An alternating voltage is applied to a series $LCR$ circuit. If the current leads the voltage by $45^{\circ}$,then $\left(\tan 45^{\circ}=1\right)$:

$A$ resistance $R$ draws power $P$ when connected to an $AC$ source. If an inductance is now placed in series with the resistance,such that the impedance of the circuit becomes $Z$,the power drawn will be

$A$ coil of inductance $0.50 \; H$ and resistance $100 \; \Omega$ is connected to a high-frequency supply $(240 \; V, 10 \; kHz)$.
$(a)$ What is the maximum current in the coil?
$(b)$ What is the time lag between the voltage maximum and the current maximum?
Hence,explain the statement that at very high frequency,an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a $DC$ circuit after the steady state?

An $AC$ circuit having supply voltage $E = 10 \sin \omega t$ consists of a resistor of resistance $R = 3 \Omega$ and an inductor of reactance $X_L = 4 \Omega$ as shown in the figure. The voltage across the inductor at $t = \pi / \omega$ is: (in $\text{ V}$)

When an inductance $L$ and resistor $R$ are connected in series to $25 \ V$, $50 \ Hz$ supply, a current of $0.5 \ A$ flows in the circuit. The current lags behind in phase from applied voltage by $(\frac{\pi}{3})$ radian. The value of $R$ is $(\cos 60^{\circ} = \frac{1}{2})$ (in $\Omega$)

When a coil is connected to an $AC$ supply of frequency $50 \, Hz$, a current of $4 \, A$ flows in it and it consumes $240 \, W$ power. If the potential difference across the coil is $100 \, V$, then the inductance value of the coil is