(Solved):     PLEASE SOLVE QUESTION 2 A \( 60.0 \)-Hz AC generator with a peak voltage of \( 110 ...

 

 

PLEASE SOLVE QUESTION 2

A \( 60.0 \)-Hz AC generator with a peak voltage of \( 110 \mathrm{~V} \) drives a series \( R L \) circuit with \( R=10.0 \Omega \) and \( L=10.0 \mathrm{mH} \). The impedance is \begin{tabular}{c} \( 3.77 \Omega \) \\ \hline \( 9.26 \Omega \) \\ \hline \( 10.0 \Omega \) \\ \hline \( 10.7 \Omega \) \\ \hline \( 13.8 \Omega \) \end{tabular} Question 2 2 pts All three circuits below have \( R=100 \Omega, C=1.0 \mathrm{mF} \), and emf \( \varepsilon=(5.0 \mathrm{~V}) \sin (377 t) \). The inductors in (B) and (C) are placed sufficiently far apart so that they do not alter one another's inductance. Such inductors add combine like resistors. Which statement regarding the angular resonance frequencies \( \omega_{\mathrm{A}}, \omega_{\mathrm{B}} \), and \( \omega_{\mathrm{C}} \) is correct? \[ \begin{array}{l} \omega_{C}>\omega_{A}=\omega_{B} \\ \omega_{C}<\omega_{A}=\omega_{B} \\ \omega_{A}=\omega_{B}=\omega_{C} \\ \omega_{B}<\omega_{A}=\omega_{C} \\ \omega_{B}>\omega_{A}=\omega_{C} \end{array} \]