You are given several identical resistances each of value -R=10Ω and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5Ω which can carry a current of 4 ampere. The minimum number of resistances of the type R that will be required for this job is

To carry a current of 4 amperes, we need four paths, 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 problemr4=5  or  r=20ΩaFor this purpose two resistances should be connected. There are four such combinations. Hence, the total number of resistances = 4x 2= 8