Solved Numerical Problem No. 16.10 ( Unit-16: Alternating Current); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

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)

Data Given:

Inductance = L = 150 μH = 150 ｘ10⁻⁶ H

Capacitance Maximum = `C_{max}` = 500 pF = 500 ｘ10⁻¹² F

Capacitance Minimum = `C_{min}` = 20 pF = 20 ｘ10⁻¹² F

To Find:

(i) Frequency Maximum = `\f_{max}` = ?

(ii) Frequency Minimum = `\f_{min}` = ?

Solution:

(i) Frequency Maximum = `\f_{max}` = ?

We know that 

`\f_{max}` = `\frac {1}{2ㄫ sqrt {LC_{min}}}`

By putting values

`\f_{max}` = `\frac {1}{2 ｘ 3.1415 sqrt {150 ｘ10⁻⁶ H ｘ 20 ｘ10⁻¹² F}}`

`\f_{max}` = `\frac {1}{6.283 sqrt {3000 ｘ10⁻¹⁸ s²}}`

`\f_{max}` = `\frac {1}{6.283 ｘ 54.772 ｘ 10⁻⁹ s}`

`\f_{max}` = `\frac {1}{344.132476 ｘ 10⁻⁹ s}`

`\f_{max}` = 0.00291ｘ 10⁹ s⁻¹

or

`\f_{max}`= 2.91ｘ 10⁶ Hz

or

`\f_{max}` = 2.91 MHz -----------------Ans.1

(ii) Frequency Minimum = `\f_{min}` = ?

We know that

`\f_{min}` = `\frac {1}{2ㄫ sqrt {LC_{max}}}`

By putting values

`\f_{min}` = `\frac {1}{2 ｘ 3.1415 sqrt {150 ｘ10⁻⁶ H ｘ 500 ｘ10⁻¹² F}}`

`\f_{min}` = `\frac {1}{6.283 sqrt {75,000 ｘ10⁻¹⁸ s²}}`

`\f_{min}` = `\frac {1}{6.283 ｘ 273.861 ｘ 10⁻⁹ s}`

`\f_{min}` = `\frac {1}{1720.668663 ｘ 10⁻⁹ s}`

`\f_{min}`= 0.000581ｘ 10⁹ s⁻¹

or

`\f_{min}` = 0.581ｘ 10⁶ Hz

or

`\f_{min}` = 0.581 MHz -----------------Ans.2

Similar Questions:

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

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