The alternating voltage is given by $\mathrm{v}=\mathrm{v}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ when will be the voltage maximum for first time?

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Aicharya · Accepted Answer

$\sin \left(\omega t+\frac{\pi}{3}ight)=1 \quad$ [For voltage to be maximum] $V=V_0 \sin \left(\omega t+\frac{\pi}{3}ight)$ $\begin{array}{ll} 	herefore \quad & \sin \left(\omega t+\frac{\pi}{3}ight)=\sin \left(\frac{\pi}{2}ight) \ & \omega t+\frac{\pi}{3}=\frac{\pi}{2} \ 	herefore \quad & \omega t=\frac{\pi}{6} \ & t=\frac{\pi}{6 \omega} \ 	herefore \quad & =\frac{\pi 	imes T}{6 	imes 2 \pi} \ & t=\frac{T}{12} \end{array}$ $\ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}ight)$

The alternating voltage is given by $\mathrm{v}=\mathrm{v}_0 \sin \left(\omega…

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