What is (a) the highest, (b) the lowest total resistance that can be secured by combinations
of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answers
Answered by
7
Answer:
The answer to this question is
(a) 48 ohm
(b) 2 ohm
Explanation:
The highest resistance is =
R = R1 + R2 + R3 + R4
R = 4 + 8 + 12 + 24
R = 48 ohm
The longest Resistance is =
1/R = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/R = 1/4 + 1/8 + 1/12 + 1/24
1/R = 1/2
R = 2 ohm.
Therefore The lowest resistance = 2 ohm.
I HOPE IT WILL HELP YOU.
Answered by
4
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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