Ciorcaid leictreacha le friotóirí i gcomhthreo agus friotaíocht inmheánach – fadhbanna agus réitigh

Ciorcaid leictreacha le friotóirí i gcomhthreo agus friotaíocht inmheánach – fadhbanna agus réitigh

1. Based on the figure below, if the source of electromotive force (emf) is 24 Volt, determine sruth leictreach I.

Ar a dtugtar:

Source of emf = 24 Volt

Internal resistance (r) = 2 Ohm

Friotóir 40 Ohm, 20 Ohm, and 20 Ohm.

Wanted : Electric current I

Réiteach:

Resistors connected in parallel, the equivalent resistor :

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 Óim

Terminal voltage :

V = emf – I r

V = 24 – I 2

Electric current I :

V = IR

24 – I 2 = I 8

24 = I 8 + I 2

24 = I (10)

I = 24/10

I = 2.4 Aimpéar

2. Based on figure below, if R₁ = 3 Ω, R₂ = 4Ω, R₃ = 4 Ω and electric current is 0.5 A. Determine voltas leictreach.

Ar a dtugtar:

Sruth leictreach (I) = 0.5 Aimpéar

Friotóir R₁ = 3 Óm

Friotóir R₂ = 4 Óm

Friotóir R₃ = 4 Óm

Ag Teastáil: Electric voltage (V)

Réiteach:

Calculate the equivalent resistance :

R₂ agus R₃ atá ceangailte go comhthreomhar. An friotóir coibhéiseach:

1/R₂₃ = 1 / R.₂ + 1 / R.₃ = 1/4 + 1/4 = 2/4

R₂₃ = 4/2 = 2 Óim

R₁ agus R₂₃ atá ceangailte i sraith. An friotóir coibhéiseach:

R=R₁ + R₂₃ = 3 + 2 = 5 Ohm

Electric voltage :

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