In the given circuit,the $AC$ source has $\omega = 100 \, rad \, s^{-1}$. Considering the inductor and capacitor to be ideal,what will be the current $I$ flowing through the circuit? (in $A$)

In an electrical circuit,$R, L, C$ and an $a.c.$ voltage source are all connected in series. When $L$ is removed from the circuit,the phase difference between the voltage and the current in the circuit is $\pi /3$. If instead,$C$ is removed from the circuit,the phase difference is again $\pi /3$. The power factor of the circuit is

What is the reason we prefer $A.C.$ voltage over $D.C.$ voltage for power transmission?

Assertion : In the purely resistive element of a series $LCR$ $AC$ circuit,the maximum value of $rms$ current increases with an increase in the angular frequency of the applied $e.m.f$.
Reason : $I_{\max} = \frac{\varepsilon_{\max}}{Z}$,where $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$ and $I_{\max}$ is the peak current in a cycle.

$A$ $\frac{2.5}{\pi} \mu F$ capacitor and $3000 \, \Omega$ resistor are joined in series to an $AC$ source of $200 \, V$ and $50 \, Hz$ frequency. The power factor of the circuit and the power dissipated in it will respectively be:

There is a $100\,\Omega$ resistance in an $L-C-R$ $AC$ circuit. An $AC$ $emf$ of $200\,V$ and $\omega = 300\,rad/s$ is applied to this circuit. When only the capacitor is removed,the current lags the voltage by $60^o$. When only the inductor is removed,the current leads the voltage by $60^o$. The current in this $L-C-R$ circuit will be.....$A$