Misali na Da'irar Wutar Lantarki Kai Tsaye

Misalan 9 na Tambayoyin Da'irar Wutar Lantarki Kai Tsaye

1. Kalli hoton tsarin resistor ɗin da ke ƙasa! ƙarfin yanzu ta hanyar R₁ shine…

A. 2,0 Ampere
B. 2,5 Ampere
C. 4,0 Amperes
D. 4,5 Amperes
E. 5,0 Ampere

Tattaunawa
An sani :
Resistor 1 (R)₁) = 4 Ω
Resistor 2 (R)₂) = 4 Ω
Resistor 3 (R)₃) = 8 Ω
Wutar lantarki (V) = Volts 40
An tambaya : Na yanzu zuwa R₁
Jawab :

Wutar lantarki tana gudana daga babban ƙarfin lantarki zuwa ƙaramin ƙarfin lantarki. Alkiblar wutar lantarki a cikin da'irar da ke sama iri ɗaya ce da alkiblar hannun agogo.

Wutar lantarki tana fitowa daga batirin
Da farko, ƙididdige resistor ɗin maye gurbin (R). Bayan haka, ƙididdige wutar lantarki ta amfani da dabarar Dokar Ohm :
V = IR ko I = V / R
Bayanin dabara: V = ƙarfin lantarki, I = halin yanzu, R = resistor na maye gurbin
Resistor mai maye gurbin:
Hitung ɗan adawa maye gurbin 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
Wutar lantarki da ke fitowa daga batirin shine 4 amperes.

Ƙarfin wutar lantarki V_ab kuma V_bc
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.
Ƙarfin wutar lantarki V_ab :
V_ab = Ni_ab R_ab = (4)(2) = Volts 8
Ƙarfin wutar lantarki V_bc :
V_bc = Ni_bc R_bc = (4)(8) = Volts 32
An shirya da'irar da ke sama a jere don haka jimlar ƙarfin lantarki shine V = V_ab +V_bc = Volts 8 + Volts 32 = Volts 40.
Wutar lantarki da ke gudana ta cikin R₁ = 4 Ω
I₁ = V_ab /r₁ = Volts 8 / Ohms 4 = Amps 2
I₂ = V_ab /r₂ = Volts 8 / Ohms 4 = Amps 2
Wutar lantarki da ke fitowa daga batirin shine Amperes 4. Idan ya isa wurin A, wutar lantarkin za ta rabu gida biyu, tare da Amperes 2 da ke gudana ta cikin resistor R.₁ kuma wutar lantarki ta 2 amperes tana gudana ta cikin resistor R₂. 2 Amperes + 2 Amperes = Amperes 4. Wannan ya yi daidai da dokar farko ta Kirchhoff.
Amsar da ta dace ita ce A.

2. Yi la'akari da da'irar lantarki mai zuwa. Girman wutar lantarki da ke gudana ta cikin resistor 4 Ω shine…
A. 1,0 A
B. 1,2 A
C. 1,6 A
D. 2,4 A
E. 3,2 A
Tattaunawa
An sani :
Resistor 1 (R)₁) = 6 Ω
Resistor 2 (R)₂) = 4 Ω
Resistor 3 (R)₃) = 1,6 Ω
Wutar lantarki (V) = Volts 16
An tambaya Wutar lantarki tana gudana ta cikin resistor 4 Ω
Jawab :
Wutar lantarki tana gudana daga babban ƙarfin lantarki zuwa ƙaramin ƙarfin lantarki. Alkiblar wutar lantarki a cikin da'irar da ke sama iri ɗaya ce da alkiblar hannun agogo.
Wutar lantarki tana fitowa daga batirin
Resistor mai maye gurbin:
Resistor R₁ da kuma juriya R₂ An shirya shi a layi ɗaya. Resistor ɗin maye gurbin shine:
1 / R₁₂ = 1/R₁ +1/R₂ = 1/6 + 1/4 = 2/12 + 3/12 = 5/12
R₁₂ = 12/5 = 2,4 Ω
Resistor R₁₂ da kuma juriya R₃ An shirya shi a jere. Resistor ɗin maye gurbin shine:
R= R₁₂ + R₃ = 2,4 + 1,6 = 4 Ω
Wutar lantarki tana fitowa daga batirin :
I = V / R = 16 / 4 = Amperes 4
Wutar lantarki da ke fitowa daga batirin shine 4 amperes.

Ƙarfin wutar lantarki V_ab kuma V_bc
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.
Ƙarfin wutar lantarki V_ab :
V_ab = Ni_ab R_ab = (4)(2,4) = Volts 9,6
Ƙarfin wutar lantarki V_bc :
V_bc = Ni_bc R_bc = (4)(1,6) = Volts 6,4
An shirya da'irar da ke sama a jere don haka jimlar ƙarfin lantarki shine V = V_ab +V_bc = Volts 9,6 + Volts 6,4 = Volts 16.
Wutar lantarki da ke gudana ta cikin R₂ = 4 Ω
I₁ = V_ab /r₁ = Volts 9,6 / Ohms 6 = Amps 1,6
I₂ = V_ab /r₂ = Volts 9,6 / Ohms 4 = Amps 2,4
Wutar lantarki da ke fitowa daga batirin shine Amperes 4. Idan ya isa wurin A, wutar lantarkin za ta rabu gida biyu, tare da Amperes 1,6 da ke gudana ta cikin resistor R.₁ kuma wutar lantarki ta 2,4 amperes tana gudana ta cikin resistor R₂. 1,6 Amperes + 2,4 Amperes = Amperes 4. Wannan ya yi daidai da dokar farko ta Kirchhoff.
Amsar da ta dace ita ce D.

3. Kula da da'irar lantarki mai zuwa!

Idan aka maye gurbin resistor 5 Ohm da ke cikin da'irar da resistor 7 Ohm, to rabon jimlar wutar lantarki da ke gudana a cikin da'irar kafin da kuma bayan maye gurbin shine...

Tattaunawa

An haɗa resistor mai 6 Ohm da resistor mai 6 Ohm a layi ɗaya. Resistor masu daidai da juna sune:

1 / R_AB = 1/6 + 1/6 = 2/6

R_AB = 6/2 = 3 Ohms

A yi amfani da dokar Kirchhoff ta biyu a cikin da'irar, idan juriya = 5 Ohm:

Zaɓi alkiblar da ake bi a yanzu a hannun agogo.

12 – 5I – 4 – 3I = 0

12 – 4 – 5I – 3I = 0

8 – 8I = 0

8 = 8I

I = 8/8

I = 1 Amperes

An yi mini alama mai kyau, ma'ana alkiblar wutar lantarki ta dogara ne akan zaɓin da aka yi, wato a gefen agogo.

A yi amfani da dokar Kirchhoff ta biyu a cikin da'irar, idan juriya = 7 Ohm:

Zaɓi alkiblar da ake bi a yanzu a hannun agogo.

12 – 7I – 4 – 3I = 0

12 – 4 – 7I – 3I = 0

8 – 10I = 0

8 = 10I

I = 8/10

I = 0,8 Amperes

An yi mini alama mai kyau, ma'ana alkiblar wutar lantarki ta dogara ne akan zaɓin da aka yi, wato a gefen agogo.

Kwatanta wutar lantarki kafin da kuma bayanta:

1: 0,8

10: 8

5: 4

4. Kula da da'irar lantarki mai zuwa!

Girman ƙarfin da ke kan resistor 6 Ω shine…

A. Watts 1,5

B. Watts 3,0

C. Watts 6,0

D. Watts 9,0

E. Watts 18,0

Tattaunawa

Tsarin wutar lantarki:

P=I² R

Kafin a ƙididdige wutar lantarki, a fara ƙididdige wutar lantarki da ke ratsawa ta juriya ta 6. Ω yana amfani da dokar Kirchhoff.

An zaɓi alkiblar da ake amfani da ita a yanzu kamar yadda aka nuna a hoton da ke gefe.

Yi amfani da dokar Kirchhoff ta farko:

I₁ + Ni₂ = Ni₃ .......... Daidaito ta 1

Aiwatar da Dokar Kirchhoff ta II ga I₁ :

6 – 12 I₁ – 6 I₃ = 0

-12 I₁ = 6 I₃ - 6

I₁ = (6 I₃ – 6) / -12 ……… Daidaito ta 2

Aiwatar da Dokar Kirchhoff ta II ga I₂ :

6 – 12 I₂ – 6 I₃ = 0

-12 I₂ = 6 I₃ - 6

I₂ = (6 I₃ – 6) / -12 ………. Daidaito ta 3

Madadin lissafi na 2 dan lissafi na 3 ke lissafi na 1:

I₃ Alamar tabbatacce tana nufin alkibla ta I₃ bisa ga alkiblar da aka zaɓa, kamar yadda yake a hoton da ke sama.

Amfani lissafi na 2 don lissafta I₁:

I₁ = (6 I₃ – 6) / -12

I₁ = (6.0,5 – 6) / -12

I₁ = (3 – 6) / -12

I₁ = -3 / -12

I₁ = 1/4 A

I₁ Alamar tabbatacce tana nufin alkibla ta I₁ bisa ga alkiblar da aka zaɓa, kamar yadda yake a hoton da ke sama.

Amfani lissafi na 3 don lissafta I₂:

I₂ = (6 I₃ – 6) / -12

I₂ = (6.0,5 – 6) / -12

I₂ = (3 – 6) / -12

I₂ = -3 / -12

I₂ = 1/4 A

I₂ Alamar tabbatacce tana nufin alkibla ta I₁ bisa ga alkiblar da aka zaɓa, kamar yadda yake a hoton da ke sama.

Yi amfani da dokar farko ta Kirchhoff don tabbatar da daidaiton sakamakon da aka samu:

I₁ + Ni₂ = Ni₃

1/4 + 1/4 = 1/2

Ƙarfin wutar lantarki da ke ratsa juriyar 6 Ω ni ne₃ = 1/2 Ampere.

Ƙarfin lantarki akan juriya 6 Ω shine P=I² R = (1/2)² (6) = 1/4 (6) = Watts 1,5

Amsar da ta dace ita ce A.

5. A cikin zane-zanen da'irar lantarki da ke gefe, girman bambancin yuwuwar wutar lantarki a kan resistor R shine₃ shine…

A. 0,5 V

B. 0,6 V

C. 0,9 V

D. 1,0 V

E. 1,3 V

Tattaunawa

An san cewa:

Shamaki na 1 (R)₁) = 2 Ω

Shamaki na 2 (R)₂) = 4 Ω

Shamaki na 3 (R)₃) = 3 Ω

Tushen emf 1 (E)₁) = Volt 6

Tushen emf 2 (E)₂) = Volt 9

An tambaya: Bambancin ƙarfin lantarki a tsakanin resistor R₃

Amsa:

Lissafa wutar lantarki (I) da ke gudana ta cikin resistor R₃

Da farko a ƙididdige wutar lantarki da ke gudana a cikin da'irar:

Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.

E₁ - IR₁ - E₂ - IR₂ - IR₃ = 0

6 - 2I - 9 - 4I - 3I = 0

6 - 9 - 2I - 4I - 3I = 0

-3 – 9I = 0

-3 = 9I

I = -3/9

I = -1/3

Wutar lantarki da ke gudana a cikin da'irar ita ce 1/3 Ampere. Wutar lantarki mai alamar korau yana nufin alkiblar wutar lantarki ita ce ba daidai da a cikin kusan amma akasin agogon.

An shirya da'irar a jere ta yadda wutar lantarki ke gudana a cikin da'irar = wutar lantarki da ke gudana a cikin juriyar R₃ = 1/3 Ampere.

Lissafa bambancin yuwuwar (V) a fadin resistor R₃

V = IR₃ = (1/3)(3) = 1 Volt

Amsar da ta dace ita ce D.

6. Kalli hoton hanyar sadarwa a ƙasa!

Don haka ƙarfin wutar lantarki tsakanin maki C da D daidai yake da volts 4, ƙimar R shine...

A. 1 Ohm

B. 2 Ohm

C. 4 Ohm

D. 8 Ohm

E. 10 Ohm

Tattaunawa

An san cewa:

Shamaki na 1 (R)₁) = 2 Ω

Shamaki na 2 (R)₂) = 2 Ω

Shamaki na 3 (R)₃) = R

Tushen emf 1 (E)₁) = Volt 8

Tushen emf 2 (E)₂) = Volt 4

Bambancin ƙarfin lantarki tsakanin maki C da D (V)_CD) = Volt 4

An tambaya: Ƙimar juriya R

Amsa:

Lissafa darajar juriya R

V_CD = IR

4 = IR

R = 4 / I

Ci gaba…..

Lissafa wutar lantarki (I) da ke gudana ta cikin resistor R

Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.

- E₁ – 2I + E₂ – 2I – IR = 0

– 8 – 2I + 4 – 2I – I (4/I) = 0

– 8 – 2I + 4 – 2I – 4 = 0

– 8 + 4 – 4 – 2I – 2I = 0

– 8 – 4I = 0

– 8 = 4I

I = -8 / 4

I = -2 Amperes

Wutar lantarki da ke gudana a cikin da'irar ita ce 2 Amperes. Wutar lantarki tana da rashin daidaito, ma'ana alkiblar wutar lantarki ita ce ba daidai da a cikin kusan amma akasin agogon.

An shirya da'irar a jere ta yadda wutar lantarki ke gudana a cikin da'irar = wutar lantarki da ke gudana a cikin juriyar R = Amperes 2.

Darajar juriya R

Ci gaba…..

R = 4 / I = 4 / 2 = 2 Ohm

Amsar da ta dace ita ce B.

7 PKalli wannan zane na da'irar lantarki! Girman ƙarfin matsewa akan resistor R shine…

A. Volt 20

B. 8 Volt

C. 5 Volt

D. 4 Volt

E. 2 Volt

Tattaunawa

EMF (E) = yuwuwar bambancin tushen ƙarfin lantarki kafin kwararar wutar lantarki

Ƙarfin matsewa (V) = bambancin yuwuwar tushen ƙarfin lantarki bayan kwararar wutar lantarki

An san cewa:

Shamaki na 1 (R)₁) = 2 Ω

Shamaki na 2 (R)₂) = 4 Ω

Shamaki na 3 (R)₃) = 4 Ω

Tushen emf 1 (E)₁) = Volt 20

Tushen emf 2 (E)₂) = Volt 15

An tambaya: Girman ƙarfin lantarki mai ɗaurewa akan resistor R (V)

Amsa:

Lissafa wutar lantarki da ke gudana a cikin da'irar (I)

Wannan matsala tana da alaƙa da dokar Kirchhooff. Matakai da yadda za a magance wannan matsala:

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.

E₁ - IR₁ - IR₂ - IR₃ - E₂ = 0

20 – I(2) – I(4) – I(4) – 15 = 0

20 – 15 – Ni (2) – Ni (4) – Ni (4) = 0

5 – 10 I = 0

5 = 10 I

I = 5/10

I = 0,5 Amperes

Wutar lantarki da ke gudana a cikin da'irar ita ce 0,5 Amperes. Wutar lantarki mai kyau tana nuna cewa alkiblar wutar tana kamar yadda ake tsammani, a gefen agogo.

Lissafa juriyar daidai (R)

Resistor 1 (R₁), resistor 2 (R₂) da kuma resistor 3 (R)₃) an haɗa shi a jere. Resistor na maye gurbin:

R=R₁ + R₂ + R₃ = 2 Ω + 4 Ω + 4 Ω = 10 Ω

Ƙarfin wutar lantarki a kan resistor R (V)

V = IR = (0,5)(10) = 5 Volt

Hanyar sauri:

Ana iya ƙayyade ƙarfin lantarki kai tsaye ta wannan hanyar:

V = E₁ - E₂ = 20 – 15 = 5 Volt

Amsar da ta dace ita ce C.

8. Kalli wannan zane na da'irar lantarki! Rage wutar lantarki a kan resistor na 3Ω shine…

A. Watts 4

B. Watts 8

C. Watts 12

D. Watts 16

E. Watts 20

Tattaunawa

Ƙarfin wargajewa = amfani da wutar lantarki

An san cewa:

Shamaki na 1 (R)₁) = 2 Ω

Shamaki na 2 (R)₂) = 3 Ω

Shamaki na 3 (R)₃) = 4 Ω

Tushen emf 1 (E)₁) = Volt 8

Tushen emf 2 (E)₂) = Volt 10

An tambaya: Warkewar wutar lantarki a juriya ta 3Ω

Amsa:

Ƙarfin wargajewa

Ana ƙididdige ƙarfin wargazawa ta amfani da dabarar:

P = VI

Bayani: P = iko, V = bambancin yuwuwar a ƙarshen duka resistor 3Ω, I = wutar lantarki da ke gudana ta hanyar juriya 3Ω.

Lissafa wutar lantarki (I) da ke gudana ta hanyar juriya 3Ω

Da farko a ƙididdige wutar lantarki da ke gudana a cikin da'irar:

Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.

E₁ - IR₁ - IR₂ - IR₃ +E₂ = 0

8 – I(2) – I(3) – I(4) + 10 = 0

18 – 9 I = 0

18 = 9 I

I = 18/9

I = 2 Amperes

Wutar lantarki da ke gudana a cikin da'irar ita ce 2 Amperes. Wutar lantarki mai kyau tana nuna cewa alkiblar wutar tana kamar yadda ake tsammani, a gefen agogo.

An shirya da'irar a jere ta yadda wutar lantarki ke gudana a cikin da'irar = wutar lantarki da ke gudana a cikin resistor 3Ω = Amperes 2.

Lissafa bambancin yuwuwar (V) a fadin juriyar 3 Ω

V = IR₂ = (2 A)(3 Ω) = 6 Volt

Lissafi na wargaza wutar lantarki akan resistor 3Ω

P = VI = (6 Volt) (2 Ampere) = 12 Volt Ampere = 12 Watt

Amsar da ta dace ita ce C.

9. Kalli wannan zane na da'irar lantarki! Bambancin da ke tsakanin maki A da B shine…

A. 1,5 volt

B. volts 3,0

C. volts 6,0

D. volts 7,5

Wutar lantarki ta E. 9,0 volts

Tattaunawa

An san cewa:

Shamaki na 1 (R)₁) = 2 Ω

Shamaki na 2 (R)₂) = 3 Ω

Shamaki na 3 (R)₃) = 4 Ω

Tushen emf 1 (E)₁) = Volt 8

Tushen emf 2 (E)₂) = Volt 10

An tambaya: Bambancin da zai iya faruwa (V) tsakanin maki A da B

Amsa:

Lissafa wutar lantarki (I) da ke gudana ta hanyar juriya 3Ω

Da farko a ƙididdige wutar lantarki da ke gudana a cikin da'irar:

Wajen magance wannan matsala, ana zaɓar alkiblar wutar lantarki ta kasance a hannun agogo.

- E₁ - IR₁ - IR₂ - E₂ - IR₃ = 0

– 8 – Ni (2) – Ni (3) – 10 – Ni (4) = 0

– 18 – 9 I = 0

– 18 = 9 I

I = -18 / 9

I = – 2 Amperes

Wutar lantarki da ke gudana a cikin da'irar ita ce 2 Amperes. Wutar lantarki tana da rashin daidaito, ma'ana alkiblar wutar lantarki ita ce ba daidai da a cikin kusan amma akasin agogon.

An shirya da'irar a jere ta yadda wutar lantarki ke gudana a cikin da'irar = wutar lantarki da ke gudana a cikin resistor 3Ω = Amperes 2.

Lissafa bambancin yuwuwar (V) a fadin juriyar 3 Ω

V = IR₂ = (2 A)(3 Ω) = 6 Volt

Amsar da ta dace ita ce C.

Tushen Tambaya:

Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a