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: (a) 3: 2 (b) 3: 1 (c) 3: 4 (d) 4: 3

Answer: (b). Heat H = Irms² R t; for dc Irms = 4 A, for ac Irms² = 4²/2 = 8. H₁ = 16 × 3 t = 48t, H₂ = 8 × 2 t = 16t H₁: H₂ = 48: 16 = 3: 1 So the answer is (b) 3: 1.

A direct current of 4 A and an alternating current of peak value 4 A… | JEE Main 2022 (27 Jul, Shift 1)

Q25JEE Main 2022 · 27 JUL · SHIFT 1Easy

AC Basics: Peak, RMS & Average Values

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

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Answer: (b)

Heat $H = I_{rms}^2 R t$ ; for dc $I_{rms} = 4$ A, for ac $I_{rms}^2 = \frac{4^2}{2} = 8$ .
$H_1 = 16 \times 3\, t = 48t$ , $H_2 = 8 \times 2\, t = 16t$
$H_1 : H_2 = 48 : 16 = 3 : 1$
So the answer is (b) 3 : 1.

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