In a simple $L-R$ circuit with an $A.C.$ source,the potential difference at any instant across the inductor and the resistor are $V_L$ and $V_R$ respectively,and the $A.C.$ source has a potential difference $V_{AC}$ at the same instant. Then:

In a series $LCR$ circuit,$R=300 \Omega$,$L=0.9 \text{ H}$,$C=2 \mu\text{F}$,and $\omega=1000 \text{ rad/s}$. The impedance of the circuit is: (in $\Omega$)

$A$ $0.07\,H$ inductor and a $12\,\Omega$ resistor are connected in series to a $220\,V, 50\,Hz$ $AC$ source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as $\frac{22}{7}$]

For the $AC$ circuit shown, the phase difference between emf and current is $\frac{\pi}{4} \text{ rad}$ as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are:

When an inductor $L$ and a resistor $R$ in series are connected across a $15 \, V, 50 \, Hz$ a.c. supply, a current of $0.3 \, A$ flows in the circuit. The current differs in phase from the applied voltage by $(\frac{\pi}{3})^c$. The value of $R$ is:
$(\sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}, \sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2})$

An e.m.f. $E = E_0 \cos \omega t$ is applied to an $L-R$ circuit. The inductive reactance is equal to the resistance $R$ of the circuit. The power consumed in the circuit is: