What is the resonant frequency of a circuit which includes a coil of inductance 2.5 H and a capacitance of 40 μF? (Ans: 15.9 Hz ) 



Data Given:


Inductance = L = 2.5 H

Capacitance of capacitor = C = 40 μF = 40 ｘ10⁻⁶ F



To Find:


Resonant Frequency  = `\f_r` = ?


Solution:


The formula for resonant frequency is

`\f_r` = `\frac {1}{2ㄫ sqrt {LC}}`

By putting values

`\f_r` = `\frac {1}{2 ｘ 3.1415 sqrt {2.5 H ｘ 40 ｘ10⁻⁶ F}}`

`\f_r` = `\frac {1}{6.283 sqrt {100 ｘ10⁻⁶ s²}}`

`\f_r` = `\frac {1}{6.283 sqrt {10⁻⁴ s²}}`

`\f_r` = `\frac {1}{6.283 ｘ 10⁻² s}`

`\f_r` = 0.159 ｘ 10² s⁻¹

or


`\f_r` = 15.9 Hz -----------------Ans.




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