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:
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:
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?
Best Answer
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 $$X = |Z|$$ $$ \omega L - \frac{1}{\omega C} = \pm100R$$