A variable capacitor with a range from 8.7 to 326 pF is used with a coil to form

Question

A variable capacitor with a range from 8.7 to 326 pF is used with a coil to form a variable-frequency LC circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from 0.53 MHz to 1.26 MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what capacitance in picofarads should be added and (c) what inductance should the coil have?

Explanation / Answer

(a) The frequency of oscillation f is related to the inductance L and capacitance C

f = 1/ 2 ( L C)

hence f max= 1/ 2 ( L Cmin) & f min = 1/ 2 ( L C max)

f max / fmin = (C max/ C min) = (326pF/ 8.7pF) =6.12

(b) Additional capacitance to meet the frequency ratio= 1.26 MHz/0.53 MHz=2.38

Since additional capacitor is in parallel to the variable capacitor the total capacitance of the arrangement will sum of the capacitance of additional capacitor and variable capacitor.

suppose the capacitance of the additional capcitor = X pF

(X+326/ X+8.7)=2.38

2.38 * 2.38 * X + 8.7 *2.38 *2.28= X + 326

4.66 X = 276.72

X= 59.38 pF

(c) We know that

f = 1/ 2 ( L C)--------------------------------------------------------------(1)

The equation to be solved for minimum frequency

C max = 59.38 pF+ 326 pF= 385.38 pF & frequency f = 0.53 MHz

From equation (1)

L=1 / (2 )2 C f2

=1/ (2 )2 385.38x10-12 (0.53 x 106)2=2.34 x 10-4 H

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.