$A$ current of $\frac{25}{\pi} \text{ Hz}$ frequency is passing through an $AC$ circuit having a series combination of $R=100 \ \Omega$ and $L=2 \text{ H}$. The phase difference between voltage and current is . . . . . . . (in $^{\circ}$)

$A$ resistor of $200 \; \Omega$ and a capacitor of $15.0 \; \mu F$ are connected in series to a $220 \; V, 50 \; Hz$ $ac$ source.
$(a)$ Calculate the current in the circuit.
$(b)$ Calculate the voltage $(rms)$ across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes,resolve the paradox.

If $E = 200\,V$,$R = 25\,\Omega$,$L = 2\,H$,and $C = 2\,\mu F$ and the frequency is variable,then the current at $f = 0$ and $f = \infty$ will be respectively:

$A$ coil has an inductance of $0.7 \, H$ and is joined in series with a resistance of $220 \, \Omega$. When an alternating $emf$ of $220 \, V$ at $50 \, Hz$ is applied to it,the phase angle by which the current lags behind the applied $emf$ and the wattless component of the current in the circuit are respectively:

In the $A$.$C$. circuit shown,keeping switch $K$ closed,if an iron rod is inserted into the coil,the bulb in the circuit:

In the circuit shown here,the voltages across $L$ and $C$ are respectively $300\, V$ and $400\, V$. The voltage $E$ of the $AC$ source is......$V$.