As shown in the figure,a series circuit connected across a $200\, V$,$60\, Hz$ line consists of a capacitor of capacitive reactance $30\,\Omega$,a non-inductive resistor of $44\,\Omega$,and a coil of inductive reactance $90\,\Omega$ and resistance $36\,\Omega$. The power dissipated in the coil is......$W$.

In a series $LCR$ circuit, $R = 10 \ \Omega$ and the impedance $Z = 30 \ \Omega$. An $r.m.s.$ voltage of $210 \text{ V}$ is applied across the circuit. The true power consumed in the $AC$ circuit is: (in $\text{ W}$)

In an $R-L-C$ series circuit, the potential difference across each element is $20 \, V$. If the value of the resistance $R$ is doubled, what will be the potential difference across $R, L$, and $C$ respectively?

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 $= 10 \sqrt{2} \ V$

$A$ 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 $^{\circ}$)

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: