An $AC$ current is given by $I = I_0 + I_1 \sin \omega t$. Then its $rms$ value will be:

$AC$ measuring instruments measure

If an $AC$ main supply is given to be $220\,V$,the average $emf$ during a positive half cycle will be.....$V$.

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

$A$ resistance of $20 \Omega$ is connected to an alternating current source of $110 V$. If the frequency of the $A.C.$ source is $50 Hz$,then the time taken by the current to change from its maximum value to the $R.M.S.$ value is

$A$ bulb of resistance $280 \Omega$ is supplied with a $200 V$ $AC$ supply. What is the peak current?