Q. An alternating current is given by $I=I_A\sin\omega t+I_B\cos\omega t$. The r.m.s current will be
A) $\frac{|I_A+I_B|}{\sqrt2}$
B) $\sqrt{\frac{I_A^2+I_B^2}2}$
C) $\sqrt{I_A^2+I_B^2}$
D) $\frac{\sqrt{I_A^2+I_B^2}}2$
Explanation
[Solution by Dr. Abubacker] $I=\sqrt{I_A^2+I_B^2}\,\sin(\omega t+\phi)$ has peak value $\sqrt{I_A^2+I_B^2}$, so $I_{rms}=\frac{\sqrt{I_A^2+I_B^2}}{\sqrt2}=\sqrt{\frac{I_A^2+I_B^2}2}$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.