An inductance of $\left(\frac{200}{\pi}\right) \text{mH}$,a capacitance of $\left(\frac{10^{-3}}{\pi}\right) \text{F}$,and a resistance of $10 \, \Omega$ are connected in series with an $AC$ source of $220 \, \text{V}, 50 \, \text{Hz}$. The phase angle of the circuit is:

An inductor of inductance $L$,a capacitor of capacitance $C$,and a resistor of resistance $R$ are connected in series to an $AC$ source of potential difference $V$ volts as shown in the figure. The potential difference across $L$,$C$,and $R$ is $40 \, V$,$10 \, V$,and $40 \, V$,respectively. The amplitude of the current flowing through the $LCR$ series circuit is $10 \sqrt{2} \, A$. The impedance of the circuit is .......... $\Omega$.

$A$ $220 \; V, 50 \; Hz$ $AC$ source is connected to a $25 \; V, 5 \; W$ lamp and an additional resistance $R$ in series (as shown in the figure) to run the lamp at its rated power. The value of $R$ (in $\Omega$) will be:

Consider the $L-C-R$ circuit given below. The circuit is driven by a $50 \,Hz, AC$ source with peak voltage $220 \,V$. If $R=400 \,\Omega, C=200 \,\mu F$ and $L=6 \,H$,the maximum current in the circuit is closest to ............ $A$.

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 $= 10 \sqrt{2} \ V$

An inductance of $\frac{300}{\pi} \text{ mH}$,a capacitance of $\frac{1}{\pi} \text{ mF}$ and a resistance of $20 \ \Omega$ are connected in series with an a.c. source of $240 \text{ V}, 50 \text{ Hz}$. The phase angle of the circuit is