Physics - EMI/AC Question with Solution | TestHub

At resonance, let the angular frequency be ${\omega }_0$. The inductive reactance is $X_L={\omega }_{0}L$ and capacitive reactance is $X_C=1/({\omega }_{0}C)$. At resonance, $X_L=X_C$. The impedance is $Z_0=R$. The power dissipated is $P_0=V_{rms}^2/R$.

When the frequency is changed to ${\omega }^{\prime }$, the new inductive reactance $X_L^{\prime }={\omega }^{\prime }L$ and new capacitive reactance $X_C^{\prime }=1/({\omega }^{\prime }C)$. Given $X_L^{\prime }=2X_L$ and $X_C^{\prime }=X_C/2$. From $X_L^{\prime }=2X_L$, we have ${\omega }^{\prime }L=2({\omega }_{0}L)$, so ${\omega }^{\prime }=2{\omega }_0$. Since $X_L=X_C$ at resonance, the new reactances are $X_L^{\prime }=2X_L$ and $X_C^{\prime }=X_L/2$. The new impedance $Z^{\prime }$ is given by $Z^{\prime 2}=R^2+(X_L^{\prime }-X_C^{\prime }{)}^2$.$$Z^{\prime 2}=R^2+(2X_L-X_L/2{)}^2=R^2+(3X_L/2{)}^2$$ The new power dissipated is $P^{\prime }=\frac{V_{rms}^2}{Z^{\prime 2}}R$. Given $P^{\prime }=P_0/2$.$$\frac{V_{rms}^2}{Z^{\prime 2}}R=\frac{1}{2}\frac{V_{rms}^2}{R}⟹\frac{R}{Z^{\prime 2}}=\frac{1}{2R}⟹Z^{\prime 2}=2R^2$$ Substituting $Z^{\prime 2}$:$$R^2+(3X_L/2{)}^2=2R^2$$ $$(3X_L/2{)}^2=R^2$$ Taking the square root (reactance and resistance are positive), $3X_L/2=R$.$$X_L=\frac{2R}{3}$$ The quality factor $Q$ of the circuit at resonance is defined as $Q=X_L/R$.$$Q=\frac{(2R/3)}{R}=\frac{2}{3}$$