MHT CET · Physics · Alternating Current
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?
- A \(\frac{\mathrm{T}}{6}\)
- B \(\frac{\mathrm{T}}{3}\)
- C \(\frac{T}{2}\)
- D \(\frac{\mathrm{T}}{12}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{T}}{12}\)
Step-by-step Solution
Detailed explanation
\(\sin \left(\omega t+\frac{\pi}{3}\right)=1 \quad\) [For voltage to be maximum]
\(V=V_0 \sin \left(\omega t+\frac{\pi}{3}\right)\)
\(\begin{array}{ll}
\therefore \quad & \sin \left(\omega t+\frac{\pi}{3}\right)=\sin \left(\frac{\pi}{2}\right) \\
& \omega t+\frac{\pi}{3}=\frac{\pi}{2} \\
\therefore \quad & \omega t=\frac{\pi}{6} \\
& t=\frac{\pi}{6 \omega} \\
\therefore \quad & =\frac{\pi \times T}{6 \times 2 \pi} \\
& t=\frac{T}{12}
\end{array}\)
\(\ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}\right)\)
\(V=V_0 \sin \left(\omega t+\frac{\pi}{3}\right)\)
\(\begin{array}{ll}
\therefore \quad & \sin \left(\omega t+\frac{\pi}{3}\right)=\sin \left(\frac{\pi}{2}\right) \\
& \omega t+\frac{\pi}{3}=\frac{\pi}{2} \\
\therefore \quad & \omega t=\frac{\pi}{6} \\
& t=\frac{\pi}{6 \omega} \\
\therefore \quad & =\frac{\pi \times T}{6 \times 2 \pi} \\
& t=\frac{T}{12}
\end{array}\)
\(\ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}\right)\)
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