Physics - EMI/AC Question with Solution | TestHub

The phase angle $\varphi$ between current and voltage in a series RLC circuit is given by $\tan \varphi =\frac{X_L-X_C}{R}=\frac{\omega L-1/(\omega C)}{R}$. At ${\omega }_1$, current leads voltage, so ${\varphi }_1=-\pi /4$. Thus $\tan (-\pi /4)=-1$.$-1=\frac{{\omega }_{1}L-1/({\omega }_{1}C)}{R}⟹-R={\omega }_{1}L-\frac{1}{{\omega }_{1}C}$ (Equation 1). At ${\omega }_2$, current lags voltage, so ${\varphi }_2=\pi /4$. Thus $\tan (\pi /4)=1$.$1=\frac{{\omega }_{2}L-1/({\omega }_{2}C)}{R}⟹R={\omega }_{2}L-\frac{1}{{\omega }_{2}C}$ (Equation 2). Adding Equation 1 and Equation 2: $0=({\omega }_1+{\omega }_2)L-\left(\frac{1}{{\omega }_{1}C}+\frac{1}{{\omega }_{2}C}\right)=({\omega }_1+{\omega }_2)L-\frac{{\omega }_1+{\omega }_2}{{\omega }_1{\omega }_{2}C}$. Since ${\omega }_1+{\omega }_2\ne 0$, we can divide by it: $L-\frac{1}{{\omega }_1{\omega }_{2}C}=0⟹{\omega }_1{\omega }_2=\frac{1}{LC}$. The resonance frequency ${\omega }_0$ is given by ${\omega }_0=\frac{1}{\sqrt{LC}}$. Thus, ${\omega }_0^2=\frac{1}{LC}={\omega }_1{\omega }_2$. The average power consumed is $P_{avg}=V_{rms}{I}_{rms}\cos \varphi =\frac{V_{rms}^2}{Z}\cos \varphi =\frac{(V_0/\sqrt{2}{)}^2}{Z}\frac{R}{Z}=\frac{V_0^{2}R}{2Z^2}$. At ${\omega }_1$, $X_L-X_C=-R$. So impedance $Z_1=\sqrt{R^2+(-R{)}^2}=\sqrt{2R^2}=R\sqrt{2}$.$P_{avg,1}=\frac{V_0^{2}R}{2(R\sqrt{2}{)}^2}=\frac{V_0^{2}R}{2(2R^2)}=\frac{V_0^2}{4R}$. At resonance (${\omega }_0$), $X_L=X_C$, so impedance $Z_0=R$.$P_{avg,0}=\frac{V_0^{2}R}{2R^2}=\frac{V_0^2}{2R}$. The ratio of average power is $\frac{P_{avg,1}}{P_{avg,0}}=\frac{V_0^2/(4R)}{V_0^2/(2R)}=\frac{1}{2}=0.5$.