$A$ 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 $^{\circ}$)

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........$V$

$A$ $750\, Hz$,$20\, V$ (rms) source is connected to a resistance of $100\, \Omega$,an inductance of $0.1803\, H$,and a capacitance of $10\, \mu F$,all in series. The time in which the resistance (heat capacity $2\, J/^{\circ}C$) will get heated by $10^{\circ}C$ (assume no loss of heat to the surroundings) is close to $.....s$.

$A$ series $LR$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2R$. $A$ capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$. The ratio of the new power factor to the old one is

In the circuit diagram shown,$X_C = 100 \, \Omega$,$X_L = 200 \, \Omega$,and $R = 100 \, \Omega$. The source voltage is $V = 200 \sin(\omega t) \, \text{V}$. The effective $(RMS)$ current through the source is:

An $LCR$ series circuit with $R = 100\,\Omega$ is connected to a $200\,V, 50\,Hz$ $a.c.$ source. When only the capacitance is removed,the current lags the voltage by $60^o$. When only the inductance is removed,the current leads the voltage by $60^o$. The current in the circuit is.....$A$