Ngā Raru Whakaoti i roto i te Ahupūngao Taketake

1. Ite mahi tahi i ngā hau i roto i te ipu kati i te tīmatanga he rōrahi V me pēhanga P. Mena ko te pēhanga whakamutunga he 4P, ā, ka noho pūmau te rōrahi, he aha te tere o te pēhanga?tio o te pūngao nekeneke tuatahi me te pūngao nekeneke whakamutunga te kaha kinetic.

Mōhiotia:

Pēhanga tuatahi (P₁) = P

Pēhanga whakamutunga (P₂) = 4P

Te rōrahi tuatahi (V₁) = V

Te rōrahi whakamutunga (V₂) = V

Hiahia: Te ōwehenga o te pūngao nekeneke tīmatanga ki te pūngao nekeneke whakamutunga (KE₁ :KE₂)

Rongoā:

Ko te whanaungatanga i waenga i te pēhanga (P), te rōrahi (V) me te pūngao nekeneke (KE) o ngā hau pai :

Ko te ōwehenga o te pūngao nekeneke tīmatanga ki te pūngao nekeneke whakamutunga:

2. He aha te pūngao nekeneke whakawhiti toharite o ngā ngota i roto i te hau pai i te 57^oC.

Mōhiotia:

Te pāmahana o te hau (T) = 57^oC + 273 = 330 Kelvin

Boltzmannte pūmau (k) = 1.38 x 10^-23 Joule/Kelvin

Hiahia: Te pūngao nekeneke whakawhiti toharite

Rongoā:

Te whanaungatanga i waenga i te pūngao nekeneke (KE) me te pāmahana o te hau (T):

Te pūngao nekeneke whakawhiti toharite:

3. He hau i 27^oC i roto i te ipu kati. Ki te piki ake te pūngao nekeneke o te hau kia 2 ngā wā i te pūngao nekeneke tīmatanga, ko te pāmahana whakamutunga o te hau ko…

Mōhiotia:

Te pāmahana tīmatanga (T₁) = 27^oC + 273 = 300 K

Pūngao nekeneke tuatahi =KE

Pūngao nekeneke whakamutunga = 4 KE

Hiahia: Te pāmahana whakamutunga (T₂)

Rongoā:

4. Kei roto i tētahi ipu kati te hau pai rawa atu, ka whakamahanatia, nō reira koinā te mutunga. tere toharite ka piki ake te tere toharite o ngā matūriki hau mā te 3 ngā wā i te tere toharite tīmatanga. Mena he 27^oC, kātahi ko te pāmahana whakamutunga o te hau tino pai ko…

Mōhiotia:

Te pāmahana tīmatanga = 27^oC + 273 = 300 Kelvin

Te tere tīmatanga = v

Te tere whakamutunga = 2v

hiahia :Tte pāmahana whakamutunga o te hau pai

Rongoā:

Te tere toharite whakamutunga = 2 x te tere toharite tuatahi

5. E toru ngā mōrehu hau kei roto i tētahi wāhi 36 rita te rahi. Ko te pūngao nekeneke o ia ngota hau he 5 x 10^-21 Joule. Pūmau hau whānui = 8.315 J/mole.K me te pūmau Boltzmann = 1.38 x 10^-23 J/K. He aha te pēhanga hau i roto i te ipu.

Mōhiotia:

Te maha o ngā mole (n) = 3 mole

Rōrahi = 36 rita = 36 dm³ = 36x10^-3 m³

Te pūmau a Boltzmann (k) = 1.38 x 10^-23 J/K

Pūngao nekeneke (KE) = 5 x 10^-21 Joule

Pūmau hau whānui (R) = 8.315 J/mole.K

hiahia Pēhanga hau (P)

Rongoā:

Tātaihia te pāmahana mā te whakamahi i te whārite o te pūngao nekeneke o te hau.

Tātaihia te pēhanga hau mā te whakamahi i te whārite o te ture hau pai (i roto i te maha o ngā ira, n):

PV = n RT

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

P (36 x 10^-3) = 6024.22

Ko te pēhanga hau he 1.67 x 10⁵ Pascal, 1.67 rānei ngā āhuarangi.

Pānuitia atu

1. Ahaute mahi tahi i ngā hau i roto i te ipu kati i te tīmatanga he rōrahi V me pāmahana Q. Ko te pāmahana whakamutunga he 5/4T, ā, ko te pāmahana whakamutunga he pēhanga ko 2P. He aha te rōrahi whakamutunga o te hau?

Mōhiotia:

Te rōrahi tīmatanga (V₁) = V

Te pāmahana tīmatanga (T₁) = T

Te pāmahana whakamutunga (T₂) = 5/4 T

Pēhanga tuatahi (P₁) = P

Pēhanga whakamutunga (P₂) = 2P

Hiahia: Rōrahi whakamutunga (V₂)

Rongoā:

2. Whakatauhia te rōrahi o te 2.00 moles o ngā hau (hau pai) i te STP. STP = Te Pāmahana me te Pēhanga Paerewa.

Mōhiotia:

Ngā ira o te hau (n) = 2 iraes

Te pāmahana paerewa (T) = 0 ^oC = 0 + 273 = 273 Kelvin

Pēhanga paerewa (P) = 1 atm = 1.013 x 10⁵ Pa

Pūmau hau whānui (R) = 8.315 Joules/moleKelvin

hiahia Te rōrahi o ngā hau (V)

Rongoā:

Whārite o Te ture hau pai rawa atu (i roto i te maha o ngā mole, n)

Ko te rōrahi o ngā mole hau e 2 he 44.8 rita.

Ko te rōrahi o te 1 mole o ngā hau he 45.4 rita / 2 = 22.4 rita.

Ko te rōrahi o te 1 mole o tētahi hau he 22.4 rita.

3. E 27°C te pāmahana o te hau hāora e 4 rita, ā, e 2 atm te pēhanga (1 atm = 10⁵ Pa) i roto i tētahi ipu kua kati. Pūmau hau whānui (R) = 8.314 J.mole^-1.K^-1 me te tau a Avogadro (N_A) = 6.02 x 10²³ ngā ngota/moleHe aha ngā ngota o ngā hau hāora i roto i te ipu?

Mōhiotia:

Te rōrahi o ngā hau (V) = 4 rita = 4 dm³ = 4x10^-3 m³

Te pāmahana o ngā hau (T) = 27^oC = 27 + 273 = 300 Kelvin

Te pēhanga o ngā hau (P) = 2 atm = 2 x 10⁵ Pa

Pūmau hau whānui (R) = 8.314 J.mole^-1.K^-1

Te tau o Avogadro (N_A) = 6.02 x 10²³

hiahia He aha ngā ngota o ngā hau hāora i roto i te ipu (N)

Rongoā:

I roto i te 1 mole o te hau hāora, kei reira te 1.93 x 10²³ ngā ngota hāora.

4. He ipu kei roto he hau neon (Ne, papatipu ngota = 20 u) i te pāmahana me te pēhanga paerewa (STP)) he 2 m te rōrahi³. Whakatauhia te papatipu o te hau neon!

Mōhiotia:

Papatipu atomika o te neon = 20 karamu/mole = 0,02 kg/mole

Te pāmahana paerewa (T) = 0^oC = 273 Kelvin

Te pehanga paerewa (P) = 1 atm = 1.013 x 10⁵ Pascal

Rōrahi (V) = 2 m³

hiahia : papatipu (M) o te hau neon

Rongoā:

I te pāmahana me te pēhanga paerewa (STP), 1 mole o ngā hau katoa, whakaurua te hau neon, he rōrahi 22.4 ritas = 22.4 dm³ = 0.0448 m³.

I roto i te rōrahi o te 2 m³, e 44.6 ngā mole o te hau neon.

Ko te papatipu ngota whanaunga o te hau neon he 20 karamu/mole.

Ko te tikanga o tēnei, i roto i te 1 ira he 20 karamu, arā, 0.02 kg o te hau neon. Nā te mea he 0.02 kg o te hau neon i roto i te 1 ira, nō reira, i roto i te 44.6 ira he 44.6 ira x 0.02 kg/ira = 0.892 kg = 892 karamu o ngā hau neon.

Pānuitia atu

1. Ingā hau kirimana i te tīmatanga pēhanga P me pāmahana T. Ka pāngia te hau e te tukanga isochoric kia 4 ngā whakarea o te pēhanga whakamutunga i te pēhanga tīmatanga. He aha te pāmahana whakamutunga o te hau?

Mōhiotia:

Pēhanga tuatahi (P₁) = P

Pēhanga whakamutunga (P₂) = 4P

Te pāmahana tīmatanga (T₁) = T

Hiahia: Te pāmahana whakamutunga (T₂)

Rongoā:

Ko te tātai o Te ture a Gay-Lussac :

E whā ngā whakarea o te pāmahana whakamutunga i te pāmahana tīmatanga.

2. I roto i tētahi ipu kua kati, ko te pāmahana o ngā hau pai i te tīmatanga he 27^oC. Mēnā Ka rua ngā whakarea o te pēhanga whakamutunga i tōna pēhanga tīmatanga, he aha te pāmahana whakamutunga?

Mōhiotia:

Pēhanga tuatahi (P₁) = P

Pēhanga whakamutunga (P₂) = 2P

Te pāmahana tīmatanga (T₁) = 27^oC + 273 = 300 K

Hiahia: Te pāmahana whakamutunga (T₂)

Rongoā:

3. Ka whakakīia te potae ki te pēhanga ine o te 2 atm i te 27°C. I muri i te taraiwa, ka piki te pāmahana i roto i te potae ki te 47°C. He aha te pēhanga i roto i te potae ināianei?

Mōhiotia:

Te pēhanga hau = 1 atm = 1 x 10⁵ Pa

Ko te pēhanga ine tuatahi = 2 atm = 2 x 10⁵ Pa

Ko te pēhanga tino tuatahi (P₁) = 1 atm + 2 atm = 3 atm = 3 x 10⁵ Pa

Te pāmahana tīmatanga (T₁) = 27^oC + 273 = 300 K

Te pāmahana whakamutunga (T₁) = 47^oC + 273 = 320 K

E hiahiatia ana: Te pāmahana whakamutunga o te ine

Rongoā:

Te pēhanga ine whakamutunga = te pēhanga tino whakamutunga – te pēhanga hauātea

Ko te pēhanga ine whakamutunga = 3.2 atm – 1 atm

Ko te pēhanga ine whakamutunga = 2.2 atm

Pānuitia atu

1. I roto i tētahi ipu kua kati, ka whakawhanui te hau kia toru ngā whakarea o te rōrahi whakamutunga i te rōrahi tīmatanga (V = rōrahi tīmatanga, T = rōrahi tīmatanga pāmahana). He aha te pāmahana whakamutunga?

Mōhiotia:

Te rōrahi tīmatanga (V₁) = V

Rōrahi whakamutunga (V₂) = 3V

Te pāmahana tīmatanga (T₁) = T

Hiahia: Te pāmahana whakamutunga (T₂)

Rongoā:

Ko te tātai o Te ture a Charles :

E toru ngā whakarea o te pāmahana whakamutunga o ngā hau i tōna pāmahana tīmatanga.

2. AhauKo te rōrahi o ngā hau whakawhitiwhiti i te tīmatanga he V me te pāmahana T. Mena ka pāngia te hau e te tukanga isobaric kia rua ngā wā ka nui ake te pāmahana i te pāmahana tīmatanga, ko te rōrahi whakamutunga o ngā hau he…

Mōhiotia:

Te rōrahi tīmatanga (V₁) = V

Te pāmahana tīmatanga (T₁) = T

Te pāmahana whakamutunga (T₂) = 2T

Hiahia: rōrahi whakamutunga (V₂)

Rongoā:

E rua ngā whakarea o te rōrahi whakamutunga o ngā hau i te rōrahi tīmatanga.

3. I roto i tētahi ipu kua kati, ko te rōrahi o ngā hau pai i te tīmatanga he 2 rita, ā, ko te pāmahana he 27^oC. Mena ka eke te rōrahi whakamutunga o ngā hau ki te 3 rita, ko te pāmahana whakamutunga ko…

Mōhiotia:

Te rōrahi tīmatanga (V₁) = 2 rita = 2 dm³ = 2x10^-3 m³

Rōrahi whakamutunga (V₂) = 3 rita = 3 dm³ = 3x10^-3 m³

Te pāmahana tīmatanga (T₁) = 27^oC + 273 = 300 K

E hiahiatia ana: Te pāmahana whakamutunga (T₂)

Rongoā:

Ko te pāmahana whakamutunga he 177^oC rānei 177 + 273 = 450 Kelvin.

Pānuitia atu

1. Ko ētahi hau pai i te tīmatanga he pēhanga P me te rōrahi V. Mena ka pāngia te hau e tukanga isothermal kia 4 ngā whakarea o te pēhanga whakamutunga i te pēhanga tīmatanga, ko te rōrahi hau whakamutunga ko…

Mōhiotia:

Pēhanga tuatahi (P₁) = P

Pēhanga whakamutunga (P₂) = 4P

Te rōrahi tuatahi (V₁) = V

Hiahia: Te rōrahi whakamutunga (V₂)

Rongoā:

Ko te tātai o Te ture a Boyle :

PV = tamau

P₁ V₁ =P₂ V₂

(P)(V) = (4P)(V₂)

V = 4 V₂

V₂ = V / 4 = ¼ V

Ko te rōrahi whakamutunga o ngā hau ko ¼ ngā whakarea o te rōrahi tīmatanga.

2. I roto i tētahi ipu kua kati, ka whakawhanui te hau kia whakamutunga rōrahi riro 2 ngā wā o te rōrahi tīmatanga (V = rōrahi tīmatanga, P = pēhanga tīmatanga). Ko te pēhanga whakamutunga o ngā hau ko…

Mōhiotia:

Pēhanga tuatahi (P₁) = P

Te rōrahi tuatahi (V₁) = V

Te rōrahi whakamutunga (V₂) = 2V

hiahia : Pēhanga whakamutunga (P₂)

Rongoā:

P₁ V₁ =P₂ V₂

PV = P₂ (2V)

P= P₂ (2)

P₂ = P / 2 = ½ P

Ka piki te pēhanga hau ki te ½ whakarea o te pēhanga tīmatanga.

3. I roto i tētahi ipu kua kati, ko ngā hau he 2 atm te pēhanga me te 1 rita te rōrahi. Mēnā ka huri te pēhanga hau ki te 4 atm, ka huri te rōrahi hau ki...

Mōhiotia:

Pēhanga tuatahi (P₁) = 2 atm = 2 x 10⁵ Pa

Pēhanga whakamutunga (P₂) = 4 atm = 4 x 10⁵ Pa

Te rōrahi tuatahi (V₁) = 1 rita = 1 dm³ = 1x10^-3 m³

hiahia : Te rōrahi whakamutunga (V₂)

Rongoā:

P₁ V₁ =P₂ V₂

(2 x 10⁵)(1 x 10^-3) = (4 x 10⁵) V₂

(1)(1 x 10^-3) = (2) V₂

1 x 10^-3 = (2) V₂

V₂ = ½ x 10^-3

V₂ = 0.5x10^-3 m³ = 0.5 dm³ = 0.5 ritas

Pānuitia atu

1. E toru ngā pūnga iahiko, C₁ = 2 µF, C₂ = 4 µF, C₃ = 4 μF, ko hono raupapa me te whakararaWhakatauhia te kaha o tētahi kaha kotahi ka rite te pānga ki te kaha o te whakakotahitanga.

Mōhiotia:

Kaiwhakahaere C₁ = 2 μF

Pūnga C₂ = 4 μF

Pūnga C₃ = 4 μF

E hiahiatia ana: Te kaha ōrite (C)

Rongoā:

Kaiwhakahaere C₂ me C₃ honoa whakarara. Ko te kaha ōrite:

C_P = C₂ + C₃ = 4 + 4 = 8 μF

Kaiwhakahaere C₁ me C_p honoa raupapa. Ko te kaha ōrite:

1/C = 1/C₁ + 1/C_P = 1/2 + 1/8 = 4/8 + 1/8 = 5/8

C = 8/5 μF

2. Rima ngā pūnga iahiko, C₁ = 2 µF, C₂ = 4 µF, C₃ = 6 µF, C₄ = 5 µF, C₅ = 10 μF, e honoa ana i roto i te raupapa me te whakarara. Whakatauhia te kaha o tētahi kaha kotahi ka rite te pānga ki te whakakotahitanga.

Mōhiotia:

Pūnga C₁ = 2 μF

Pūnga C₂ = 4 μF

Pūnga C₃ = 6 μF

Pūnga C₄ = 5 μF

Pūnga C₅ = 10 μF

E hiahiatia ana: Te kaha ōrite (C)

Rongoā:

Kaiwhakahaere C₂ me C₃ e honoa ana i te taha whakarara. Ko te kaha ōrite:

C_P = C₂ + C₃

C_P = 4 + 6

C_P = 10 μF

Kaiwhakahaere C₁, C_P, C₄ me C₅ e honoa ana i roto i te raupapa. Ko te kaha ōrite:

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 me C₃ = 3 μKo ngā F, he mea hono raupapa, he mea hono whakarara hoki. Whakatauhia te pūngao hiko i runga i te awheawhe.

Mōhiotia:

Pūnga C₁ = 3 μF

Pūnga C₂ = 4 μF

Pūnga C₃ = 3 μF

E hiahiatia ana: Te kaha ōrite (C)

Rongoā:

Kaiwhakahaere C₂ me C₃ e honoa ana i te taha whakarara. Ko te kaha ōrite:

C_P = C₂ + C₃

C_P = 4 + 3

C_P = 7 μF

Kaiwhakahaere C₁ a C_P e honoa ana i roto i te raupapa. Ko te kaha ōrite:

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

Te pūngao hiko i runga i ngā ara iahiko:

E = ½ CV²

E = ½ (2.1 x 10^-6)(12²)

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

Pānuitia atu

1. E whā ngā pūnga iahiko, C₁ = 2 µF, C₂ = 1 µF, C₃ = 3 µF, C₄ = 4 μF, kua honoa ki roto raupapaWhakatauhia te kaha o te kotahi pūkaha ka rite te pānga ki te whakakotahitanga.

Mōhiotia:

Pūnga C₁ = 2 μF

Pūnga C₂ = 1 μF

Pūnga C₃ = 3 μF

Pūnga C₃ = 4 μF

E hiahiatia ana: Te kaha ōrite

Rongoā:

Te kaha ōrite:

1/C = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

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

Te kaha ōrite o te huinga katoa is 0.48 μF.

2. Whakatauhia te utu i runga i te pūnga C₁ ki te rerekētanga pūmanawa i waenganui i P a Q is 12 Ngāohiko…

Mōhiotia:

Pūnga C₁ = 10 μF= 10 x 10^-6 F

Pūnga C₂ = 20 μF= 20 x 10^-6 F

Te rerekētanga pūmanawa (V) = 12 wae ngaohiko

E hiahiatia ana: te utu i runga i te pūnga C₁ (Q₁)

Rongoā:

Te kaha ōrite:

1/C = 1/C₁ + 1/C₂

1/C = 1/10 + 1/20 = 2/20 + 1/20 = 3/20

C = 20/3 μF = (20/3) x 10^-6 F

Te utu hiko i runga i te pūnga taurite:

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

Q = 80 μC

Ka honoa ngā pūnga iahiko i roto i te raupapa kia rite te utu hiko i runga i ngā pūnga rite ki te utu hiko i runga i ngā pūnga C₁ = te utu hiko i runga i te pūnga C₂.

Te utu hiko i runga i te pūnga C₁ he 80 μC.

3. E rua ngā pūnga iahiko, C₁ = 2 μF me C₂ = 4 µF, kua honoa raupapa. Kua utua ngā pūnga. Ko te rerekētanga pūmanawa i runga i te pūnga C₁ he 2 Volts. Ko te utu hiko i runga i te pūnga C₂ ko ...

Mōhiotia:

Cahacitor C₁ = 2 μF = 2 x 10^-6 F

Cahacitor C₂ = 4 μF = 4 x 10^-6 F

Te rerekētanga pūmanawa i runga i te pūnga C₁ (V₁) = 2 Ngāohiko

hiahia : Utu hiko i runga i te pūnga C₂.

Rongoā:

Utu hiko i runga i te pūnga C₁ :

Q₁ = C₁ V₁ = (2 x 10^-6)(2) = 4 x 10^-6 C

Q₁ = 4 μC

Ka honoa ngā pūnga hiko i roto i te raupapa kia puta ai he utu hiko i runga i te pūnga hiko C₁ = te utu hiko i runga i te pūnga C₂.

Te utu i runga i te pūnga C₂ is 4 μC.

Pānuitia atu

1. Whā pūnga iahiko, C₁ = 2 µF, C₂ = 1 µF, C₃ = 3 µF, C₄ = 4 μF, e honoa ana i te taha whakarara. Tātaihia te kaha o tētahi kaha kotahi ka rite te pānga ki te whakakotahitanga.

Mōhiotia:

Kaiwhakahaere C₁ = 2 μF

Kaiwhakahaere C₂ = 1 μF

Kaiwhakahaere C₃ = 3 μF

Kaiwhakahaere C₃ = 4 μF

E hiahiatia ana: Te kaha ōrite

Rongoā:

Te kaha ōrite:

C = C₁ + C₂ + C₃

C = 4 μF+2 μF+3 μF = 9 μF

Ko te kaha ōrite o te huinga katoa he 9 μF.

2. Whakahuahia te tenei i runga i te pūngahiko C₂ ki te rerekētanga pūmanawa kei waenganui i te pūwāhi A me te pūwāhi B 9 Ngā Waohiko…

Mōhiotia:

Cahacitor C₁ = 20 μF = 20 x 10^-6 F

Cahacitor C₂ = 30 μF = 30 x 10^-6 F

Te rerekētanga pūmanawa i waenga i ngā pūwāhi A a B (V_AB) = 9 Ngāohiko

hiahia : te utu i runga i te pūnga C₂ (Q₂)

Rongoā:

Rerekētanga pūmanawa:

Ka honoa ngā pūnga iahiko i roto i te whakarara kia rite ai te rerekētanga pūmanawa i waenga i a A me B (V)_AB) = te rerekētanga pūmanawa i runga i te pūnga C₁ (V₁) = te rerekētanga pūmanawa i runga i te pūnga C₂ (V₂) = 9 Ngā Āhuatanga.

Utu hiko i runga i te pūnga C₂ :

Q₂ = C₂ V₂ = (30 x 10^-6)(9) = 270 x 10^-6 C

Q₂ = 270 μC

Te utu hiko i runga i te pūnga C₂ is 270 μC.

3. e toru ngā pūnga iahiko, C₁ = 4 µF, C₂ = 2 µF, C₃ = 3 μF, e honoa ana i te taha whakarara. Ka utua ngā pūnga. Ko te rerekētanga pūmanawa i runga i te pūnga C₂ he 4 Volts. Whakatauhia

(a) Utu hiko i runga i te pūnga C₁, C₂ me C₃

(b) Te utu hiko i runga i te pūnga rite o te huinga katoa

Mōhiotia:

Cahacitor C₁ = 4 μF = 4 x 10^-6 F

Cpūnga iahiko C₂ = 2 μF = 2 x 10^-6 F

Cpūnga iahiko C₃ = 3 μF = 3 x 10^-6 F

Te rerekētanga pūmanawa i runga i te pūnga C₂ (V₂) = 4 Ngāohiko

hiahia : Utu hiko i runga i te pūnga C₃ (Q₃)

Rongoā:

(a) Utu hiko i runga i te pūnga C₃

Te rerekētanga pūmanawa i runga i te pūnga C₃ :

Ka honoa ngā pūnga iahiko i te taha whakarara kia puta ai te rerekētanga pūmanawa i runga i te pūnga iahiko C₃ (V₃) = te rerekētanga pūmanawa i runga i ngā pūnga C₂ (V₂) = te rerekētanga pūmanawa i runga i ngā pūnga C₁ (V₁) = te rerekētanga pūmanawa i runga i ngā pūnga taurite (V) = 4 Ngāohiko

Utu hiko i runga i te pūnga C₁ :

Q₁ = C₁ V₁ = (4 x 10^-6)(4) = 16 x 10^-6 C

Q₁ = 16 μC

Utu hiko i runga i te pūnga C₂ :

Q₂ = C₂ V₂ = (2 x 10^-6)(4) = 8 x 10^-6 C

Q₂ = 8 μC

Utu hiko i runga i te pūnga C₃ :

Q₃ = C₃ V₃ = (3 x 10^-6)(4) = 12 x 10^-6 C

Q₃ = 12 μC

(B) Utu hiko i runga i te pūnga rite

Q = Q₁ +Q₂ +Q₃

Q = 16 μC + 8 μC + 12 μC=36 μC

Otinga kē atu:

Te kaha ōrite:

C = C₁ + C₂ + C₃

C=4 μF+2 μF+3 μF = 9 μF

C = 9 x 10^-6 F

Te rerekētanga pūmanawa i runga i te pūnga taurite:

V₁ =V₂ =V₃ = V = 4 Ngāohiko

Ko te utu hiko i runga i te pūnga rite:

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

Q = 36 μC

Pānuitia atu