An inductor of reactance $100 \ \Omega$, a capacitor of reactance $50 \ \Omega$, and a resistor of resistance $50 \ \Omega$ are connected in series with an $AC$ source of $10 \ V, 50 \ Hz$. Average power dissipated by the circuit is . . . . . . $W$.

$A$ sinusoidal voltage with a frequency of $50 \ Hz$ is applied to a series $LCR$ circuit with a resistance of $5 \ \Omega$,inductance of $20 \ mH$ and a capacitance of $500 \ \mu F$. The magnitude of impedance of the circuit is close to (in $Omega$)

In a series $LR$ circuit,$X_L = 3R$. Now,a capacitor with $X_C = R$ is added in series. What is the ratio of the new power factor to the old power factor?

In a series $LR$ circuit,power of $400 \ W$ is dissipated from a source of $250 \ V, 50 \ Hz$. The power factor of the circuit is $0.8$. In order to bring the power factor to unity,a capacitor of value $C$ is added in series to the $L$ and $R$. Taking the value of $C$ as $(\frac{n}{3 \pi}) \mu F$,then the value of $n$ is $......$

When the $rms$ voltages $V_L, V_C$ and $V_R$ are measured respectively across the inductor $L$,the capacitor $C$ and the resistor $R$ in a series $LCR$ circuit connected to an $AC$ source,it is found that the ratio $V_L : V_C : V_R = 1 : 2 : 3$. If the $rms$ voltage of the $AC$ source is $100 \, V$,the $V_R$ is close to........$V$

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 $= 10 \sqrt{2} \ V$