JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 20)
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.
The r.m.s. value of of this current is ................. A.
Answer
11
Explanation
$$f_{rms}^2 = f_{1\,rms}^2 + f_{2\,rms}^2$$
$$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$$
$$ = 121 \Rightarrow {f_{rms}}$$ = 11 A
$$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$$
$$ = 121 \Rightarrow {f_{rms}}$$ = 11 A
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