To carry a current of 4 ampere, 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 tothe given problem
r
4
=5 or r=20Ω For this purpose two resistances should be connected. There are four such combinations.Hence, the total number of resistance =4×2=8