Question

What is
(i) the highest,
(ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?​

Answer 1

(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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Answer 2

Answer:

Ans: Given resistances R1 =4 Ω, R2=8 Ω, R3=12Ω, and R4=24Ω. ∴ 2 Ω is the lowest total resistance

Explanation:

(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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