Angular Frequency

Find angular frequency of source for resonance

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

@Steven Chase have a look. Thanks in advance.

Talulah Riley - 8 months, 1 week ago

The overall impedance is:

$Z = \frac{(R + j \omega L) (-j / \omega C)}{R + j \omega L - j / \omega C }$

Do a little trick to make a new expression for $Z$ , with the denominator being a real number. This works because when the denominator is multiplied by its complex conjugate, it becomes a real number. We are effectively just creatively multiplying the original expression by $1$ .

$Z = \frac{(R + j \omega L) (-j / \omega C)}{R + j \omega L - j / \omega C } \,\, \frac{R - j \omega L + j / \omega C}{R - j \omega L + j / \omega C}$

For resonance, the imaginary part of the new numerator must be zero.

$Im\Big[ (R + j \omega L) (-j / \omega C) (R - j \omega L + j / \omega C) \Big ] = 0$

Solving for the resonant $\omega$ yields:

$\omega = \sqrt{\frac{1}{L C} - \frac{R^2}{L^2}}$

Note that for $R = 0$ , this reduces to the familiar $L C$ resonance frequency.

Steven Chase - 8 months, 1 week ago

@Steven Chase Thanks for the solution.
I am getting the same answer by doing the imaginary part of $\large \frac{1}{Z}=0$ as well.

Talulah Riley - 8 months, 1 week ago

@Steven Chase what is the meaning of oscillating term?

Talulah Riley - 8 months, 1 week ago

The part that varies with time

Steven Chase - 8 months, 1 week ago

@Steven Chase – @Steven Chase Thank you so much sir for such giving a precious information :) :)

Talulah Riley - 8 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$