Current Electricity Ques 47
47. 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
[1990]
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Answer:
Correct Answer: 47.(c)
Solution:
- (c) 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 to the given
problem $\frac{r}{4}=5$ or $r=20 \Omega$.
For this propose two resistances should be connected. There are four such combinations. Hence, the total number of resistance $=4 \times 2=8$.
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