The peak voltage of the $ac$ source is equal to:

  • A
    the $rms$ value of the $ac$ source
  • B
    $\sqrt{2}$ times the $rms$ value of the $ac$ source
  • C
    $\frac{1}{\sqrt{2}}$ times the $rms$ value of the $ac$ source
  • D
    the value of voltage supplied to the circuit

Explore More

Similar Questions

$A$ $40 \ \Omega$ electric heater is connected to a $200 \ V, 50 \ Hz$ mains supply. The peak value of electric current flowing in the circuit is approximately......$A$

The mean value of current for half cycle for a current variation shown by the graph is

The average value of an $A$.$C$. voltage given by $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is:

$A$ resistance of $20 \, \Omega$ is connected to a source of an alternating potential $V = 220 \sin(100 \pi t)$. The time taken by the current to change from its peak value to its r.m.s. value is:

An alternating voltage is given by: $e = e_1 \sin \omega t + e_2 \cos \omega t$. Then the root mean square $(RMS)$ value of the voltage is given by:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo