor ku xiran taxane-isbarbar socda sida sawirka hoose. Haddii C₁ = 2 μF, C₂ = 4 μF, C₃ = 4 μF, markaa awoodda beddelka waa…
Dood
Waa la ogyahay in:
Kondenser C₁ = 2 μF
Kondenser C₂ = 4 μF
Kondenser C₃ = 4 μF
La weydiiyay: Awoodda beddelka (C)
Jawaab:
Kondenser C₂ iyo C₃ isku xiran si is barbar socda. Awoodda beddelka waa:
C_P = C₂ +C₃ = 4 + 4 = 8 μF
Kondenser C₁ iyo C_P Ku xiran taxane. Awoodda beddelka waa:
1/C = 1/C₁ + 1/C_P = 1/2 + 1/8 = 4/8 + 1/8 = 5/8
C = 8/5 μF

μF = micro Farad (cutub awoodda korantada). 1 μF = 10^-6 Farad

2. Saddex kaabayaal ayaa isku xiran taxane ahaan iyo is barbar socda sida ku cad jaantuska hoose. Haddii C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF, C₄ = 5 μF iyo C₅ = 10 μF, markaa awoodda beddelka waa…
Dood
Waa la ogyahay in:
Kondenser C₁ = 2 μF
Kondenser C₂ = 4 μF
Kondenser C₃ = 6 μF
Kondenser C₄ = 5 μF
Kondenser C₅ = 10 μF
La weydiiyay: Awoodda beddelka (C)
Jawaab:

Kondenser C₂ iyo C₃ isku xiran si is barbar socda. Awoodda beddelka waa:

C_P = C₂ +C₃
C_P = 4 + 6
C_P = 10 μF

Kondenser C₁, C_P, C₄ iyo C₅ Ku xiran taxane. Awoodda beddelka waa:

1/C = 1/C₁ + 1/C_P + 1/C₄ + 1/C₅
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 μF

3 C₁ = 3 μF, C₂ = 4 μF iyo C₃ = 3 μF. Saddexda kapsul waxay ku xiran yihiin taxane iyo is barbar socda. Go'aami tamarta korantada ee wareegga!
Dood
Waa la ogyahay in:
Kondenser C₁ = 3 μF
Kondenser C₂ = 4 μF
Kondenser C₃ = 3 μF
La weydiiyay: Awoodda beddelka (C)
Jawaab:

Kondenser C₂ iyo C₃ isku xiran si is barbar socda. Awoodda beddelka waa:

C_P = C₂ +C₃
C_P = 4 + 3
C_P = 7 μF

Kondenser C₁ iyo C_P Ku xiran taxane. Awoodda beddelka waa:

1/C = 1/C₁ + 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 μF
C = 2,1 x 10^-6 F

Tamarta korontada ee wareegga:

E = ½ CV²
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

4. Fiiri jaantuska wareegga kaabsoodhka ee dhinaca ku yaal! Wadarta dallacaadda wareegga kaabsoodhka waa… (1 μF = 10^-6 F)

A. 1,50 μC

B. 2,75 μC

C. 3,50 μC

D. 4,50 μC

E. 4,75 μC

Dood

Waa la ogyahay in:

Kondenser 1 (C)₁= 3 μF

Kondenser 2 (C)₂= 3 μF

Kondenser 3 (C)₃= 3 μF

Kondenser 4 (C)₄= 2 μF

Kondenser 5 (C)₅= 3 μF

Danab (V) = 3 Volts

La weydiiyay: Wadarta dallacaadda wareegga kaabsootarka (Q)

Jawaab:

Kaaliyaha beddelka ah

Kondenser C₁, C₂ dC ah₃ Ku xiran taxane. Kaaliyaha beddelka ah:

1 / C₁₂₃ = 1/C₁ + 1/C₂ + 1/C₃ = 1/3 + 1/3 + 1/3 = 3/3

C₁₂₃ = 3/3 = 1 μF

Kondenser C₁₂₃ iyo C₄ isku xiran si is barbar socda. Kaaliyaha beddelka ah:

C₁₂₃₄ = C₁₂₃ +C₄ = 1 + 2 = 3 μF

Kondenser C₁₂₃₄ iyo C₅ Ku xiran taxane. Kaaliyaha beddelka ah:

1/C = 1/C₁₂₃₄ + 1/C₅ = 1/3 + 1/3 = 2/3

C = 3/2 μF

C = 3/2 x 10^-6 F

Wadarta kharashka korontada

Dakhli koronto oo ku jira kondensarka beddelka ah = mwadarta guud ee dallacaadda wareegga capacitor-ka :

Q = VC = (3 Volts)(3/2 x 10^-6 Farad) = 9/2 x 10^-6 Coulomb

Q = 9/2 microCoulomb = 9/2 μC

Q = 4,5 μC

Jawaabta saxda ah waa D.

5. Fiiri jaantuska wareegga capacitor-ka ee dhinaca! Haddii qiimaha capacitor kasta C uu yahay qiimaha capacitor kasta!₁ = C₂ = 2 μF, C₃ = C₄ = 1 μF iyo C₅ = 4 μF, markaa wadarta qiimaha dallacaadda ee wareegga waa… (1 μF = 10^-6 F)

A. 1/7 μC

B. 2/7 μC

C. 4/7 μC

D. 6/7 μC

E. 1 μC

Dood

Waa la ogyahay in:

Kondenser 1 (C)₁= 2 μF

Kondenser 2 (C)₂= 2 μF

Kondenser 3 (C)₃= 1 μF

Kondenser 4 (C)₄= 1 μF

Kondenser 5 (C)₅= 4 μF

Danab (V) = 1,5 Volts

La weydiiyay: Wadarta dallacaadda wareegga (Q)

Jawaab:

Kaaliyaha beddelka ah

Kondenser C₃ iyo C₄ isku xiran si is barbar socda. Kaaliyaha beddelka ah:

C₃₄ = C₃ +C₄ = 1 + 1 = 2 μF

Kondenser C₅, C₁, C₂ iyo C₃₄ Ku xiran taxane. Kaaliyaha beddelka ah:

1/C = 1/C₅ + 1/C₁ + 1/C₂ + 1/C₃₄

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

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

1/C = 7/4

C = 4/7 μF

C = 4/7 x 10^-6 F

Wadarta kharashka korontada

Dakhli koronto oo ku jira kondensarka beddelka ah = mawoodda guud ee wareegga :

Q = VC = (1,5 Volts)(4/7 x 10^-6 Farad) = 6/7 x 10^-6 Coulomb

Q = 6/7 microCoulomb

Q = 6/7 μC

Jawaabta saxda ah waa D.

6. Ka fiirso wareegga capacitor-ka soo socda! Wadarta dallacaadda wareegga waa…

A. 20 μC

B. 40 μC

C. 60 μC

D. 80 μC

E. 160kii μC

Dood

Waa la ogyahay in:

Kondenser 1 (C)₁= 3 μF

Kondenser 2 (C)₂= 3 μF

Kondenser 3 (C)₃= 4 μF

Kondenser 4 (C)₄= 4 μF

Kondenser 5 (C)₅= 8 μF

Danab (V) = 10 Volts

La weydiiyay: Wadarta dallacaadda wareegga (Q)

Jawaab:

Kaaliyaha beddelka ah

Kondenser C₁ iyo C₂ isku xiran si is barbar socda. Kaaliyaha beddelka ah:

C₁₂ = C₁ +C₂ = 3 + 3 = 6 μF

Kondenser C₃ dC ah₄ Ku xiran taxane. Kaaliyaha beddelka ah:

1 / C₃₄ = 1/C₃ + 1/C₄ = 1/4 + 1/4 = 2/4

C₃₄ = 4/2 = 2 μF

Kabasitor C₁₂, capacitor C₃₄ iyo capacitors C₅ isku xiran si is barbar socda. Kaaliyaha beddelka ah:

C = C₁₂ +C₃₄ +C₅ = 6 + 2 + 8 = 16 μF = 16 x 10^-6 Farad

Wadarta kharashka korontada

Dakhli koronto oo ku jira kondensarka beddelka ah = mwadarta guud ee dallacaadda wareegga capacitor-ka :

Q = VC = (10 Volts)(16 x 10^-6 Farad) = 160 x 10^-6 Coulomb

Q= 160 microCoulomb = 160 μC

Jawaabta saxda ah waa E.

7. Fiiri jaantuska wareegga kaabsootarka ee dhinaca ku yaal! Qadarka tamarta korantada ee wareegga waa… (1 µF = 10^-6 F)

A. 65 x 10^-6 Joule
B. 52 x 10^-6 Joule
C. 39 x 10^-6 Joule
D. 26 x 10^-6 Joule
E. 13 x 10^-6 Joule

Dood
Waa la garanayaa :
Kondenser 1 (C)₁) = 3 µF
Kondenser 2 (C)₂) = 1 µF
Kondenser 3 (C)₃) = 2 µF
Kondenser 4 (C)₄) = 6 µF
Kondenser 5 (C)₅) = 4 µF
Danabka korontada (V) = 5 volts
La weydiiyay Tamarta korontada ee wareegga
Jawab :

Qalabka beddelka ee wareegga capacitor-ka ee taxanaha-isku midka ah
Kondenser-yada 2 iyo 3 waxay ku xiran yihiin is barbar socda. Kondenser-ka beddelka ah:
C_A = C₂ +C₃ = 1 + 2 = 3 µF
Kondenser 1, kondenser A iyo kondenser 4 ayaa si taxane ah isugu xiran. Kondenser-ka beddelka ah:
1 / C_B = 1/C₁ + 1/C_A + 1/C₄ = 1/3 + 1/3 + 1/6 = 2/6 + 2/6 + 1/6 = 5/6
C_B = 6/5 µF
Kondenser-ka B iyo kondenser-ka 5 ayaa isku xiran si is barbar socda. Kondenser-ka beddelka ah:
C = C_B +C₅ = 6/5 + 4 = 6/5 + 16/4 = 24/20 + 80/20 = 104/20 = 5,2 µF
C = 5,2 x 10^-6 Farad
Tamarta korontada ee wareegga
E = ½ CV² = ½ (5,2 x 10)^-6(5)²) = (2,6 x 10^-6) (25)
E = 65 x 10^-6 Joule
Jawaabta saxda ah waa A.

8. Fiiri jaantuska wareegga kaabsoodhka ee soo socda! Qadarka tamarta korontada ee wareegga kaabsoodhka ee isku dhafan waa… (1 µF = 10^-6 F)

A. 0,6 x 10^-3 Joule
B. 1,2 x 10^-3 Joule
C. 1,8 x 10^-3 Joule
D. 2,4 x 10^-3 Joule
E. 3,0 x 10^-3 Joule

Dood
Waa la garanayaa :
Kondenser 1 (C)₁) = 4 µF
Kondenser 2 (C)₂) = 6 µF
Kondenser 3 (C)₃) = 12 µF
Kondenser 4 (C)₄) = 2 µF
Kondenser 5 (C)₅) = 2 µF
Danabka korontada (V) = 40 volts
La weydiiyay Tamarta korontada ee wareegga
Jawab :
Qalabka beddelka ee wareegga capacitor-ka ee taxanaha-isku midka ah
Kondenser 1, kondenser 2 iyo kondenser 3 ayaa si taxane ah isugu xiran. Kondenser beddelka ah:
1 / C_A = 1/C₁ + 1/C₂ + 1/C₃ = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12
C_A = 12/6 = 2 µF
Kondenser-yada 4 iyo 5 waxay ku xiran yihiin taxane. Kondenser-ka beddelka ah:
1 / C_B = 1/C₄ + 1/C₅ = 1/2 + 1/2 = 2/2
C_B = 2/2 = 1 µF
Kondenser-ka A iyo kondenser-ka B ayaa isku xiran si is barbar socda. Kondenser-ka beddelka ah:
C = C_A +C_B = 2 + 1 = 3 µF
C = 3 x 10^-6 Farad
Tamarta korontada ee wareegga
E = ½ CV² = ½ (3 x 10)^-6(40)²) = (1,5 x 10^-6) (1600)
E = 2400 x 10^-6 = 2,4x10^-3 Joule
Jawaabta saxda ah waa D.

9. Eeg jaantuska wareegga awoodda ee soo socda. Tamarta ku kaydsan wareegga korantada ee kor ku xusan waa…
A. 576 Joule
B. 288 Joule
C. 144 Joule
D. 72 Joule
E. 48 Joule
Dood
Waa la garanayaa :
Kondenser 1 (C)₁) = 4 F
Kondenser 2 (C)₂) = 4 F
Kondenser 3 (C)₃) = 4 F
Kondenser 4 (C)₄) = 4 F
Kondenser 5 (C)₅) = 2 F
Danabka korontada (V) = 12 volts
La weydiiyay Tamarta korontada ee wareegga
Jawab :
Jawab :
Qalabka beddelka ee wareegga capacitor-ka ee taxanaha-isku midka ah
Kondenser-ka 1, kondenser-ka 2 iyo kondenser-ka 3 ayaa isku xiran. Kondenser-ka beddelka ah:
C_A = C₁ +C₂ +C₃ = 4 + 4 + 4 = 12 F
Kondenser-yada 4 iyo 5 waxay ku xiran yihiin is barbar socda. Kondenser-ka beddelka ah:
C_B = C₄ +C₅ = 4 + 2 = 6 F
Kondenser-ka A iyo kondenser-ka B waxay ku xiran yihiin taxane. Kondenser-ka beddelka ah:
1/C = 1/C_A + 1/C_B = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
C = 12/3 = 4 Farad
Tamarta korontada ee wareegga
E = ½ CV² = ½ (4)(12)²) = (2)(144)
E = 288 Joule
Jawaabta saxda ah waa B.

10. Fiiri wareegga hoose! Cabbirka dallacaadda ee capacitor C₅ waa….

A. 36 Coulombs
B. 24 Coulombs
C. 12 Coulombs
D. 6 Coulombs
E. 4 Coulombs

Dood
Waa la ogyahay in:
Kondenser 1 (C)₁) = 6 F
Kondenser 2 (C)₂) = 6 F
Kondenser 3 (C)₃) = 3 F
Kondenser 4 (C)₄) = 12 F
Kondenser 5 (C)₅) = 6 F
Danab (V) = 12 Volts
Su'aal: Kharashkii ku jiray capacitor-ka (C)₅)
Jawaab:

Kabasitor
Kondenser C₂ iyo capacitor C₃ Ku xiran taxane. Kaaliyaha beddelka ah:
1 / C_A = 1/C₂ + 1/C₃ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6
C_A = 6/3 = 2 Farad
Kondenser C₄ iyo capacitor C₅ Ku xiran taxane. Kaaliyaha beddelka ah:
1 / C_B = 1/C₄ + 1/C₅ = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
C_B = 12/3 = 4 Farad

Kondenser C_A iyo capacitor C_B isku xiran si is barbar socda. Kaaliyaha beddelka ah:
C_C = C_A +C_B = 2 + 4 = 6 Farad
Kondenser C₁ iyo capacitor C_C ku xiran taxane:
1/C = 1/C₁ + 1/C_C = 1/6 F + 1/6 F = 2/6
C = 6/2 = 3 Farad

Kharashka Korontada
Kharashka korontada ee ku jira capacitor-ka beddelka ah ee C:
q = VC = (12 Volts)(3 Farad) = 36 Coulombs
Kondenser 1 iyo kondenser CC waxay ku xiran yihiin taxane si lacagta korontada ee kondenser-ka beddelka ah ee C = lacagta korontada ee kondenser-ka C₁ = dallac koronto oo ku yaal capacitor C_C = 36 Coulombs.
Kondenser C_C wuxuu leeyahay awood dhan 6 Farad wuxuuna leeyahay awood dhan 36 Coulombs. Farqiga suurtagalka ah ee ku dhex jira capacitor C_C waa: V = q/C_C = 36 Coulombs / 6 Farad = 6 Volts
Kondenser C_C waa kapastarka beddelka u ah kapastarka C_A iyo capacitor C_B kuwaas oo isku xiran si is barbar socda. Sidaa darteed, farqiga suurtagalka ah ee ku dhex jira capacitor C_C (V_C) = farqiga suurtagalka ah ee ku dhex jira capacitor C_A (V_A) = farqiga suurtagalka ah ee ku dhex jira capacitor C_B (V_B) = 6 Volts.

Kharashka korontada ee capacitor C_B :
q_B =V_B C_B = (6 Volts)(4 Farad) = 24 Coulombs
Kaaliyaha CB waa kaaliyaha beddelka u ah kaaliyaha C₄ iyo C₅ kuwaas oo ku xiran taxane. Sababtoo ah waxay ku xiran yihiin taxane, dallacaadda korantada ee ku taal capacitor C_B (q_B) = koronto ku shaqeeya capacitor C₄ (q₄) = koronto ku shaqeeya capacitor C₅ (q₅) = 24 Coulombs.
Jawaabta saxda ah waa B.

11. Shan kaabsoodh oo isku mid ah oo midkiiba ah 20 µF ayaa loo habeeyay sida ku cad jaantuska, iyagoo ku xiran ilo danab oo 6 volt ah. Wadarta dallacaadda lagu kaydiyay kaabsoodhayaasha waa C₅ waa….

A. 12 µC
B. 24 µC
C. 60 µC
D. 120 µC
E. 600 µC

Dood
Waa la ogyahay in:
Kondenser C₁ = C₂ = C₃ = C₄ = C₅ = 20 µF
Farqiga suurtagalka ah (V) = 6 Volts
Su'aal: Wadarta guud ee kharashka lagu kaydiyay capacitor C5
Jawaab:
Kabasitor
Kondenser C₁ iyo capacitor C₂ Ku xiran taxane. Kaaliyaha beddelka ah:
1 / C_A = 1/C₁ + 1/C₂ = 1/20 + 1/20 = 2/20
C_A = 20/2 = 10 µF
Kondenser C₃ iyo capacitor C₄ Ku xiran taxane. Kaaliyaha beddelka ah:
1 / C_B = 1/C₃ + 1/C₄ = 1/20 + 1/20 = 2/20
C_B = 20/2 = 10 µF
Kondenser C_A iyo capacitor C_B isku xiran si is barbar socda. Kaaliyaha beddelka ah:
C_C = C_A +C_B = 10 + 10 = 20 µF
Kondenser C_C iyo capacitor C₅ ku xiran taxane:
1/C = 1/C_C + 1/C₅ = 1/20 + 1/20 = 2/20
C = 20/2 = 10 µF

Kharashka Korontada
Kharashka korontada ee ku jira capacitor-ka beddelka ah ee C:
q = VC = (6 Volts)(10 x 10^-6 Farad) = 60 x 10^-6 Coulomb = 60 µC
Kondenser C_C iyo capacitor 5 waxaa lagu xiraa taxane si lacagta korontada ee capacitor-ka beddelka ah ee C = lacagta korontada ee capacitor-ka C_C = dallac koronto oo ku yaal capacitor C₅ = 60 µC.
Jawaabta saxda ah waa C.

Isha su'aasha:

Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda