$(a)$ The peak voltage of an $ac$ supply is $300 \; V$. What is the $rms$ voltage?
$(b)$ The $rms$ value of current in an $ac$ circuit is $10 \; A$. What is the peak current?

The alternating current is given by $i = \left\{\sqrt{42} \sin \left(\frac{2 \pi}{T} t\right) + 10\right\} \text{ A}$. The $r.m.s.$ value of this current is $\text{A}$.

For an $AC$ current $I = 50 \cos(100t + 45^{\circ}) \ A$. The value of $I_{rms} =$ . . . . . . $A$.

For a domestic $AC$ supply of $220 \ V$ at $50 \ \text{cycles per sec}$, the potential difference between the terminals of a two-pin electric outlet in a room is given by

An $AC$ current is given by $I = I_{1} \sin \omega t + I_{2} \cos \omega t$. $A$ hot wire ammeter will give a reading of:

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