The alternating voltage is given by v=v_0 sin | vedclass

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(D) For the voltage to be maximum,the sine function must equal $1$.$\sin \left(\omega t+\frac{\pi}{3}\right) = 1$Since $\sin \left(\frac{\pi}{2}\right

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$\frac{T}{12}$

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(D) For the voltage to be maximum,the sine function must equal $1$.$\sin \left(\omega t+\frac{\pi}{3}ight) = 1$Since $\sin \left(\frac{\pi}{2}ight) = 1$,we have:$\omega t + \frac{\pi}{3} = \frac{\pi}{2}$$\omega t = \frac{\pi}{2} - \frac{\pi}{3}$$\omega t = \frac{\pi}{6}$Substituting $\omega = \frac{2\pi}{T}$:$\left(\frac{2\pi}{T}ight) t = \frac{\pi}{6}$$t = \frac{\pi}{6} 	imes \frac{T}{2\pi}$$t = \frac{T}{12}$

The alternating voltage is given by $v=v_0 \sin \left(\omega t+\frac{\pi}{3}\right)$. When will the voltage be maximum for the first time?

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