Misalai 5 na Tambayoyin Shari'a 2 Kirchhoff
Haka kuma a yi nazarin kayan da ke kan Dokar Kirchhoff da kuma Tattaunawa kan dokokin Kirchhoff
1. Da'irar lantarki ta ƙunshi resistors guda huɗu, kowannensu yana da R1= 6 Ohm, R2 = 6 Ohm, R3 = 9 Ohm da R4 = 3 Ohm da aka haɗa da E1= Volts 6, E2 = Volts 12 kamar yadda aka nuna a hoton da ke ƙasa. Wutar lantarki abin da ke gudana shine….
A. 1/5 A
B. 2/5 A
C. 3/5 A
D. 4/5 A
E. 1 A
Tattaunawa
An san cewa:
Resistor 1 (R)1) = 6 Ω
Resistor 2 (R)2) = 6 Ω
Resistor 3 (R)3) = 9 Ω
Resistor 4 (R)4) = 3 Ω
Tushen emf 1 (E)1) = Volt 6
Tushen emf 2 (E)2) = Volt 12
An tambaya: Wutar lantarki da ke gudana a cikin da'irar (I)
Amsa:
Resistor 1 (R)1) da kuma resistor 2 (R)2) an haɗa shi a layi ɗaya. Resistor ɗin maye gurbin shine:
1 / R12 = 1/R1 +1/R2 = 1/6 + 1/6 = 2/6
R12 = 6/2 = 3 Ω
Matakai don magance wannan matsalar:
Na farko, zaɓi alkiblar da kake so a yanzu. Zaka iya zaɓar akasin agogo ko agogo.
Na biyu, lokacin da wutar lantarki ta ratsa ta cikin juriya ko juriya (R) akwai raguwar yuwuwar hakan yasa V = IR ya zama mara kyau.
Na uku, idan wutar lantarki ta motsa daga ƙaramin ƙarfin lantarki zuwa babban ƙarfin lantarki (- zuwa +) to tushen emf (E) an yi masa alama mai kyau saboda akwai cajin makamashi a tushen emf. Idan wutar lantarki ta motsa daga babban ƙarfin lantarki zuwa ƙaramin ƙarfin lantarki (+ zuwa -) to tushen emf (E) an yi masa alama mara kyau saboda akwai fitar da makamashi a tushen emf.
Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.
- IR12 - E1 - IR3 - IR4 +E2 = 0
– 3I – 6 – 9I – 3I + 12 = 0
– 3I – 9I – 3I = 6 – 12
– 15I = – 6
I = -6 /-15
I = 2/5 Ampere
2. Kalli hoton tsarin resistor a ƙasa! Girman wutar lantarki ta hanyar R3 shine…
Tattaunawa
An sani:
R1 = 4 Ω
R2 = 4 Ω
R3 = 8 Ω
An tambaya: I3
Amsa:
Alkiblar wutar lantarki a cikin da'irori na lantarki di saman iri ɗaya ne da alkiblar hannun agogo, daga babban ƙarfin aiki zuwa ƙarancin ƙarfin aiki.
Na farko, ƙididdige resistor mai maye gurbin (R). Bayan haka, ƙididdige wutar lantarki ta amfani da Tsarin dokar Ohm : I = V / R
Resistor mai maye gurbin:
Lissafin maye gurbin resistor a cikin da'irar da ke sama.
Resistor R1 da kuma juriya R2 An shirya shi a layi ɗaya. Resistor ɗin maye gurbin shine:
1 / R12 = 1/R1 +1/R2 = 1/4 + 1/4 = 2/4
R12 = 4/2 = 2 Ω
Resistor R12 da kuma juriya R3 An shirya shi a jere. Resistor ɗin maye gurbin shine:
R=R12 + R3 = 2 + 8 = 10 Ω
Wutar lantarki tana fitowa daga batirin :
I = V / R = 40 / 10 = Amperes 4
Dokar Farko ta Kirchhoff ya bayyana cewa adadin wutar lantarki da ke shiga wani reshe iri ɗaya ne da adadin wutar lantarki da ke barin wannan reshe. Bisa ga dokar farko ta Kirchhoff, an kammala da cewa idan wutar lantarki da ke fitowa daga batirin ta kai Amperes 4, to wutar lantarki da ke ratsa ab daidai take da Amperes 4, haka nan wutar lantarki da ke ratsa bc ta kai Amperes 4.
3. Duba da'irar lantarki a cikin hoton da ke ƙasa. Lissafi I1, na2 ni kuma3.
Tattaunawa
Wutar lantarki tana motsawa daga babban ƙarfin lantarki zuwa ƙaramin ƙarfin lantarki; wutar lantarki tana motsawa daga tabbataccen ƙarfin lantarki zuwa ga mummunan ƙarfin lantarki. Saboda haka, an zaɓi alkiblar wutar lantarki ta I.1 ni kuma2 kamar yadda yake a hoton, yayin da alkiblar halin yanzu I3 an zaɓa a bazata.
Na farko, yi amfani da dokar Kirchhoff ta farko a wurin reshe
I3 = Ni1 + Ni2 .......... lissafi na 1
Na biyu, yi amfani da dokar Kirchhoff ta biyu don madauki 1 (madauki na hagu)
32 – 4 I1 – 10 I3 = 0
8 – Ni1 – 2,5 I3 = 0
.......... lissafi na 2
Na uku, yi amfani da dokar Kirchhoff ta biyu don madauki 2 (madauki na dama)
15 – 5 I2 – 10 I3 = 0
3 – Ni2 – 2 I3 = 0
.......... lissafi na 3
Sauya daidaito 2 da 3 zuwa lissafi 1:
Alamar korau tana nufin alkibla ta I2 akasin zaɓin shugabanci da ke sama.
4. Da'irar lantarki ta ƙunshi resistors guda huɗu, kowannensu yana da R1 = 12 Ohm, R2 = 12 Ohm, R3 = 3 Ohm da R4 = 6 Ohm da aka haɗa da E1 = Volts 6, E2 = Volts 12 kamar yadda aka nuna a hoton da ke ƙasa. Wutar lantarki da ke gudana ita ce….
A. 1/5 A
B. 2/5 A
C. 3/5 A
D. 4/5 A
E. 1 A
Tattaunawa
An san cewa:
Resistor 1 (R)1) = 12 Ω
Resistor 2 (R)2) = 12 Ω
Resistor 3 (R)3) = 3 Ω
Resistor 4 (R)4) = 6 Ω
Tushen emf 1 (E)1) = Volt 6
Tushen emf 2 (E)2) = Volt 12
Ana so: Ƙarfin wutar lantarki da ke gudana a cikin da'irar (I)
Amsa:
Wannan matsala tana da alaƙa da dokar Kirchhooff. Matakai da yadda za a magance wannan matsala:
Da farko, zaɓi alkiblar da kake so a halin yanzu. Za ka iya zaɓar akasin agogo ko agogo.
Na biyu, idan wutar lantarki ta ratsa ta cikin juriya ko resistor (R) akwai raguwar yuwuwar hakan yasa V = IR ya zama mara kyau.
Na uku, idan wutar lantarki ta motsa daga ƙarancin ƙarfin lantarki zuwa babban ƙarfin lantarki (- zuwa +) to tushen emf (E) ana nuna shi da kyau saboda akwai cikar kuzari a tushen emf. Idan wutar lantarki ta motsa daga babban ƙarfin lantarki zuwa ƙaramin ƙarfin lantarki (+ zuwa -) to tushen emf (E) ana nuna shi da mummunan rauni saboda akwai fitar da makamashi a tushen emf.
Resistor 1 (R)1) da kuma resistor 2 (R)2) an haɗa shi a layi ɗaya. Resistor ɗin maye gurbin shine:
1 / R12 = 1/R1 +1/R2 = 1/12 + 1/12 = 2/12
R12 = 12/2 = 6 Ω
Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.
- IR12 - E1 - IR3 - IR4 +E2 = 0
– 6 I – 6 – 3I – 6I + 12 = 0
– 6I – 3I – 6I = 6 -12
– 15I = – 6
I = -6/-15
I = 2/5 Ampere
Wutar lantarki da ke gudana a cikin da'irar ita ce 2/5 Ampere. Wutar lantarki mai kyau tana nuna cewa alkiblar wutar lantarki tana tafiya kamar yadda ake tsammani, a gefen agogo.
Amsar da ta dace ita ce B.
5. Kalli da'irar lantarki mai zuwa! Girman wutar lantarki (I) da ke gudana a cikin da'irar shine….
A. 0,1 A
B. 0,2 A
C. 0,5 A
D. 1,0 A
E. 5,0 A
Tattaunawa
An san cewa:
Resistor 1 (R)1) = 10 Ω
Resistor 2 (R)2) = 6 Ω
Resistor 3 (R)3) = 5 Ω
Resistor 4 (R)4) = 20 Ω
Tushen emf 1 (E)1) = Volt 8
Tushen emf 2 (E)2) = Volt 12
Ana so: Ƙarfin wutar lantarki da ke gudana a cikin da'irar (I)
Amsa:
Resistor 3 (R)3) da kuma resistor 4 (R)4) an haɗa shi a layi ɗaya. Resistor ɗin maye gurbin shine:
1 / R34 = 1/R3 +1/R4 = 1/5 + 1/20 = 4/20 + 1/20 = 5/20
R34 = 20/5 = 4 Ω
Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.
- IR1 - IR2 - E1 - IR34 +E2 = 0
– 10I – 6I – 8 – 4I + 12 = 0
– 10I – 6I – 4I = 8 – 12
– 20I = – 4
I = -4/-20
I = 1/5 Ampere
I = 0,2 Amperes
Wutar lantarki da ke gudana a cikin da'irar ita ce 1/5 Ampere. Wutar lantarki mai kyau tana nuna cewa alkiblar wutar lantarki tana tafiya kamar yadda ake tsammani, a gefen agogo.
Amsar da ta dace ita ce B.
