Question

Consider the network shown in Fig. 12.20. Let us determine the resonant frequency, the voltage across each element at resonance, and the value of the quality factor

 

Answer 1

\( Solution: \) The resonant frequency is obtained from the expression 

\( \omega_0=\frac{1}{\sqrt {LC}}=\frac{1}{\sqrt {(25)(10^{-3})(10)(10^{-6})}}=2000 rad/s \)

At this resonant frequency

\( I=\frac{V}{Z}=\frac{V}{R}=5\angle0°A \)

Therefore, 

\( V_R=(5\angle0°)(2)=10\angle0°V \)

\( V_L=j\omega_0LI=250\angle90°V \)

\( V_C=\frac{I}{j\omega_0 C}=250\angle-90°V \)

Note the magnitude of the voltages across the inductor and capacitor with respect to the input voltage. Note also that these voltages are equal and are 180° out of phase with one another. Therefore, the phasor diagram for this condition is shown in Fig. 12.19 for ω = \( \omega_0 \) The quality factor Q derived from Eq. (12.13) is 

\( Q=\frac{\omega_0L}{R}=25 \)

The voltages across the inductor and capacitor can be written in terms of Q as 

\( |V_L|=\omega_0L|I|=\frac{\omega_0L}{R}|V_S|=Q|V_S| \) 

and \( |V_L|+\frac{|I|}{\omega_0C}=\frac{1}{\omega_0CR}|V_S|=Q|V_S| \)

This analysis indicates that for a given current there is a resonant voltage rise across the inductor and capacitor that is equal to the product of Q and the applied voltage.

Answer

Therefore,

\( V_R=(5\angle0°)(2)=10\angle0°V \)

\( V_L=j\omega_0LI=250\angle90°V \)

\( V_C=\frac{I}{j\omega_0 C}=250\angle-90°V \)