Question

An alternating voltage is \(E=100\,\sin \left(\omega t+\frac{\pi }{6}\right)\) V. The voltage will be maximum for the first time when t is
\(\left(\sin \frac{\pi }{2}=1\right)\) (T = Periodic time)

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

Hint:

The voltage is maximum when the sine term equals 1, that is when the phase equals pi/2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(E\) is maximum (100 V) when \(\sin\left(\omega t+\dfrac{\pi}{6}\right)=1\).

Step 2: Set the Phase
The first time this happens is when \(\omega t+\dfrac{\pi}{6}=\dfrac{\pi}{2}\), so \(\omega t=\dfrac{\pi}{3}\).

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

Final Answer:
The voltage first reaches its maximum at t = T/6. \[ \boxed{\text{(C) }\dfrac{T}{6}} \]