An alternating voltage is given by E = 100 si | vedclass

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(C) Given,the alternating voltage is $E = 100 \sin \left(\omega t + \frac{\pi}{6}ight) 	ext{ V}$.For the voltage to be maximum,the sine term must b

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$\frac{T}{6}$

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(C) Given,the alternating voltage is $E = 100 \sin \left(\omega t + \frac{\pi}{6}ight) 	ext{ V}$.For the voltage to be maximum,the sine term must be equal to $1$.$\sin \left(\omega t + \frac{\pi}{6}ight) = 1$Since $\sin \frac{\pi}{2} = 1$,we have:$\omega t + \frac{\pi}{6} = \frac{\pi}{2}$$\omega t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$Using the relation $\omega = \frac{2\pi}{T}$,we substitute $\omega$:$\left(\frac{2\pi}{T}ight) t = \frac{\pi}{3}$$t = \frac{\pi}{3} 	imes \frac{T}{2\pi} = \frac{T}{6}$Thus,the voltage will be maximum for the first time at $t = \frac{T}{6}$.

An alternating voltage is given by $E = 100 \sin \left(\omega t + \frac{\pi}{6}\right) \text{ V}$. The voltage will be maximum for the first time when $t$ is $(T = \text{periodic time})$:

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