Saved Bookmarks
| 1. |
A radio can tune over the frequency range of a portion of MW broadcast band: ( 800 kHz to 1200kHz ) . If its LC circuit has an effective inductance of 200 mu H. What must be the range of its variable capacitor ? |
|
Answer» Solution :For tunning a particular station in a radio, the frequency of free oscillations `f_(0) = ( 1)/(2pi sqrt( LC ))` is changed( by changing L or C ) and then when its is made equal to frequency f of radio waves transmitted by that station, the conditionfor resonance `f= f_(0)` is satisfied which gives RISE flow of maximum current in the CIRCUIT, producing sound with maximum clarity which ensures that particular radio station is tuned with our radio Now, `f_(0) = ( 1)/(2pi sqrt( LC ))` `:. f_(0)^(2) = ( 1)/( 4pi^(2) (LC))` `:. C = ( 1)/( 4pi^(2) f_(0)^(2) L )`...(1) For `(f_(0))_(max) = 1200 KHZ`, we have `C_("min") = ( 1)/( 4pi^(2) ( f_(0))_("max")^(2)L)` `= ( 1)/( ( 4) ( 3.14)^(2) ( 1200 xx 10^(3)) ( 200 xx 10^(-6)))` `= 88 xx10^(-12) F` `= 88 pF `...(2) For `(f_(0))_("min") = 800 kHz`, we have `C_(max) = ( 1)/( 4pi^(2) ( f_(0))_("max")^(2) L )` ` = ( 1)/(( 4) ( 3.14 ) ( 800 xx 10^(3))^(2) ( 200 xx 10^(-6)))` `= 198 xx 10^(-12) F` `= 198 pF `...(3) Final answer `:` From results (2) and ( 3), the required range of variable capacitor should be 88 pF to 198 pF . |
|