Dokokin Kirchhoff - matsaloli da mafita
1. Menene tashar irin ƙarfin lantarki na batirin da ke cikin kewaye kasa?
Magani
emf = ƙarfin lantarki = bambancin yuwuwar tsakanin tashoshi idan babu yanzu kwarara zuwa da'irar waje.
Ƙarfin wutar lantarki (V) = bambancin da ke tsakanin tashoshin wutar lantarki lokacin da wutar lantarki ke gudana daga batirin.
Idan babu wutar lantarki da aka cire daga batirin, ƙarfin wutar lantarki na ƙarshe zai yi daidai da emf.
An sani:
Tsayayya 1 (R1) = 2 Ω
Tsayayya 2 (R2) = 4 Ω
Resistor 3 (R)3) = 4 Ω
emf 1 (E1) = Volt 20
emf 2 (E)2) = Volt 15
Ana so: Ƙarfin wutar lantarki (V)
Magani 1:
Ƙarfin wutar lantarki:
V = E1 - E2 = 20 – 15 = 5 Volt
Magani 2:
Lissafa kwararar wutar lantarki a cikin da'irar (I)
Da farko, Zaɓi alkiblar kowace wutar lantarki. Ana iya zaɓar alkiblar ba tare da son rai ba: idan wutar lantarki a zahiri tana akasin alkiblar, za ta fito da alamar cirewa a cikin maganin.
Sannan, farawa daga kowane lokaci a cikin da'irar, muna tunanin tafiya a kusa da madauki, ƙara kalmomin emfs da IR yayin da muke zuwa gare su.
Na biyuIdan muka yi tafiya ta cikin tushe daga – zuwa +, ana ɗaukar emf a matsayin mai kyau; idan muka yi tafiya daga + zuwa -, ana ɗaukar emf a matsayin mara kyau.
Na ukuIdan muka yi tafiya ta cikin resistor a daidai hanyar da ake ɗauka a matsayin wutar lantarki, kalmar IR ba ta da kyau domin wutar lantarki tana tafiya ne a matsayin hanyar da za ta rage ƙarfin lantarki. Idan muka yi tafiya ta cikin resistor a alkiblar da aka saba da wacce ake ɗauka a matsayin wutar lantarki, kalmar IR tana da kyau domin wannan tana wakiltar ƙaruwar ƙarfin lantarki.
Ana zaɓar alkiblar wutar lantarki kamar yadda ake yi a gefen 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 Ampere. Layin wutar lantarki mai alamar positive yana nufin alkiblar wutar lantarki ita ce daidai da alkiblar agogo.
Lissafa juriya mai daidai (R):
Resistor 1 (R1), resistor 2 (R2) da kuma juriya 3 (R)3) an haɗa su cikin jerin. Resistor mai kama da haka:
R=R1 + R2 + R3 = 2 Ω + 4 Ω + 4 Ω = 10 Ω
Ƙarfin wutar lantarki a cikin resistor R (V)
V = IR = (0.5)(10) = 5 Volt
2. A cikin da'irar kamar yadda aka nuna a cikin hoton da ke ƙasa, nemo wutar da aka watsa a ciki da Resistor mai ƙarfin 3-Ω.
Magani:
An sani:
Resistor 1 (R1) = 2 Ω 
Resistor 2 (R2) = 3 Ω
Tsayayya 3 (R3) = 4 Ω
emf 1 (E1) = Volt 8
emf 2 (E2) = Volt 10
SE busca: Ƙarfin ya ɓace a cikin da Resistor 3-Ω
Magani:
The iko an wargaza a cikin da Resistor 3-Ωr:
P = VI
P= iko, V = da voltage a fadin da Resistor 3-Ω, na = wutar lantarki tana ratsawa ta cikin Resistor 3-Ω
Lissafa wutar lantarki (I) wucewa da Resistor 3-Ω
Ana zaɓar alkiblar wutar lantarki kamar yadda ake yi a gefen 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 tana 2 Ampere. Lantarkin lantarki mai alamar positive yana nufin alkiblar wutar lantarki ita ce daidai da alkiblar agogo.
Ana haɗa da'irori a jere don haka wutar lantarki da ke gudana a cikin da'irar = wutar lantarki ta ratsa ta 3-Ω ɗan adawa = Amperes 2.
Lissafi ƙarfin lantarki (V) a fadin 3-Ω ɗan adawa
V = IR2 = (2 A)(3 Ω) = 6 Volt
Ƙarfin ya ɓace a cikin da Resistor na 3-Ω:
P = VI = (6 Volt) (2 Ampere) = 12 Volt Ampere = 12 Watt
3. Dangane da da'irar kamar yadda aka nuna a cikin hoton da ke ƙasa, menene bambancin yuwuwar tsakanin maki A da B?
An sani:
Tsayayya 1 (R1) = 2 Ω 
Tsayayya 2 (R2) = 3 Ω
Tsayayya 3 (R3) = 4 Ω
emf 1 (E1) = Volt 8
emf 2 (E2) = Volt 10
Ana so: Bambancin da zai iya faruwa (V) tsakanin aya ta A da B
Magani:
Lissafa wutar lantarki (I) yana gudana a cikin 3-Ω ɗan adawa
Ana zaɓar alkiblar wutar lantarki kamar yadda ake yi a gefen 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 tana 2 An sanya hannu kan wutar lantarki ta Ampere. korau yana nufin alkiblar wutar lantarki counteralkiblar agogo.
Ana haɗa da'irori a jere don haka wutar lantarki da ke gudana a cikin da'irar = wutar lantarki ta ratsa ta 3-Ω ɗan adawa = Amperes 2.
Lissafa bambancin da zai yiwu (V) a fadin 3-Ω juriya:
V = IR2 = (2 A)(3 Ω) = 6 Volt
4. Dangane da da'irar kamar yadda aka nuna a hoton da ke ƙasa, menene bambancin yuwuwar da ke tsakanin R3 juriya.
An sani:
Tsayayya 1 (R1) = 2 Ω 
Tsayayya 2 (R2) = 4 Ω
Tsayayya 3 (R3) = 3 Ω
emf 1 (E1) = Volt 6
emf 2 (E)2) = Volt 9
Ana so: Bambancin da zai iya faruwa a fadin R3 ɗan adawa
Magani:
Lissafa wutar lantarki (I) wucewa ta cikin R3 ɗan adawa
Ana zaɓar alkiblar wutar lantarki kamar yadda ake yi a gefen 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 tana 1/3 An sanya hannu kan wutar lantarki ta Ampere. korau yana nufin alkiblar wutar lantarki counteralkiblar agogo.
Ana haɗa da'irori a jere don haka wutar lantarki da ke gudana a cikin da'irar = wutar lantarki ta ratsa ta R3 ɗan adawa = 1/3 Ampere.
Lissafa bambancin da zai yiwu (V) a fadin R3 ɗan adawa
V = IR3 = (1/3)(3) = 1 Volt
5. Wutar lantarki mai yuwuwa tsakanin maki C da D = 4 Volts, sami R!
An sani:
Tsayayya 1 (R1) = 2 Ω 
Tsayayya 2 (R2) = 2 Ω
Tsayayya 3 (R3) = R
emf 1 (E1) = Volt 8
emf 2 (E2) = Volt 4
Bambancin da ke tsakanin C da kuma D (V)CD) = Volt 4
Ana so: R
Magani:
VCD = IR
4 = IR
R = 4 / I
...............
Lissafa wutar lantarki (I) yana wucewa ta cikin resistor na R
Ana zaɓar alkiblar wutar lantarki kamar yadda ake yi a gefen 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 tana 2 An sanya hannu kan wutar lantarki ta Ampere. korau yana nufin alkiblar wutar lantarki counteralkiblar agogo.
Ana haɗa da'irori a jere don haka wutar lantarki da ke gudana a cikin da'irar = wutar lantarki ta ratsa ta R ɗan adawa = 2 Ampere.
R:
R = 4 / I = 4 / 2 = 2 Ohm
