Skip to main content
added 206 characters in body
Source Link
jjstcool
  • 494
  • 4
  • 16

looks like your guess is right. Imagine frequency = 0, then the inductor becomes a conductor and only \$R_L\$ is responsible for the impedance. on the other end of the spectrum, \$\omega \rightarrow \infty\$ only \$R_c\$ makes for the impedance. So at \$\omega_0\$ the \$L\$ and \$C\$ termes most likely dominate and are much greater than the corresponding resistors. This is due to the fact that at \$\omega_0\$ the overall-impedance is the highest and if \$C\$/\$L\$ do not dominate the resistances at this frequency, then they don't at all and all we have is the resistances anyway.

These resistors probably represent the parasitic resistances which have an effect by decreasing the Q-factor (and therefore are small compared to the corresponding Inductance/Capacitance) So if you assume just that your calculations yield your given formula.

looks like your guess is right. Imagine frequency = 0, then the inductor becomes a conductor and only \$R_L\$ is responsible for the impedance. on the other end of the spectrum, \$\omega \rightarrow \infty\$ only \$R_c\$ makes for the impedance. So at \$\omega_0\$ the \$L\$ and \$C\$ termes most likely dominate and are much greater than the corresponding resistors. These resistors probably represent the parasitic resistances which have an effect by decreasing the Q-factor (and therefore are small compared to the corresponding Inductance/Capacitance) So if you assume just that your calculations yield your given formula.

looks like your guess is right. Imagine frequency = 0, then the inductor becomes a conductor and only \$R_L\$ is responsible for the impedance. on the other end of the spectrum, \$\omega \rightarrow \infty\$ only \$R_c\$ makes for the impedance. So at \$\omega_0\$ the \$L\$ and \$C\$ termes most likely dominate and are much greater than the corresponding resistors. This is due to the fact that at \$\omega_0\$ the overall-impedance is the highest and if \$C\$/\$L\$ do not dominate the resistances at this frequency, then they don't at all and all we have is the resistances anyway.

These resistors probably represent the parasitic resistances which have an effect by decreasing the Q-factor (and therefore are small compared to the corresponding Inductance/Capacitance) So if you assume just that your calculations yield your given formula.

Source Link
jjstcool
  • 494
  • 4
  • 16

looks like your guess is right. Imagine frequency = 0, then the inductor becomes a conductor and only \$R_L\$ is responsible for the impedance. on the other end of the spectrum, \$\omega \rightarrow \infty\$ only \$R_c\$ makes for the impedance. So at \$\omega_0\$ the \$L\$ and \$C\$ termes most likely dominate and are much greater than the corresponding resistors. These resistors probably represent the parasitic resistances which have an effect by decreasing the Q-factor (and therefore are small compared to the corresponding Inductance/Capacitance) So if you assume just that your calculations yield your given formula.