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.

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 $\omega = 10^6 \text{ rad } s^{-1}$.

Match List-$I$ with List-$II$:

List-$I$	List-$II$
$A$. $AC$ generator	$I$. Detects the presence of current in the circuit
$B$. Galvanometer	$II$. Converts mechanical energy into electrical energy
$C$. Transformer	$III$. Works on the principle of resonance in $AC$ circuit
$D$. Metal detector	$IV$. Changes an alternating voltage for smaller or greater value

Choose the correct answer from the options given below:

In a series $LR$ circuit with $X_L = R$. Power factor is $P_1$. If a capacitor of capacitance $C$ with $X_c = X_L$ is added to the circuit the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be :

At very high frequencies,the effective impedance of the given circuit will be $.... \Omega$.

$A$ 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: $t_0 \gg \frac{L}{R}$)