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.....$ms$

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

The average value of potential difference $V$ shown in figure is

The $AC$ voltage across a resistance can be measured using a

Match the following

Currents                                     $r.m.s.$ values

(1)${x_0}\sin \omega \,t$                               (i)$ x_0$

(2)${x_0}\sin \omega \,t\cos \omega \,t$                  (ii)$\frac{{{x_0}}}{{\sqrt 2 }}$

(3)${x_0}\sin \omega \,t + {x_0}\cos \omega \,t$        (iii) $\frac{{{x_0}}}{{(2\sqrt 2 )}}$

An $ac$ source is rated at $220V, 50 Hz.$ The time taken for voltage to change from its peak value to zero is.....$sec$