The minimum capacitance of a variable capacitor in a radio is 4.19 pF . A)What i

Question

The minimum capacitance of a variable capacitor in a radio is 4.19 pF .

A)What is the inductance of a coil connected to this capacitor if the oscillation frequency of the L-C circuit is 1.53 MHz , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance?..H

B)The frequency at the other end of the broadcast band is 0.535 MHz . What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?..F

Explanation / Answer

:  PART A:

f = 1 [2••(L•C)]

1.53 ×10^6 = 1 [2••[L • (4.19×10^-12)]

(1.53 ×10^6) • [2••[L • (4.19×10^-12)] = 1

L = 1 { (4.19×10^-12) • [2••(1.53 ×10^6)]² }

L = 1.103215.64 mH = 0.0011032 Henrys

PART B:

f = 1 [2••(L•C)]

0.535 ×10^6 = 1 [2••[(1.11 ×10^-3) • C]

(0.534 ×10^6) • [2••[(1.11 ×10^-3) • C] = 1

C = 1 { (1.11 ×10^-3) • [2••(0.534 ×10^6)]² }

C = 80.026 pF = 80.02 ×10^-12Farads

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