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A tank circuit uses a 0.09 uH inductor and a 0.4-uF capacitor. What would the resonat...
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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
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.
Posted by valentin68 on September 7, 2013 at 5:54 PM (Answer #1)
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