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.
Answer (integer)
11
Solution
$f_{rms}^2 = f_{1\,rms}^2 + f_{2\,rms}^2$<br><br>$= {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$<br><br>$= 121 \Rightarrow {f_{rms}}$ = 11 A
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
This question is part of PrepWiser's free JEE Main question bank. 115 more solved questions on Alternating Current are available — start with the harder ones if your accuracy is >70%.