- #1

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Q=Lw(sub 0)/R. So far I know that Q=w(sub 0)/delta but my professor used this formula to continue the derivation delta w=R/L and I don't know where it comes from. Can anyone give me a hint?

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- Thread starter gazepdapi1
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- #1

- 54

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Q=Lw(sub 0)/R. So far I know that Q=w(sub 0)/delta but my professor used this formula to continue the derivation delta w=R/L and I don't know where it comes from. Can anyone give me a hint?

- #2

marcusl

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Quality factor is defined by

Q= 2*pi*(energy stored in resonator)/(energy lost per cycle).

The peak stored energy is [tex]LI^2 / 2.[/tex]

The power dissipated in the resistance is [tex]I^2R/2[/tex], in terms of the peak current.

The time in one cycle is 1/f or, in terms of angular frequency, [tex]2\pi/\omega.[/tex] Putting these into the definition of quality factor gives

[tex]Q=\frac{\omega L}{R}.[/tex]

EDIT: clean up equation rendering and add following:

Note that this is true for a series LCR circuit. For a parallel circuit Q is one over the expression on the right!

Q= 2*pi*(energy stored in resonator)/(energy lost per cycle).

The peak stored energy is [tex]LI^2 / 2.[/tex]

The power dissipated in the resistance is [tex]I^2R/2[/tex], in terms of the peak current.

The time in one cycle is 1/f or, in terms of angular frequency, [tex]2\pi/\omega.[/tex] Putting these into the definition of quality factor gives

[tex]Q=\frac{\omega L}{R}.[/tex]

EDIT: clean up equation rendering and add following:

Note that this is true for a series LCR circuit. For a parallel circuit Q is one over the expression on the right!

Last edited:

- #3

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thanks a lot

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