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(C) Let $n$ be the total number of resistors used. Let the equivalent resistance be $R_{eq} = 5\,\Omega$ and the maximum current capacity be $I_{total

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$8$

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(C) Let $n$ be the total number of resistors used. Let the equivalent resistance be $R_{eq} = 5\,\Omega$ and the maximum current capacity be $I_{total} = 4\,A$.Each resistor has $R = 10\,\Omega$ and $I_{max} = 1\,A$.The power dissipated by the combination is $P = I_{total}^2 	imes R_{eq} = 4^2 	imes 5 = 16 	imes 5 = 80\,W$.The power dissipated by each individual resistor is $P_{res} = I_{max}^2 	imes R = 1^2 	imes 10 = 10\,W$.Since the total power is the sum of power dissipated by each resistor,we have $n = P / P_{res} = 80 / 10 = 8$.Thus,a minimum of $8$ resistors is required.

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