Question

A 2.0 mH inductor is connected in parallel with a variable capacitor. The capacitor can be varied from 100 pF to 200 pF. What is the minimum oscillation frequency for this circuit?

Answer 1

The oscillating frequency is from 251613.43 Hz to 388835.1 Hz

Step-by-step explanation:

L = 2.0 mH = 2 × 10⁻³ H

c = 100 pF to 200 pf

π = 3.142

oscillating frequency
`f=(1)/(2*3.142√(LC) )`

When c = 100 pf (100 × 10⁻¹²) :

oscillating frequency
`f=(1)/(2*3.142√(LC) )`

Substituting values

`f=\frac{1}{2*3.142\sqrt{2*10^(-3)*100*10^(-12)} }\\f=(1)/(2.81*10^(-6))\\f= 355835.1`

When c = 100 pf, f = 388835.1 Hz

When c = 200 pf (200 × 10⁻¹²) :

oscillating frequency
`f=(1)/(2*3.142√(LC) )`

Substituting values

`f=\frac{1}{2*3.142\sqrt{2*10^(-3)*200*10^(-12)} }\\f=(1)/(3.97*10^(-6))\\f= 251613.43`

When c = 200 pf, f = 251613.43 Hz

The oscillating frequency is from 251613.43 Hz to 388835.1 Hz

Answer 2

The minimum oscillation frequency for this circuit is 251,613.43 Hz

Step-by-step explanation:

Given;

inductance, L = 2.0 mH

capacitor is varied from 100 pF to 200 pF

Oscillating frequency is given as;

`F = (1)/(2\pi√(LC) )`

where;

F is the oscillating frequency

L is the inductance

C is the capacitance

When the capacitor, C = 100 pF

`F = (1)/(2\pi √(LC) ) \\\\F = \frac{1}{2\pi \sqrt{2*10^(-3)*100*10^(-12)} }\\\\F =355,835.13 \ HZ`

When the capacitor, C = 200 pF

`F = (1)/(2\pi √(LC) ) \\\\F = \frac{1}{2\pi \sqrt{2*10^(-3)*200*10^(-12)} } \\\\F = 251,613.43\ HZ`

Therefore, the minimum oscillation frequency for this circuit is 251,613.43 Hz