In an electrical circuit $R,$ $L,$ $C$ and an AC voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\tan^{-1}\sqrt{3}$. If instead, $C$ is removed from the circuit, the phase difference is again $\tan^{-1}\sqrt{3}$. The power factor of the circuit is:

An inductor of $20~\text{mH}$, a capacitor of $100~\mu \text{F}$, and a resistor of $50~\Omega$ are connected in series across a source of emf, $V=10 \sin (314 t)$. What is the power loss in this circuit?
1. $0.79 ~\text{W}$
2. $0.43 ~\text{W}$
3. $2.74 ~\text{W}$
4. $1.13 ~\text{W}$

An AC source rated $100~\text{V}$ (rms) supplies a current of $10~\text{A}$ (rms) to a circuit. The average power delivered by the source:

An AC source given by $V=V_m\sin(\omega t)$ is connected to a pure inductor $L$ in a circuit and $I_m$ is the peak value of the AC current. The instantaneous power supplied to the inductor is:
1. $\dfrac{V_mI_m}{2}\mathrm{sin}(2\omega t)$

2. $-\dfrac{V_mI_m}{2}\mathrm{sin}(2\omega t)$

3. ${V_mI_m}\mathrm{sin}^{2}(\omega t)$

4. $-{V_mI_m}\mathrm{sin}^{2}(\omega t)$

For a series $\mathrm{LCR}$ circuit, the power loss at resonance is:
1. $\frac{V^2}{\left[\omega L-\frac{1}{\omega C}\right]}$

2. $\mathrm{I}^2 \mathrm{~L} \omega$

3. $I^2 R$

4. $\frac{\mathrm{V}^2}{\mathrm{C} \omega}$