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A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 $Ω$ and 2 $Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be :

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3 : 2

3 : 1

3 : 4

4 : 3

answer is B.

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Detailed Solution

$\begin{array}{l} H_{1} = i^{2} R_{1} t H_{2} = i_{rms}^{2} R_{2} t \{i_{rms} = \frac{i_{0}}{\sqrt{2}}\} \\ H_{1} = 16 (3) t H_{2} = \frac{i_{0}^{2}}{2} R_{2} t \\ H_{2} = 16 t \\ H_{1} : H_{2} = 3 : 1 \end{array}$

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