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:

In an $AC$ circuit, the instantaneous current is zero when the instantaneous voltage is maximum. In this case, the source may be connected to:
$A$. Pure inductor.
$B$. Pure capacitor.
$C$. Pure resistor.
$D$. Combination of an inductor and capacitor.
Choose the correct answer from the options given below:

When an inductor of inductance $L = \frac{6}{\pi} \ H$, a capacitor of capacitance $C = \frac{50}{\pi} \ \mu F$ and a resistor of resistance $R$ are connected in series with an $AC$ supply of rms voltage $V_{rms} = 220 \ V$ and frequency $f = 50 \ Hz$, the rms current through the circuit is $I_{rms} = 440 \ mA$. Match the inductive reactance $X_L$, the capacitive reactance $X_C$, the resistance $R$, and the impedance $Z$ of the circuit given in List-$I$ with the corresponding values given in List-$II$.

List-$I$	List-$II$
$(A) \ X_L$	$(i) \ 200 \ \Omega$
$(B) \ X_C$	$(ii) \ 300 \ \Omega$
$(C) \ R$	$(iii) \ 500 \ \Omega$
$(D) \ Z$	$(iv) \ 600 \ \Omega$

An $LCR$ circuit is equivalent to a damped oscillator. $A$ capacitor is charged to a charge $Q_0$ as shown and then connected to $L$ and $R$. If a student plots graphs of the square of the maximum charge on the capacitor $(Q^2_{max})$ versus time for two different values of inductors $L_1$ and $L_2$ $(L_1 > L_2)$, which of the following graphs will correctly represent it? (The diagram is schematic and not drawn to scale.)

$A$ coil has inductance of $0.4 \text{ H}$ and resistance of $8 \Omega$. It is connected to an $AC$ source with peak emf $4 \text{ V}$ and frequency $\frac{30}{\pi} \text{ Hz}$. The average power dissipated in the circuit is (in $\text{ W}$)

In the circuit shown, the charge $q$ varies with time $t$ as $q = t^2 - 5$, where $q$ is in coulomb and $t$ is in second. At time $t = 3$ second, the voltage $V_{AB}$ in volt will be: