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, $\cos 0^{\circ }=\sin 90^{\circ }=1$)

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 shift view
The voltage already starts at a phase of $\dfrac{\pi}{4}$ and needs another $\dfrac{\pi}{4}$ to reach $\dfrac{\pi}{2}$.

Step 2: Convert
A phase of $2\pi$ takes time $T$, so $\dfrac{\pi}{4}$ takes $\dfrac{T}{8}$.

Step 3: Result
The first maximum occurs at $t=\dfrac{T}{8}$.

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

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, \(\cos 0^{\circ }=\sin 90^{\circ }=1\))

Show Hint

The Correct Option is C

Solution and Explanation

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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, \(\cos 0^{\circ }=\sin 90^{\circ }=1\))