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

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Aicharya · Accepted Answer

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

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…

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