The AM band extends from approximately 500 kHz to 1,600 kHz. If a 1.6-μH inductor is used in a tuning circuit for a radio, what are the extremes that a capacitor must reach in order to cover the complete band of frequencies? Cmax = nF Cmin = nF

Respuesta :

Answer:

a

[tex]C_{max} =   63.3 n  F[/tex]

b

[tex]C_{min} = 6.182 nF  [/tex]

Explanation:

From the question we are told that

  The minimum frequency is [tex]f_{min} = 500kHz = 500*10^{3} \  Hz[/tex]

  The maximum frequency is [tex]f_{max} = 1600 kHz =  1600*10^{3} \  Hz[/tex]

  The inductance is  [tex]L  =  1.6 \mu H  =  1.60 *10^{-6} \  H[/tex]

Generally the low band frequency is mathematically represented as

     [tex]f_{min} =  \frac{1}{2\pi \sqrt{LC_{max}} }[/tex]

=>  [tex]C_{max} =  \frac{1}{4 \pi^2 *  f_{min}^2 *  L }[/tex]

=>  [tex]C_{max} =  \frac{1}{4 * 3.142 ^2 *  500*10^{3} *  1.60 *10^{-6}  }[/tex]

=>  [tex]C_{max} =   6.33 * 10^{-8} \  F[/tex]

=>  [tex]C_{max} =   63.3 n  F[/tex]  

Generally the high  band frequency is mathematically represented as

     [tex]f_{max} =  \frac{1}{2\pi \sqrt{LC_{min}} }[/tex]

=>  [tex]C_{min} =  \frac{1}{4 \pi^2 *  f_{max}^2 *  L }[/tex]

=>  [tex]C_{min} =  \frac{1}{4 * 3.142^2 *  ( 1600*10^{3})^2 *   1.60 *10^{-6} }[/tex]

=> [tex]C_{min} =  \frac{1}{4 * 3.142^2 *  ( 1600*10^{3})^2 *   1.60 *10^{-6} }[/tex]

=> [tex]C_{min} = 6.182 *10^{-9} \  m  [/tex]

=> [tex]C_{min} = 6.182 nF  [/tex]