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Quiz : Résistance équivalente

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Quiz : Résistance équivalente

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About this activity

Déterminez R_eq dans des circuits simples

Created by

Radouane Mefouaj
Radouane Mefouaj
Morocco

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Quiz : Résistance équivalente
 

Quiz : Résistance équivalenteOnline version

Déterminez R_eq dans des circuits simples

by Radouane Mefouaj
1

Deux résistances de 5Ω et 10Ω sont en série : R_eq = ?

2

Deux résistances de 5Ω et 10Ω sont en parallèle : R_eq ≈ ?

3

R_eq = 6Ω en série avec (8Ω // 4Ω) : R_eq ≈ ?

4

Trois résistances de 2Ω en parallèle donnent R_eq ?

5

12Ω en série avec (6Ω // 3Ω) donne R_eq ?

6

R_eq d’un réseau: 4Ω en parallèle avec (8Ω // 4Ω) ?

7

Qu’en résulte R_eq de quatre résistances égales de 10Ω en parallèle ?

8

R_eq d’un réseau: 9Ω en parallèle avec 6Ω puis en série avec 3Ω ?

9

R_eq d’un trio en série: 5Ω, 5Ω, 5Ω ?

10

Un réseau: 1Ω en parallèle avec (1Ω en série avec 1Ω) donne R_eq ?

Retour

En série, les résistances s’ajoutent: 5+10=15 Ω.

Pour le parallèle: R_eq = (5×10)/(5+10) = 50/15 ≈ 3.33 Ω.

Parallèle 8//4 = (8×4)/12 = 32/12 ≈ 2.67 Ω; puis en série avec 6Ω: 6 + 2.67 ≈ 8.67 Ω.

En parallèle: R_eq = 2/3 ≈ 0.67 Ω.

Parallèle 6//3 = (6×3)/9 = 18/9 = 2 Ω; total 12+2 = 14 Ω.

8//4 = (8×4)/12 = 32/12 ≈ 2.67 Ω; en parallèle avec 4Ω donne ≈ 6.67 Ω.

Pour n égal à 4, R_eq = 10/4 = 2.5 Ω.

Parallèle: (9×6)/15 = 54/15 = 3.6 Ω; ajoute 3Ω → 6.6 Ω.

R_eq = 5+5+5 = 15 Ω en série.

1Ω || (1+1)Ω = 1Ω || 2Ω = (1×2)/(1+2) = 2/3 Ω.

 
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