Question

You are given several identical resistances each of value $R=10\, \Omega$ and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of $5\, \Omega$ 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

Answer 1

Answer: (C)

8

Answer 2

To carry a current of 4 ampere, we need four path, 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 $\frac{r}{4}=5$ or $r = 20\, \Omega.$
For this purpose two resistances should be connected. There are four such combinations.
Hence, the total number of resistance $= 4 \times 2 = 8$