Q154 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Physics

If the emf of an AC source is given by $6 \sin \omega t+4 \sin 2 \omega t \mathrm{~V}$, then the rms value of the emf is

  • A
    $\sqrt{10} \mathrm{~V}$
  • Correct answer
    B
    $\sqrt{26} \vee$
  • C
    $\sqrt{32} \mathrm{~V}$
  • D
    $\sqrt{20} \mathrm{~V}$

Solution

Given, emf of an AC source, $$ \begin{aligned} & \quad E=6 \sin \omega t+4 \sin 2 \omega t \\ & \text { Now, } \quad E^2=(6 \sin \omega t+4 \sin 2 \omega t)^2 \\ & \therefore E^2=36 \sin ^2 2 \omega t+16 \sin ^2 2 \omega t+48 \sin \omega t \cdot \sin 2 \omega t \\ & =36 \sin ^2 \omega t+16 \sin ^2 2 \omega t+48\left(\cos \frac{\omega t}{2}-\cos \frac{3 \omega t}{2}\right) \end{aligned} $$ Now, the average value of $E^2$, $$ \begin{aligned} & \therefore \quad E_{\text {avg }}=\frac{1}{T} \int_0^T\left[E^2 d t\right] \\ & =\frac{1}{T} \int_0^T\left[36 \sin ^2 \omega t+16 \sin ^2 2 \omega t+48\left(\frac{\cos \omega t}{2}-\frac{\cos 3 \omega t}{2}\right)\right] \\ & =\frac{1}{T}\left[36 \cdot \frac{T}{2}+16 \times \frac{T}{2}+48(0-0)\right] \\ & \text { or } \quad E_{\text {avg }}=18+8=26 \mathrm{~V} \\ & \therefore \quad E_{\mathrm{rms}}=\sqrt{E_{\mathrm{avg}}}=\sqrt{26} \mathrm{~V} \\ & \end{aligned} $$ So, the rms value of emf is $\sqrt{26} \mathrm{~V}$.

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

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