"An alternating current is given by the equation i = i1 sin $\\omega$t + i2 cos $\\omega$t. The rms current will be :"

"${I_0} = \\sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\\cos \\theta }$ ${I_0} = \\sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\\cos 90^\\circ }$ ${I_0} = \\sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}(0)} = \\sqrt {I_1^2 + I_2^2}$ We know that, ${I_{rms}} = {{{I_0}} \\over {\\sqrt 2 }}$ So, ${I_{rms}} = {{\\sqrt {I_1^2 + I_2^2} } \\over {\\sqrt 2 }}$"

Medium MCQ +4 / -1 PYQ · JEE Mains 2021

An alternating current is given by the equation i = i₁ sin $\omega$t + i₂ cos $\omega$t. The rms current will be :

A ${1 \over {\sqrt 2 }}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}$ Correct answer
B ${1 \over {\sqrt 2 }}({i_1} + {i_2})$
C ${1 \over {\sqrt 2 }}{({i_1} + {i_2})^2}$
D ${1 \over 2}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}$

Solution

${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos \theta }$ ${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos 90^\circ }$ ${I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}(0)} = \sqrt {I_1^2 + I_2^2}$ We know that, ${I_{rms}} = {{{I_0}} \over {\sqrt 2 }}$ So, ${I_{rms}} = {{\sqrt {I_1^2 + I_2^2} } \over {\sqrt 2 }}$

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Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C

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An alternating current is given by the equation i = i1 sin $\omega$t + i2 cos $\omega$t. The rms current will be :

Solution

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An alternating current is given by the equation i = i₁ sin $\omega$t + i₂ cos $\omega$t. The rms current will be :