Q179 / 180 · KCET 2022 · Physics

An alternating current is given by $i=i_1 \sin \omega t+i_2 \cos \omega t$. The rms current is given by

  • A
    $\frac{i_1-i_2}{\sqrt{2}}$
  • Correct answer
    B
    $\sqrt{\frac{i_1^2+i_2^2}{2}}$
  • C
    $\sqrt{\frac{i_1^2+i_2^2}{\sqrt{2}}}$
  • D
    $\frac{i_1+i_2}{\sqrt{2}}$

Solution

Equation of alternating current, $$ \begin{aligned} & i=i_1 \sin \omega t+i_2 \cos \omega t \\ & \text { Let } i_1=I \cos \phi \\ & \text { and } i=I \sin \phi \\ & i=I \cos \phi \sin \omega t+I \sin \phi \cos \omega t \\ & =I(\sin \omega t \cos \phi+\cos \omega t \sin \phi) \\ & \Rightarrow \quad i=I \sin (\omega t+\phi) \\ & \therefore \text { rms value of } i_1 \text {, } \\ & i_{\mathrm{rms}}=\frac{I}{\sqrt{2}} \\ & \end{aligned} $$ From Eqs. (i) and (ii), we get $$ \begin{gathered} i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \\ =I^2 \\ \Rightarrow \quad \sqrt{i_1^2+i_2^2}=I \end{gathered} $$ $$ i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \sin ^2 \phi=I^2\left(\cos ^2 \phi+\sin ^2 \phi\right) $$ From Eqs. (iii) and (iv), we get $$ i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}} $$

Hence, the correct answer is option: 'B'.

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