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 :

Solution

Physics sir · 3 steps

For heating, an AC behaves like a DC equal to its rms value, so square I₀/√2, not I₀.

1. Heat from DC. Joule heating is H = I² R t. For the steady current, H₁ = (4)² × 3 × t = 48t.
2. Use rms for AC. An alternating current heats a resistor like a steady current equal to its rms value. Here Iᵣₘₛ = I₀/√2 = 4/√2 A, so Iᵣₘₛ² = 8 A².

   H₂ = Iᵣₘₛ² R t = 8 × 2 × t = 16t

3. So. H₁ : H₂ = 48t : 16t = 3 : 1, which is option (b).

The idea to keep. Whenever an AC question asks about heat or average power, switch to the rms current first: Iᵣₘₛ = I₀/√2, so the heat is half of what the peak value would give.

Answer from the NTA final answer key. Solution by PrepWiser.