QJEEM21Ph5d33: JEE Main 2021 Physics - Class 12 JEEM21Ph5d33 Solution: The alternating current is given by The r.m.s. value of of this …

Question

Verified Solution • 2025–26 Session moderate

The alternating current is given by

i = \left\{ {\sqrt {42} \sin \left( {{{2\pi } \over T}t} \right) + 10} \right\}A

The r.m.s. value of of this current is ................. A.

Accepted Answer

f_{rms}^2 = f_{1\,rms}^2 + f_{2\,rms}^2

= {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}

= 121 \Rightarrow {f_{rms}}

= 11 A

Step-by-Step Verified Solution