The rms value of the emf given by E = (8 sin | vedclass

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Question

(A) The given emf is $E = 8 \sin \omega t + 6 \cos \omega t$. We can express this in the form $E = A \sin(\omega t + \phi)$,where $A$ is the peak ampl

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

Vedclass · Answer

(A) The given emf is $E = 8 \sin \omega t + 6 \cos \omega t$. We can express this in the form $E = A \sin(\omega t + \phi)$,where $A$ is the peak amplitude. Comparing $8 \sin \omega t + 6 \cos \omega t$ with $A \sin(\omega t + \phi) = A \sin \omega t \cos \phi + A \cos \omega t \sin \phi$,we get: $A \cos \phi = 8$ and $A \sin \phi = 6$. Squaring and adding these equations: $A^2(\cos^2 \phi + \sin^2 \phi) = 8^2 + 6^2$ $A^2 = 64 + 36 = 100$ $A = 10 	ext{ V}$. The rms value $E_{rms}$ is related to the peak value $A$ by the formula $E_{rms} = \frac{A}{\sqrt{2}}$. $E_{rms} = \frac{10}{\sqrt{2}} = \frac{10 	imes \sqrt{2}}{2} = 5 \sqrt{2} 	ext{ V}$.

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

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