The peak value of an alternating e.m.f. $E$ given by $E = E_0 \cos \omega t$ is $10 \ V$ and its frequency is $50 \ Hz$. At time $t = \frac{1}{600} \ s$,the instantaneous e.m.f. is:

The frequency of $AC$ mains in India is ....... $Hz$.

The instantaneous values of current and emf in an $AC$ circuit are $I = 1/\sqrt{2} \sin(314t) \, A$ and $E = \sqrt{2} \sin(314t - \pi/6) \, V$ respectively. The phase difference between $E$ and $I$ 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

Alternating voltage:

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