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 inductor of $10 \text{ mH}$, capacitor of $0.1 \text{ } \mu\text{F}$ and a resistor of $100 \text{ } \Omega$ are connected in series across an $a.c.$ power supply of $220 \text{ V}$, $70 \text{ Hz}$. The power factor of the given circuit is $0.5$. The difference between the inductive reactance and capacitive reactance is $\sqrt{3}\alpha \text{ } \Omega$. The value of $\alpha$ is . . . . . . .

In a series $LCR$ circuit,the voltages across $R, L$ and $C$ are shown in the figure. The voltage of the applied source is ........ $V$. (in $, V$)

The net impedance of the circuit (as shown in the figure) will be $...........\,\Omega$.

$A$ 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

$A$ $50 \,W$ bulb connected in series with a heater coil is put to an $AC$ mains. Now the bulb is replaced by a $100 \,W$ bulb. The heater output will ...........