JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 2)
An alternating current is given by the equation i = i1 sin $$\omega$$t + i2 cos $$\omega$$t. The rms current will be :
$${1 \over {\sqrt 2 }}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}$$
$${1 \over {\sqrt 2 }}({i_1} + {i_2})$$
$${1 \over {\sqrt 2 }}{({i_1} + {i_2})^2}$$
$${1 \over 2}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}$$
Explanation
$${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos \theta } $$
$${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos 90^\circ } $$
$${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}(0)} = \sqrt {I_1^2 + I_2^2} $$
We know that,
$${I_{rms}} = {{{I_0}} \over {\sqrt 2 }}$$
So, $${I_{rms}} = {{\sqrt {I_1^2 + I_2^2} } \over {\sqrt 2 }}$$
$${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos 90^\circ } $$
$${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}(0)} = \sqrt {I_1^2 + I_2^2} $$
We know that,
$${I_{rms}} = {{{I_0}} \over {\sqrt 2 }}$$
So, $${I_{rms}} = {{\sqrt {I_1^2 + I_2^2} } \over {\sqrt 2 }}$$
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