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You are given several identical resistances each of value R = 10Ω and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5Ω 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
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- 4
- 10
- 8
- 20
Correct Option: 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
| r | = 5 or r = 20 Ω | |
| 4 |
For this propose two resistances should be connected. There are four such combinations. Hence, the total number of resistance = 4 × 2 = 8.
