Verified Solution • 2025–26 Session moderate

QJEEM21Phf075: JEE Main 2021 Physics - Class 12 JEEM21Phf075 Solution: An alternating current is given by the equation i = i 1 sin t + …

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

Multiple Choice Options

  • A 12(i12+i22)12{1 \over {\sqrt 2 }}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}
  • B 12(i1+i2){1 \over {\sqrt 2 }}({i_1} + {i_2})
  • C 12(i1+i2)2{1 \over {\sqrt 2 }}{({i_1} + {i_2})^2}
  • D 12(i12+i22)12{1 \over 2}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}

Step-by-Step Verified Solution

I0=I12+I22+2I1I2cos⁡θ{I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos \theta }

I0=I12+I22+2I1I2cos⁡90∘{I_0} = \sqrt {I_1^2 + I_2^2 + 2{I_1}{I_2}\cos 90^\circ }

I0=I12+I22+2I1I2(0)=I12+I22{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,

Irms=I02{I_{rms}} = {{{I_0}} \over {\sqrt 2 }}

So, Irms=I12+I222{I_{rms}} = {{\sqrt {I_1^2 + I_2^2} } \over {\sqrt 2 }}

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