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)

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 $=4\times 2=8$