If ${E_0}$ represents the peak value of the voltage in an ac circuit,the r.m.s. value of the voltage will be

The $r.m.s.$ value of an $AC$ of $50 \ Hz$ is $10 \ A$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

The instantaneous voltages at three terminals marked $X, Y$ and $Z$ are given by
$V_x = V_0 \sin \omega t$
$V_y = V_0 \sin \left(\omega t + \frac{2 \pi}{3}\right)$
$V_z = V_0 \sin \left(\omega t + \frac{4 \pi}{3}\right)$
An ideal voltmeter is configured to read the $rms$ value of the potential difference between its terminals. It is connected between points $X$ and $Y$ and then between $Y$ and $Z$. The reading$(s)$ of the voltmeter will be:
$[A]$ $V_{XY}^{rms} = V_0 \sqrt{\frac{3}{2}}$
$[B]$ $V_{YZ}^{rms} = V_0 \sqrt{\frac{1}{2}}$
$[C]$ $V_{XY}^{rms} = V_0$
$[D]$ independent of the choice of the two terminals

$A$ sinusoidal voltage produced by an $AC$ generator at any instant $t$ is given by the equation $V = 311 \sin(314t)$. The $rms$ value of voltage and frequency are respectively:

An electric bulb rated as $100 \ W-220 \ V$ is connected to an $ac$ source of $rms$ voltage $220 \ V$. The peak value of current through the bulb is: (in $A$)

The mean and $rms$ value of an alternating voltage for a half cycle,as shown in the figure,are respectively: