$A$ 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 an $LCR$ circuit,$R = 100 \ \Omega$. When the capacitor $C$ is removed,the current lags behind the voltage by a phase of $\pi / 3$. When the inductor $L$ is removed,the current leads the voltage by a phase of $\pi / 3$. What is the impedance of the circuit in $\Omega$?

$A$ resistor of resistance $40 \Omega$,a capacitor of capacitive reactance $20 \Omega$,and an inductor of inductive reactance $50 \Omega$ are connected in series to an $ac$ source of $100 \ V$. The current through the circuit is (in $A$)

What will be the impedance of the circuit shown below in $\Omega$?

In an $LCR$ series $ac$ circuit,the voltage across each of the components,$L, C$ and $R$ is $50\,V$. The voltage across the $LC$ combination will be........$V$.

$A$ series $LCR$ circuit consists of an inductor $L$,a capacitor $C$,and a resistor $R$ connected across a source of emf $\varepsilon = \varepsilon_0 \sin \omega t$. When $\omega L = \frac{1}{\omega C}$,the current in the circuit is $I_0$. If the angular frequency of the source is changed to $\omega^{\prime}$,the current in the circuit becomes $\frac{I_0}{2}$. Then,the value of $\left|\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right|$ is