Solved Numerical Problem No. 16.9 ( Unit-16: Alternating Current); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

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.

Similar Questions:

An inductor of inductance 150 μH is connected in parallel with a variable capacitor whose capacitance can be changed from 500 pF to 20 pF. Calculate the maximum frequency and minimum frequency for which the circuit can be tuned. (Ans : 2.91 MHz, 0.58 MHz)

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