In an $A.C.$ circuit,the instantaneous current and $E.M.F.$ are represented as $i = i_0 \sin(\omega t - \frac{\pi}{6})$ and $E = E_0 \sin(\omega t + \frac{\pi}{3})$ respectively. The voltage leads the current by

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 $\text{A}$.

An alternating current is given by $i = 2 \sin \omega t + 6 \cos \omega t$. The $R.M.S.$ current in amperes is,

$A$ $280 \ \Omega$ electric bulb is connected to a $200 \ V$ electric line. The peak value of current in the bulb 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

The output of a step-down transformer is measured to be $24 \ V$,when connected to a $12 \ W$ light bulb. The value of peak current is $...... \ A$. (in $A$)