You are given several identical resistances each of value R = 10W and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5W 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(a) 4(b) 10(c) 8(d) 20

(c)

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 problem

 r/4=5

r=20

For this propose two resistances should be connected. There are four such combinations. Hence, the total number of resistance = 4 x 2 =8