Question

The alternating voltage is given by \(V=V_{0}\sin \left(\omega t+\frac{\pi }{6}\right)\). When will the voltage be maximum for the first time?
\(\{\sin (90)^{\circ }=1\}\)

• A. \(\frac{T}{12}\)
• B. \(\frac{T}{6}\)
• C. \(\frac{T}{3}\)
• D. \(\frac{T}{2}\)

Hint:

The voltage is maximum when the phase first reaches a quarter of a cycle. Subtract the initial phase and convert the remaining phase to time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(V=V_{0}\sin\left(\omega t+\dfrac{\pi}{6}\right)\) is maximum when \(\omega t+\dfrac{\pi}{6}=\dfrac{\pi}{2}\).

Step 2: Calculation
\(\omega t=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{\pi}{3}\)
With \(\omega=\dfrac{2\pi}{T}\): \(t=\dfrac{\pi/3}{2\pi/T}=\dfrac{T}{6}\)

Step 3: Final answer
The voltage first becomes maximum at \(t=\frac{T}{6}\).

Final Answer:
The voltage first becomes maximum at \(t=\frac{T}{6}\). \[ \boxed{\text{(B) }\dfrac{T}{6}} \]