Question

An alternating voltage is given by \(V=V_{0}\sin \left(\omega t+\frac{\pi }{4}\right)\). The time when the voltage is maximum for the first time is
(\(T\) = periodic time of the alternating voltage, \(\cos 0^{\circ }=\sin 90^{\circ }=1\))

• A. \(\frac{T}{4}\)
• B. \(\frac{T}{6}\)
• C. \(\frac{T}{8}\)
• D. \(\frac{T}{12}\)

Hint:

The voltage is maximum when the sine of the phase equals one, so the phase must equal pi over 2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(V=V_0\) when \(\sin\left(\omega t+\dfrac{\pi}{4}\right)=1\).

Step 2: Find the Phase
The first time this happens is when \(\omega t+\dfrac{\pi}{4}=\dfrac{\pi}{2}\), so \(\omega t=\dfrac{\pi}{4}\).

Step 3: Calculation
With \(\omega=\dfrac{2\pi}{T}\): \(t=\dfrac{\pi}{4}\times\dfrac{T}{2\pi}=\dfrac{T}{8}\)

Step 4: Compare the options
This is the third option.

Final Answer:
The voltage first becomes maximum at T/8. \[ \boxed{\text{(C) }\dfrac{T}{8}} \]