You are given several identical resistances,each of value `R= 10 Omega` and each capable of carrying a maximum current of 1A. It is required to make a suitable combination of these resistances to obtain a resistance of `5 Omega` which can carry a current of 4A, FInd the mninimum number of resistances of the type R that will be required.

Explanation

The resistances have to be connected in a series parallel combination,Since each resistance can carry current of 1A, so to pass 4 A current, we need four paths in parallel.
Let r be the resistance of each path.
The equivalent resistance of 4 parallel paths will be `r/4`.
According to the given problem,
`r/4=5`
`therefore r=5 times 4=20 Omega`
In order to have `20 Omega` resistance in each path, two resistances each of resistance `10 Omega` have to be connected in series.
Since there are four paths, the total number of resistances required =2 `times `4=8.