At very high frequencies, the current $(i)$ in the given circuit is (in $A$)

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 $A.C.$ circuit,the instantaneous $e.m.f.$ and current are given by:
$e = 100 \sin(20t)$
$i = 20 \sin(30t - \frac{\pi}{4})$
In one cycle of $A.C.$,the average power consumed by the circuit and the wattless current are,respectively:

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 in the figure,the $ac$ source gives a voltage $V = 20\cos(2000t)$. Neglecting source resistance,the voltmeter and ammeter readings will be:

The following figure shows an $AC$ generator connected to a 'black box' through a pair of terminals. The box contains possible $R, L, C$ or their combination,whose elements and arrangements are not known to us. Measurements outside the box reveal that $e = 75 \sin(\omega t) \text{ V}$ and $i = 1.5 \sin(\omega t + 45^\circ) \text{ A}$. Then,the wrong statement is: