The alternating voltage is given by $v=v_0 \sin \left(\omega t+\frac{\pi}{3}\right)$. When will the voltage be maximum for the first time?
- A$\frac{T}{6}$
- B$\frac{T}{3}$
- C$\frac{T}{2}$
- D$\frac{T}{12}$
Explore More
Similar Questions
Match the following current $\text{r.m.s.}$ values:
$(A) \ x_0 \sin \omega t$ $(i) \ x_0$ $(B) \ x_0 \sin \omega t \cos \omega t$ $(ii) \ \frac{x_0}{\sqrt{2}}$ $(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ $(iii) \ \frac{x_0}{2 \sqrt{2}}$
| $(A) \ x_0 \sin \omega t$ | $(i) \ x_0$ |
| $(B) \ x_0 \sin \omega t \cos \omega t$ | $(ii) \ \frac{x_0}{\sqrt{2}}$ |
| $(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ | $(iii) \ \frac{x_0}{2 \sqrt{2}}$ |
Medium
View SolutionAn alternating voltage is given by $E = 100 \sin \left(\omega t + \frac{\pi}{6}\right) \text{ V}$. The voltage will be maximum for the first time when $t$ is $(T = \text{periodic time})$:
The output voltage of a step-down transformer is measured to be $24 \text{ V}$, when connected to a $12 \text{ W}$ light bulb. The value of the peak current is . . . . . . .
DifficultGUJCET 2025
View SolutionIn an $ac$ circuit,the $r.m.s.$ value of current,$I_{rms}$ is related to the peak current,$I_0$ by the relation:
Easy
View SolutionAn $AC$ source is rated $222 \,V, 60 \,Hz$. The average voltage is calculated in a time interval of $16.67 \,ms$. It
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo