A periodic voltage $V$ varies with time $t$ as shown in the figure. $T$ is the time period. The $r.m.s$. value of the voltage is :-

Three alternating voltage sources $V_1$ = $3 sin \omega t $ volt , $V_2= 5  sin(\omega t + \phi _1)$ volt and $V_3 = 5  sin(\omega t -\phi_2 )$ volt connected across a resistance $R= \sqrt {\frac{7}{3}} \Omega $ as shown in the figure (where $ \phi_1$  and $ \phi_2$   corresponds to $30^o $ and $127^o $ respectively). Find the peak current (in Amp) through the resistor

The peak value of an Alternating current is $ 6$ amp, then r.m.s. value of current will be

If instantaneous current is given by $i = 4\cos \,(\omega \,t + \phi )$ amperes, then the $r.m.s$. value of current is

An alternating voltage $v\left( t \right) = 220\,\sin \,100\pi l\,volt$ is applied to a purely resistive load of $50\,\Omega $ . The time taken for the current to rise from half of the peak value of the peak value is.....$ms$

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