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Series LC circuit, connected across a constant voltage AC source. If the capacitor had the greatest reactance, could the circuit be simplified to simply a capacitor (by resolving the reactances)? How would this all work with the relevant phasors?

Thanks!

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Of course.

For example, at low frequencies, the series LC circuit looks like an open circuit, because the capacitor produces a high reactance at those frequencies.

And at high frequencies, the series LC circuit looks like an open circuit because the inductor produces a high reactance at those frequencies.

Whenever two elements are in series, their impedances add

$$Z_{\mathrm{series}}=Z_1 + Z_2$$

In the case of inductors and capacitors (i.e. when no real component is present) this becomes

$$X_{\mathrm{series}}=X_1 + X_2$$

And there are many many cases in physics and engineering where if you have a sum with one element much larger than the other (for example \$X_1\gg X_2\$) then you can approximate with \$X_1+X_2\approx{}X_1\$.

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You don't need phasors to solve this but you can look at a phasor diagram and realize that (say) equal amounts of inductive and capacitive reactance in series produce zero ohms. This is called resonance and of course only exactly happens at one frequency BUT it sets the scene for the very simple case of one type of reactance mathematically cancelling another.

Theoretically a negative resistance can totally cancel a normal resistance but of course you cannot buy a component that has negative resistance but, you can make one with op-amps and, within the power constraints of those op-amps, you can cancel a positive resistance with a negative one.

Capacitors and inductors are the nearest thing you can get. Look at the maths when they have equal numerical reactance: -

\$jwL = -\dfrac{1}{jwC}\$

\$-j^2w^2LC = 1\$

Therefore \$w = \sqrt{\dfrac{1}{LC}} \$ or more commonly seen as f = \$\dfrac{1}{2\pi\sqrt{LC}}\$

I.e. this is the formula for resonance of a tuned circuit and for a parallel tuned circuit the impedance becomes infinite and for series the impedance becomes zero. Note that if you are into power electrical circuits, the parallel tuned circuit is one which is used for power factor correction. I remember being taught power factor correction using phasor diagrams and then quite separately taught about tuned circuits using the math. For a short length of time I never actually realized both guys were talking about the same thing.

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Sure!

Like The Photon mentioned, you can calculate the impedance of a series LC circuit as $$ Z_{series} = Z_L + Z_C $$ The impedance of an inductor is a pure reactance, ie: $$ Z_L = j\omega L $$ The impedance of a capacitor is a pure negative reactance, ie: $$ Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C} $$ In your case, \$|Z_C| > |Z_L|\$, so the total impedance is $$ Z_{series} = j\omega L - \frac{j}{\omega C} = -j\left(\frac{1}{\omega C} - \omega L \right) $$ and since this is negative, you can treat this impedance as if it's simply a capacitance.

Likewise, if the inductor had a higher reactance, you could simplify the circuit to a single inductor.

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