In a series $LCR$ circuit,the inductance,capacitance,and resistance are $L = 100 \, mH$,$C = 100 \, \mu F$,and $R = 10 \, \Omega$ respectively. They are connected to an $AC$ source of voltage $220 \, V$ and frequency $50 \, Hz$. The approximate value of current in the circuit will be.....$A$.

To an $AC$ power supply of $220 \ V$ at $50 \ Hz$,a resistor of $20 \ \Omega$,a capacitor of reactance $25 \ \Omega$,and an inductor of reactance $45 \ \Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage are,respectively:

The figure shows an $LCR$ series $AC$ circuit. If $L$ is removed from the circuit,the current leads the voltage by $45^o$,while if $C$ is removed,the current lags the voltage by $45^o$. The current passing in the original circuit is......$A$.

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:

In a series $RLC$ circuit,the $r.m.s.$ voltages across the resistor and the inductor are respectively $400 \,V$ and $700 \,V$. If the equation for the applied voltage is $\varepsilon = 500 \sqrt{2} \sin \omega t$,then the peak voltage across the capacitor is ........... $V$.

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?