A tank circuit uses a 0.09 uH inductor and a 0.4-uF capacitor. What would the resonat frequency be?

Expert Answers
valentin68 eNotes educator| Certified Educator

No matter if the resonant (or other said, tank) circuit is parallel or series the resonance condition is the same, that is the currents (or voltages) on the inductor and capacitor cancel each other. This, in turn leads to the equality of component reactances X(C) = X(L)

Because `X(C) = 1/(omega*C)` and `X(L) = omega*L` the above equality writes as (at resonance)

`1/(omega_r*C) = omega_r*L`  or `omega_r = 1/sqrt(L*C)`

Now, the general relation between angular frequency `omega` and frequency F itself is

`omega =2*pi*F`

which gives a value for the resonant frequency

`F_r = 1/(2*pi*sqrt(L*C)) = 1/(2*pi*sqrt(9*10^(-8)*4*10^(-7)))= `

`=838.82*10^3 Hz =838.82 KHz`

The resonant frequency is 838.82 KHz.