1
\$\begingroup\$

Assuming that we have a simplified equation from a simple series RLC circuit where the total resistance/reactance in the series RLC circuit is equal to 100 times the resistance R:

equation 1

How do we modify the equation in such a way that we can eliminate the R^2 on the LHS of the equation? An online source gives me an explanation in this form:

equation 2

My question is, what is the reason behind taking the LHS of the equation within the square root and equating its inductance and capacitance variables as greater than the resistance variable? How is this exactly implied?

\$\endgroup\$
0
\$\begingroup\$

Here's the impedance magnitude plot of a very commonly encountered RLC circuit (a multilayer ceramic capacitor with parasitic inductance and resistance):

enter image description here

You can see that from DC to about 3 MHz, the curve nearly ideally matches the reactance of an ideal 1 uF capacitor, and from about 30 MHz out as far as the graph goes it nearly ideally matches the reactance of an ideal 300 pH inductor. Only near resonance, when the capacitive and inductive effects nearly cancel each other out, does the resistive component play a major role in determining the magnituce of impedance.

This kind of behavior is fairly common, because there will be some frequency below which the capacitive effect is dominant, and there will be some frequency above which the inductive effect is dominant, no matter what the resistance value is. (It is, of course, possible to have a much wider band where the resistance dominates than in this example, but there will always be upper and lower limits to the resistive band)

\$\endgroup\$
0
\$\begingroup\$

Your equation: $$(R^2 + (\omega L - \frac{1}{\omega C})^2)^{\frac{1}{2}} = 100R $$ has the form of: $$\sqrt{R^2 + X^2} = |Z| = Z $$ which defines the MAGNITUDE of Z, where X is reactance and R is resistance. So we can say 100R is the magnitude of the impedance. Notice the equation above looks alot like the pythagorean theorem where resistance is in the horizontal direction and reactance is in the vertical direction; it forms a triangle with Z as the hypotenuse. Just focusing on the LHS, reactance (X) takes an $$\omega$$ variable which is the frequency of the circuit whereas R, L, and C are constants of the circuit and therefore do not change over the operating lifetime of the circuit. The above implication: $$(\omega L - \frac{1}{\omega C})^2 \gg R^2$$ is just analyzing the edge case when the frequency of the circuit is massive. You could also take the frequency really close to zero and you'll see the implication above still holds! This is just a method to analyze edge cases of a circuit.

For instance, fix R, L, C and let frequency (omega) get larger and larger to infinity. The inductance term balloons to infinity, the capacitance terms goes to zero. IF $$\omega \rightarrow \infty$$ THEN $$(\infty L - \frac{1}{\infty C})^2 \gg R^2$$ $$(\infty L - 0)^2 \gg R^2 $$ $$(\infty L)^2 \gg R^2 $$ $$\infty ^2 \gg R^2 $$ $$\infty \gg R^2 $$ So the implication holds true: as frequency gets large, the reactance will be so much larger than resistance that we can essentially disregard resistance in the impedance equation.

The other case: IF $$ \omega \rightarrow 0 $$ THEN $$(0L - \frac{1}{0C})^2 \gg R^2 $$ $$(0 - \frac{1}{0})^2 \gg R^2 $$ $$(-\infty)^2 \gg R^2 $$ $$\infty \gg R^2 $$ Again the implication holds true. We can for sure say as frequency gets small (or large), reactance becomes so much larger than resistance that resistance is a negligible term in the impedance equation and thus can be omitted. Stated another way: IF $$ X^2 \gg R^2 $$ THEN $$ Z \approx \sqrt{X^2} $$

So to tie into the triangle visualization, when making the frequency massive or super small, the impedance triangle starts to look like a straight vertical line. Thus $$X = |Z|$$ $$ \omega L - \frac{1}{\omega C} = \pm100R$$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.