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Question
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?
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Solution
There are four coils of resistances: 4 Ω, 8 Ω, 12 Ω and 24 Ω, respectively.
(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by,
`1/(1/4+1/8+1/12+1/24)=1/((6+3+2+1)/24)=24/12=2 Omega`
Therefore, 2 Ω is the lowest total resistance.
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Silver | 1.60 × 10−8 |
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| Manganese | 1.84 × 10−6 | |
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