An alternating current is given by I=I_A sin ω t+I_B cos ω t. The r.m.s current will be

Solution

Physics sir · 3 steps

Combine sine and cosine into one sine wave first; then its rms is just the peak divided by root 2.

1. One sine wave. A sine and a cosine of the same ω add up to a single sine wave, I = I₀ sin(ω t + φ). The two terms are 90° out of phase, so their amplitudes add like perpendicular vectors.

   I₀ = √(I_A² + I_B²)

2. Check by averaging. Squaring gives I² = I_A²sin² ω t + I_B²cos² ω t + 2I_AI_B sin ω t cos ω t. Over one cycle, sin² and cos² each average to 1/2 and the cross term averages to 0, so ⟨ I² ⟩ = (I_A² + I_B²)/2.
3. So. Iᵣₘₛ = √(⟨ I² ⟩) = √((I_A² + I_B²)/2), which is option (b).

The idea to keep. For any single sinusoid, Iᵣₘₛ = I₀/√2. When sine and cosine terms are mixed, square, average over a cycle and let the cross term vanish.

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