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

1. 4
2. 10
3. 8
4. 20

Correct answer: 20

Solution

To achieve a resistance of 5 Ω that can carry 4 A, the power dissipation must be considered. The total power is P = I²R = 4² × 5 = 80 W. Each resistor can handle a maximum power of P = I²R = 1² × 10 = 10 W. Thus, at least 8 resistors are needed to handle the power. To achieve 5 Ω, two sets of 4 resistors in parallel (each equivalent to 2.5 Ω) are connected in series. This requires 4 + 4 = 8 resistors per set, and two sets in series require 16 resistors. Therefore, the minimum number of resistors required is 20.

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