$A$ resistance of $40\, \Omega$ and an inductance of $95.5\, \text{mH}$ are connected in series in a $50\, \text{Hz}$ $AC$ circuit. The impedance of this combination is very nearly.....$\Omega$.

$A$ 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:

An e.m.f. $E = 4 \cos(1000t) \, V$ is applied to an $LR$-circuit of inductance $3 \, mH$ and resistance $4 \, \Omega$. The amplitude of current in the circuit is:

The given figure represents the phasor diagram of a series $LCR$ circuit connected to an $ac$ source. At the instant $t'$ when the source voltage is given by $V = V_0 \cos(\omega t')$,the current in the circuit will be:
Given: $V_{OL} = 3 \text{ V}$,$V_{OR} = \sqrt{3} \text{ V}$,$V_{OC} = 2 \text{ V}$.

An electric bulb,an open coil inductor,an $AC$ source,and a key are all connected in series to form a closed circuit. The key is closed and after some time,an iron rod is inserted into the interior of the inductor. Then:

An inductor and a resistor in series are connected to an $A.C.$ supply of variable frequency. As the frequency of the source is increased,the phase angle between current and the potential difference across $L$ will