Comments on: A tuning circuit in an antique radio receiver has a fixed inductance of 0.20 mH and a variable capacitor (Fig. http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm Thu, 09 Feb 2012 07:50:01 +0000 http://wordpress.org/?v=2.7.1 hourly 1 By: Gary H http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm/comment-page-1#comment-511 Gary H Wed, 01 Jul 2009 08:25:38 +0000 http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm#comment-511 The coil and cap form a resonant circuit in order to tune the station. The formula is f = 1/(2PI(LC)^0.5) Solving for C, we get (LC)^0.5 = 1/2PI(f) => LC = 1/(4PI^2 (f^2)) => C = 1/(4pi^2(f^2)L) => 1/(4(9.87)(7.74)(10^8)(0.20)) = 163pF, if I entered it right - check, ok? The coil and cap form a resonant circuit in order to tune the station. The formula is

f = 1/(2PI(LC)^0.5)

Solving for C, we get
(LC)^0.5 = 1/2PI(f) =>
LC = 1/(4PI^2 (f^2)) =>
C = 1/(4pi^2(f^2)L) =>
1/(4(9.87)(7.74)(10^8)(0.20))

= 163pF, if I entered it right - check, ok?

]]> By: Omar I http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm/comment-page-1#comment-512 Omar I Wed, 01 Jul 2009 08:25:38 +0000 http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm#comment-512 show the Fig. 21.15 try 10pf with air isolation show the Fig. 21.15

try 10pf with air isolation

]]> By: rrabbit http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm/comment-page-1#comment-513 rrabbit Wed, 01 Jul 2009 08:25:38 +0000 http://antiquerecordplayers.info/a-tuning-circuit-in-an-antique-radio-receiver-has-a-fixed-inductance-of-020-mh-and-a-variable-capacitor-fig.htm#comment-513 Frequency of resonant circuit (whether series or parallel) is f = 1 / (2pi x sqrt(LC)) Hz C = 1 / ((2pi x f)^2 x L) F Here we have L = 0.2 x 10^(-3) and f = 880 x 10^3, so C = 1 / ((2pi x 880 x 10^3)^2 x 0.2 x 10^(-3)) F = 0.16355 x 10^(-9) F = 0.164 nF. Frequency of resonant circuit (whether series or parallel) is

f = 1 / (2pi x sqrt(LC)) Hz
C = 1 / ((2pi x f)^2 x L) F

Here we have L = 0.2 x 10^(-3) and f = 880 x 10^3, so

C = 1 / ((2pi x 880 x 10^3)^2 x 0.2 x 10^(-3)) F = 0.16355 x 10^(-9) F = 0.164 nF.

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