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?
$\{\sin (90)^{\circ }=1\}$

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 fraction
The extra phase needed is $\dfrac{\pi}{3}$, which is $\dfrac{1}{6}$ of a full cycle $2\pi$.

Step 2: Result
So the time is one-sixth of the period, $\dfrac{T}{6}$.

Step 3: Final answer
One-sixth of a cycle is needed, so $t=\frac{T}{6}$.

Final Answer:
One-sixth of a cycle is needed, so $t=\frac{T}{6}$. \[ \boxed{\text{(B) }\dfrac{T}{6}} \]

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?
\(\{\sin (90)^{\circ }=1\}\)

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The Correct Option is B

Solution and Explanation

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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?\(\{\sin (90)^{\circ }=1\}\)