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

Answer:

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

Explanation:

Given;

inductance, L = 2.0 mH

capacitor is varied from  100 pF to 200 pF

Oscillating frequency is given as;

$F = \frac{1}{2\pi\sqrt{LC} }$

where;

F is the oscillating frequency

L is the inductance

C is the capacitance

When the capacitor, C = 100 pF

$F = \frac{1}{2\pi \sqrt{LC} } \\\\F = \frac{1}{2\pi \sqrt{2*10^{-3}*100*10^{-12}} }\\\\F =355,835.13 \ HZ$

When  the capacitor, C = 200 pF

$F = \frac{1}{2\pi \sqrt{LC} } \\\\F = \frac{1}{2\pi \sqrt{2*10^{-3}*200*10^{-12}} } \\\\F = 251,613.43\ HZ$

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

Answer 2

Answer:

The oscillating frequency is from 251613.43 Hz to 388835.1 Hz

Explanation:

L = 2.0 mH = 2 × 10⁻³ H

c = 100 pF to 200 pf

π = 3.142

oscillating frequency $f=\frac{1}{2*3.142\sqrt{LC} }$

When c = 100 pf (100 × 10⁻¹²) :

oscillating frequency $f=\frac{1}{2*3.142\sqrt{LC} }$

Substituting values

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

When c = 100 pf, f = 388835.1 Hz

When c = 200 pf (200 × 10⁻¹²) :

oscillating frequency $f=\frac{1}{2*3.142\sqrt{LC} }$

Substituting values

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

When c = 200 pf, f = 251613.43 Hz

The oscillating frequency is from 251613.43 Hz to 388835.1 Hz