Saketi za umeme zenye vipingamizi katika upinzani sambamba na wa ndani - matatizo na suluhisho

Saketi za umeme zenye vipingamizi katika upinzani sambamba na wa ndani - matatizo na suluhisho

1. Kulingana na mchoro ulio hapa chini, ikiwa chanzo cha nguvu ya kielektroniki (emf) ni Volti 24, tambua mkondo wa umeme I.

Inajulikana:

Chanzo cha emf = 24 Volti

Upinzani wa ndani (r) = 2 Ohm

Vipinga 40 Ohm, 20 Ohm, na 20 Ohm.

Wante d: Mkondo wa umeme I

Suluhisho:

Vipingamizi vilivyounganishwa sambamba, kipingamizi sawa:

1/R = 1/40 + 1/20 + 1/20

1/R = 1/40 + 2/40 + 2/40

1/R = 5/40

R = 40/5

R = 8 Ohm

Volti ya terminal:

V = emf – I r

V = 24 – I 2

Mkondo wa umeme I:

V = IR

24 – Mimi 2 = Mimi 8

24 = Mimi 8 + Mimi 2

24 = Mimi (10)

Mimi = 24/10

I = Ampea 2.4

2. Kulingana na mchoro ulio hapa chini, ikiwa R ₁ = 3 Ω, R ₂ = 4 Ω, R ₃ = 4 Ω na mkondo wa umeme ni 0.5 A. Tambua volteji ya umeme.

Inajulikana:

Mkondo wa umeme (I) = 0.5 Ampea

Kipingamizi R ₁ = 3 Ohm

Kipingamizi R ₂ = 4 Ohm

Kipingamizi R ₃ = 4 Ohm

Inayohitajika: Volti ya umeme (V)

Suluhisho:

Hesabu upinzani sawa:

R2 na R3 _{zimeunganishwa} sambamba. Kipingamizi sawa _:

1 /R ₂₃ = 1/R ₂ + 1/R ₃ = 1/4 + 1/4 = 2/4

R ₂₃ = 4/2 = 2 Ohm

R ₁ na R ₂₃ zimeunganishwa mfululizo. Kipingamizi sawa:

R = R ₁ + R ₂₃ = 3 + 2 = 5 Ohms

Volti ya umeme:

V = IR = (0.5)(5) = Volti 2.5