]What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
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Answer:
(a)Highest will be by series =4+12+8+24=48ohms
(b)Lowest will be by parallel =1/R=1/4+1/8+1/24+1/12=12/24=1/2
so R=2
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GIVEN:
- R₁ = 4 Ω
- R₂ = 8 Ω
- R₃ = 12 Ω
- R₄ = 24 Ω
TO FIND:
- Highest total resistance that can be secured by combinations of the four coils of resistances.
- Lowest total resistance that can be secured by the combinations of the four coils of resistances.
SOLUTION:
Highest total resistance that can be secured by combinations of the four coils of resistances. [Series combination]
⇒ R = R₁ + R₂ + R₃ + R₄
⇒ R = 4 + 8 + 12 + 24
⇒ R = 48 Ω
Lowest total resistance that can be secured by the combinations of the four coils of resistances. [Parallel combination]
⇒ 1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
⇒ 1/R = 1/4 + 1/8 + 1/12 + 1/24
⇒ 1/R = 12/24
⇒ 1/R = 1/2
⇒ R = 2 Ω
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