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An MRI machine needs to detect signals that oscillate at very high frequencies. It does so with an LC circuit containing a 15mH coil. To what value should the capacitance be set to detect a 450 MHz signal?

Respuesta :

okpalawalter8

Answer:

The capacitance is  [tex]C = 3.2 9 *10^{-16} \ F[/tex]

Explanation:

From the question we are told that

   The  induction of the LC circuit is  [tex]L = 15 mH = 15 *10^{-3} \ H[/tex]

    The  frequency is   [tex]w = 450 \ MHz = 450 *10^{6} \ Hz[/tex]

The natural frequency is mathematically represented as

          [tex]w = \frac{1}{\sqrt{LC} }[/tex]

Where C is the capacitance So  

=>      [tex]C = \frac{1}{L * w^2}[/tex]

substituting values

         [tex]C = \frac{1}{15 *10^{-3} * [450 *10^{6}]^2}[/tex]

         [tex]C = 3.2 9 *10^{-16} \ F[/tex]

Lanuel

The value for which the capacitance should be set to detect a 450 MHz signal is [tex]8.34 \times 10^{-24} \;F[/tex]

Given the following data:

  • Inductance = 15 mH = [tex]15 \times 10^{-3}\;H[/tex]
  • Frequency = 450 MHz = [tex]450 \times 10^6 \;Hz[/tex]

To determine the value for which the capacitance should be set to detect a 450 MHz signal:

Mathematically, natural frequency is given by the formula:

[tex]f_o = \frac{1}{2\pi \sqrt{LC} }[/tex]

Where:

  • L is the inductance.
  • C is the capacitance.

Making C the subject of formula, we have:

[tex]C = \frac{1}{(2\pi f_o)^2L} \\\\C = \frac{1}{(2\;\times \;3.142 \times \;450 \;\times\; 10^9)^2 \; \times \;15 \times 10^{-3}}\\\\C = \frac{1}{8 \times 10^{24} \;\times \;15 \times 10^{-3} } \\\\C = \frac{1}{1.2 \times 10^{23}} \\\\C= 8.34 \times 10^{-24} \;F[/tex]

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