You are given several identical resistors each of value `10Omega` and each capable of carrying a maximum current of 1 A. It is required to make a suitable combination of these to resistances to produce a resistance of `5Omega` which can carry a current of 4 A. The minimum number of resistors required for this job is

Answer

`8`

Explanation

Since each part can carry current of one ampere, therefore to pass `x` A current we need four paths in parallel .Let `r` be the resistance of each path .The equivalent resistance of a parallel paths will be `r//4` According to given problem
`r//4 = 5 or r = 5xx 4 = 20 Omega`
In order to have `20 Omega` resistance in each path two resistance each of resistance `10 Omega` be connected in series. Since there are four paths the total number of resistances required `= 2 xx 4 = 8`