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)

Question

In an a.c. circuit, voltage and current are given by: \[ V = 100 \sin(100t) \, \text{V} \quad \text{and} \quad I = 100 \sin\left(100t + \frac{\pi}{3}\right) \, \text{mA}. \] The average power dissipated in one cycle is:

Answer 1

Step 1: Phase Difference
The phase must increase from $\dfrac{\pi}{6}$ to $\dfrac{\pi}{2}$, an increase of $\dfrac{\pi}{3}$.

Step 2: Fraction of a Cycle
A full cycle is $2\pi$, so $\dfrac{\pi/3}{2\pi}=\dfrac{1}{6}$ of a cycle.

Step 3: Time
That fraction takes $\dfrac{T}{6}$, which is the third option.

Final Answer:
The first maximum occurs at T/6. \[ \boxed{\text{(C) }\dfrac{T}{6}} \]

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)

Show Hint

The Correct Option is C

Solution and Explanation

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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)