You are given several identical resistors each of value 10ohm and each capable of 1A. It is required to make a suitable combination of these two resistances to produce a resistance of 5ohm which can carry a current of4A . The minimum number of resistors required for this job?

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Solution

To carry a current of 4 amp,we need four path,each carrying a current of one ampere.
Let r be the resistance of each path.These are connected in parallel.Hence their equivalent resistance will be r/4.
according to the given problem , r/4 = 5 or r=20 ohm
For this purpose two resistance should be connected .
there are four such combinations.
Hence.the total number of resistance = 4*2 = 8

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