In the series $L-C-R$ circuit shown,the impedance is (in $\Omega$)

An a.c. source is applied to a series $LR$ circuit with $X_L = 3R$ and the power factor is $X_1$. Now,a capacitor with $X_C = R$ is added in series to the $LR$ circuit and the power factor is $X_2$. The ratio $X_1$ to $X_2$ is

In the given circuit,the $AC$ source has $\omega = 100 \ rad/s$. Considering the inductor and capacitor to be ideal,the correct choice$(s)$ is(are):
$(A)$ The current through the circuit,$I$ is $0.3 \ A$
$(B)$ The current through the circuit,$I$ is $0.3 \sqrt{2} \ A$
$(C)$ The voltage across $100 \ \Omega$ resistor $= 10 \sqrt{2} \ V$
$(D)$ The voltage across $50 \ \Omega$ resistor $= 10 \sqrt{2} \ V$

With an increase in the frequency of an $A.C.$ supply,the impedance of an $L-C-R$ series circuit

The reading of the ammeter in the circuit is ............ $A$.

$A$ capacitance of $\left(\frac{10^{-3}}{2 \pi}\right) F$,an inductance of $\left(\frac{100}{\pi}\right) mH$,and a resistance of $10 \Omega$ are connected in series with an $AC$ voltage source of $220 V, 50 Hz$. The phase angle of the circuit is (in $^{\circ}$)