In $ac$ circuit when $ac$ ammeter is connected it reads $i$ current if a student uses $dc$ ammeter in place of $ac$ ammeter the reading in the $dc$ ammeter will be:

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 process by which $ac$ is converted into $dc$ is known as

An alternating voltage $\mathrm{V}(\mathrm{t})=220 \sin 100 \ \pi \mathrm{t}$ 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 to the peak value is:

The effective value of current $i = 2\, sin\, 100\, \pi\, t + 2 \,sin(100\, \pi \,t + 30^o)$ is :

If a direct current of $'a'$ amp is superimposed with an alternating current $'I = b\,sin \,\omega t',$ then effective value of resulting current is