Physics - EMI/AC Question with Solution | TestHub

Initially, at resonance, ${\omega }_0=1/\sqrt{LC}$. The inductive reactance $X_L={\omega }_{0}L$ and capacitive reactance $X_C=1/({\omega }_{0}C)={\omega }_{0}L$. So $X_L=X_C$.

When the frequency is doubled to ${\omega }^{\prime }=2{\omega }_0$: The new inductive reactance $X_L^{\prime }={\omega }^{\prime }L=2{\omega }_{0}L$. The new capacitive reactance $X_C^{\prime }=1/({\omega }^{\prime }C)=1/(2{\omega }_{0}C)=\frac{1}{2}({\omega }_{0}L)=\frac{1}{2}{X}_L$. The net reactance is $X^{\prime }=X_L^{\prime }-X_C^{\prime }=2{\omega }_{0}L-\frac{1}{2}{\omega }_{0}L=\frac{3}{2}{\omega }_{0}L$. The impedance is $Z^{\prime }=\sqrt{R^2+(X^{\prime }{)}^2}=\sqrt{R^2+{\left(\frac{3}{2}{\omega }_{0}L\right)}^2}=\sqrt{R^2+\frac{9}{4}{\omega }_0^2{L}^2}$. Since ${\omega }_0^2=1/(LC)$, $Z^{\prime }=\sqrt{R^2+\frac{9L}{4C}}$. A) The power factor is $\cos {\varphi }^{\prime }=R/Z^{\prime }=\frac{R}{\sqrt{R^2+\frac{9L}{4C}}}$. So, option A is correct.

B) At resonance, impedance $Z_0=R$. Since $Z^{\prime }=\sqrt{R^2+\frac{9L}{4C}}>R$, the impedance increases. The maximum current $I_{max}^{\prime }=V_0/Z^{\prime }$ will therefore decrease compared to resonance ($I_{max,0}=V_0/R$). So, option B is correct.

C) The voltage across the inductor $V_L^{\prime }=I_{max}^{\prime }{X}_L^{\prime }$ and across the capacitor $V_C^{\prime }=I_{max}^{\prime }{X}_C^{\prime }$. Since $X_L^{\prime }=2{\omega }_{0}L$ and $X_C^{\prime }=\frac{1}{2}{\omega }_{0}L$, we have $X_L^{\prime }>X_C^{\prime }$. Thus, $V_L^{\prime }>V_C^{\prime }$. So, option C is correct.

D) Since $X_L^{\prime }>X_C^{\prime }$, the circuit is inductive. This means the voltage leads the current, or equivalently, the current lags the voltage. If the phase difference $\varphi$ is defined as the phase of current relative to voltage, then current lagging voltage implies $\varphi <0$. So, option D is correct.