You are given several identical resistors each of value 10 $Ω$ and each capable of carrying a maximum current of 1 A. It is required to make a suitable combination of these resistances to produce a resistance of 5 $Ω$ which can carry a current of 4 A. The minimum number of resistors required for this job is

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answer is B.

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Detailed Solution

No of rows, $n = \frac{i_{T}}{i}; n = \frac{4}{1} = 4$
$∴$ m be the number of resistances in a row
$m = \frac{R_{T} \times n}{R}; m = \frac{5 \times 4}{10} = 2$
Total number of resistance = mn = 2 × 4 = 8

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