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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial