An $LCR$ series circuit with $100 \,\Omega$ resistance is connected to an $AC$ source of $200 \,V$ and angular frequency $300 \,rad/s$. When only the capacitance is removed,the current leads the voltage by $60^o$. When only the inductance is removed,the current lags the voltage by $60^o$. Then the current and power dissipated in the $LCR$ circuit are respectively:

$A$ series $LR$ circuit is connected to a voltage source with $V(t) = V_0 \sin \omega t$. After a very large time,how does the current $I(t)$ behave? (Given: $t_0 \gg \frac{L}{R}$)

Three alternating voltage sources $V_1 = 3 \sin \omega t \text{ V}$,$V_2 = 5 \sin(\omega t + \phi_1) \text{ V}$,and $V_3 = 5 \sin(\omega t - \phi_2) \text{ V}$ are connected in series with a resistor $R = \sqrt{\frac{7}{3}} \, \Omega$ as shown in the figure (where $\phi_1 = 30^\circ$ and $\phi_2 = 127^\circ$). Find the peak current (in Ampere) through the resistor.

An $AC$ source of $20\,V$ (assuming standard frequency $f = 50\,Hz$ for calculation) is connected to an inductance of $100\,mH$,a capacitance of $100\,\mu F$,and a resistance of $120\,\Omega$ as shown in the figure. The time in which the resistance,having a thermal capacity of $2\,J/^{\circ}C$,will get heated by $16^{\circ}C$ is ..........$s$.

Consider the infinite ladder circuit shown below. For which angular frequency $\omega$ will the circuit behave like a pure inductance?

In the circuit shown,$L = 1 \mu H$,$C = 1 \mu F$,and $R = 1 k\Omega$. They are connected in series with an $a.c.$ source $V = V_0 \sin \omega t$ as shown. Which of the following options is/are correct?
[$A$] The frequency at which the current will be in phase with the voltage is independent of $R$.
[$B$] At $\omega \sim 0$,the current flowing through the circuit becomes nearly zero.
[$C$] At $\omega \gg 10^6 \text{ rad } s^{-1}$,the circuit behaves like a capacitor.
[$D$] The current will be in phase with the voltage if $\omega = 10^6 \text{ rad } s^{-1}$.