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 $\Omega$?

In the given $LCR$ $AC$ circuit,the effective current flowing through the circuit will be:

$A$ circuit has a resistance of $11\,\Omega$,an inductive reactance of $25\,\Omega$,and a capacitive reactance of $18\,\Omega$. It is connected to an $AC$ source of $260\,V$ and $50\,Hz$. The current through the circuit (in amperes) is:

In the circuit shown,$C = \frac{\sqrt{3}}{2} \times 10^{-3} \, F$,$R_2 = 20 \, \Omega$,$L = \frac{\sqrt{3}}{10} \, H$,and $R_1 = 10 \, \Omega$. The current in the $L-R_1$ branch is $I_1$ and in the $C-R_2$ branch is $I_2$. The voltage of the $A.C.$ source is given by $V = 200\sqrt{2} \sin(100t) \, V$. The phase difference between $I_1$ and $I_2$ is:

If a resistance of $100\, \Omega$,an inductance of $0.5\, H$,and a capacitance of $10 \times 10^{-6}\, F$ are connected in series to a $50\, Hz$ $AC$ supply,then the impedance is:

$A$ series $R-L-C$ circuit consists of an $8.0\ \Omega$ resistor,a $5.0\ \mu F$ capacitor,and a $50.0\ mH$ inductor. $A$ variable frequency source applies an $emf$ of $400\ V\ (rms)$ across the combination. The power delivered to the circuit when the frequency is equal to one-half the resonance frequency is.....$W$