In an L–C–R series circuit, under the resonance condition, identify the correct options.

1) At resonance condition, ${\mathrm{V}}_{\text{applied}}={\mathrm{V}}_{\mathrm{R}}$.

2) Quality factor, $\mathrm{Q}=\frac{{\mathrm{\omega }}_0\mathrm{L}}{\mathrm{R}}=\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}=\frac{{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\mathrm{R}}}=\frac{{\mathrm{V}}_{\mathrm{L}}}{{\mathrm{V}}_{\text{applied}}} \Rightarrow {\mathrm{V}}_{\mathrm{L}}={\mathrm{QV}}_{\text{applied}}$

3) Voltage across the inductor leads by 90° with current whereas voltage across capacitor lags by 90° with current. Hence total phase difference between voltage across capacitor and inductor is 180°.

4) Under resonance condition, power factor of the circuit is 1, i.e., $\mathrm{\varphi }={\cos }^{-1}\left(1\right)=0°$. Hence, wattless current $={\mathrm{I}}_{\mathrm{rms}} \sin \mathrm{\varphi }$ is zero.