The RMS value of emf given by E = (8 sin omeg | vedclass

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(A) The given emf is $E = 8 \sin \omega t + 6 \cos \omega t$. We can rewrite this in the form $E = E_0 \sin(\omega t + \phi)$. To do this,multiply and

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$5 \sqrt{2} 	ext{ V}$

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(A) The given emf is $E = 8 \sin \omega t + 6 \cos \omega t$. We can rewrite this in the form $E = E_0 \sin(\omega t + \phi)$. To do this,multiply and divide by $\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$. $E = 10 \left( \frac{8}{10} \sin \omega t + \frac{6}{10} \cos \omega t ight)$. Let $\cos \phi = \frac{8}{10} = 0.8$ and $\sin \phi = \frac{6}{10} = 0.6$. Then $E = 10 (\sin \omega t \cos \phi + \cos \omega t \sin \phi) = 10 \sin(\omega t + \phi)$. Comparing this with $E = E_0 \sin(\omega t + \phi)$,we get the peak value $E_0 = 10 	ext{ V}$. The $RMS$ value is given by $E_{RMS} = \frac{E_0}{\sqrt{2}}$. $E_{RMS} = \frac{10}{\sqrt{2}} = 5 \sqrt{2} 	ext{ V}$.

The $RMS$ value of emf given by $E = (8 \sin \omega t + 6 \cos \omega t) \text{ V}$ is

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