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I've been searching around the internet to find out how to derive the reactance formula for capacitors and inductors. But I couldn't really find anything, so I thought why not make a post about it.

I gave it a try myself though, but I could not get rid of the \$sin(\omega t)\$ in the numerator and the \$cos(\omega t)\$ in the denominator. This is what I was left with:

\$X_c = sin(\omega t)/(C\cdot\omega\cdot cos(\omega t))\$

I tried to convert the \$v(t) = V_{peak} * sin(\omega t)\$ to \$V_{RMS}\$, and the \$i(t)\$ by doing the same procedure, and it sure worked. But I feel like this isn't the right way of doing it. Wikipedia mentioned the use of phasors, but I couldn't really figure out a way to do it.

Thank you in advance, Mr.Mongoloid

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    \$\begingroup\$ Take the Laplace transform of the differential equation relating current to voltage. Use the derivative properties. \$\endgroup\$
    – user110971
    Nov 10, 2016 at 20:11

5 Answers 5

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The phasor approach is the easiest imo. You simply let V and I become a phasor. Then you replace all of the differential operators by algebraic expressions. Ive done the capacitor for you .enter image description here

Note: in this form do not forget that both I and V are now phasors and V/I is a complex number ofcourse.

Edit: you can notice i made the phase phi = 0. I did this to make things clear, but its value makes no difference. When you take dV/dt the phase phi is never part of that expression anyways...

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  • \$\begingroup\$ I don't know, if phi != 0, then you will have e^(j*(wt + phi)) = e^(jwt) * e^(j* phi) = e^(jwt) * (cos(phi) + j sin(phi) ). And since the cos/sin part is a constant, it will remain unchanged after the derivation. Unless you mean to take the absolute value of e^(j*phi)? Which in that case is always equal to 1. Thank you for your comment. \$\endgroup\$ Nov 11, 2016 at 9:37
  • \$\begingroup\$ The key is being able to relplace dV/dt by jw.V You can do this regardless of what the initial phase is: V = e^j(wt+phi), dV/dt = jw.e^j(wt+phi) = jw.V! So it still works. \$\endgroup\$
    – MAM
    Nov 11, 2016 at 13:27
  • \$\begingroup\$ Yes, most definitely. My bad! But one thing though, can you regard the voltage/current as a phasor of the format: e^j*(wt+phi)? I thought you had the use the other formula, that sin(wt) = (1/2) * (e^j*(wt+phi) - e^-j*(wt+phi)). A quick statement would be much appreciated. \$\endgroup\$ Nov 12, 2016 at 11:08
  • \$\begingroup\$ You could do, but then you cannot claim that your derived formula is strictly general. Remember, that the phasor e^j*(wt+phi) is the most general input you can excite a circuit with, ie it is a complex number and so it could represent any real number and/or complex number with it. So if you use that and derive the formulas, you have a formula that is valid for a much wider range of inputs. However, note that a "real world" signal can never be complex, so your choice of just sin(wt) will be practically sufficient. \$\endgroup\$
    – MAM
    Nov 24, 2016 at 11:49
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Let current through inductor be \$i=I\:sin (\omega\:t)\$, then voltage across will be \$v=L \:\dfrac{di}{dt}= I\: \omega L\: cos (\omega t)\$. The reactance is \$X_L =\dfrac{|v|}{|i|}=\omega L\$, where the magnitude signs indicate either amplitude or RMS value, as appropriate.

Similar analysis for capacitive reactance, but this time: \$v=\frac{1}{C}\large\int \small i\:\small dt=-\dfrac{I}{\omega C}cos\:(\omega t)\$

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  • \$\begingroup\$ But haven't you made the assumption that |sin(wt)| = 1? The way I see it is that |sin(wt)| <= 1. Same with cos(wt) of course. Besides, is it allowable to use the absolute value? And why? Anyways, I appreciate the comment. Thank you. \$\endgroup\$ Nov 11, 2016 at 9:27
  • \$\begingroup\$ The magnitude of a reactance is the amplitude (or RMS) value of the voltage divided by the amplitude (or RMS) value of the current. Expressing this as an imaginary number takes care of the 90 degree phase difference between voltage and current, but, strictly, the ratio of amplitudes is the reactance in ohms. Thus \$jX_L\$ for the inductor and \$-jX_C\$ for a capacitor gives the magnitude and phase information. \$\endgroup\$
    – Chu
    Nov 11, 2016 at 10:24
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I'm currently reading The Art of Electronics and had some real trouble understanding the derivation for the impedance of a capacitor \$\boldsymbol Z_C\$ at a frequency \$\omega\$ in section 1.7.4. I finally figured it out so I figured I'd add it here.

The text starts out with a sinusoidal voltage:

$$ V(t) = \cos \omega t\ $$

We can then notice from Euler's formula \$re^{j\theta}=r\cos\theta + jr\sin \theta\$ that our function \$V(t)\$ is the real part of the left hand side:

$$ V(t) = \Re(V_0 e^{j \omega t}) $$

Now, we know that \$I = C(dV(t) /dt)\$, so we can substitute to find

$$ \begin{eqnarray*} I(t) &=& C \frac{d\left(\Re(V_0e^{j\omega t})\right)}{dt}\\ &=& \Re \left( V_0C\frac{de^{j\omega t}}{dt} \right) \\ &=& \Re \left( V_0Cj\omega e^{j\omega t}\right) \end{eqnarray*} $$

Now, this is where I got thrown for a bit of a loop. The trick to finding the impedance is to consider Ohm's law: \$V = IZ\$. We want to rewrite the above equation so it's in the form:

$$ I(t) = \frac{V(t)}{Z} $$

We know \$V(t) = \Re(V_0 e^{j \omega t})\$, so if we multiply by \$j/j\$ we get:

$$ \begin{eqnarray*} I(t) &=& \Re \left( V_0Cj \omega e^{j \omega t} \cdot \frac{j}{j}\right)\\ &=& \Re \left( \frac{V_0 C j^2 \omega e^{j \omega t}}{j} \right)\\ &=& \Re \left( \frac{V_0 C \omega e^{j \omega t}}{-j} \right)\\ &=& \Re \left( \frac{V_0 e^{j\omega t}}{-j / \omega C} \right)\\ &=& \Re \left( \frac{V(t)}{-j / \omega C} \right) \end{eqnarray*} $$

Now that we have it in this form, it's clear that \$Z_C = -j/\omega C\$.

I'm obviously still a beginner so I would appreciate it if anybody could point out any errors here, but it seems to match up with the text pretty well.

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Capacitor:
I=C * dv/dt
d/dt=jw=s (shortcut Laplace transform)
I=C * sV
C=I/sV
1/sC=V/I=Xc

Inductor:
V=L * di/dt
V=L * sI
L=V/sI
sL=V/I=Xl

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The equation I grew up with is this :

Xc = 1 / ( 2 pi f C ) expressed in ohms

If this is the equation you are trying to derive, forget it... this equation is not ( in math terms ) "rigorously correct", it is actually a crude "rule of thumb", which has some built-in assumptions that are NEVER identified, anywhere. You were on the right track but didn't go far enough... The term SIN wt / COS wt can be re-written as TAN wt, which evaluates to a quantity that varies from +infinity to -infinity, twice during each cycle.

The equation you created actually expresses the INSTANTANEOUS RESISTANCE of a capacitor, driven with a sine wave. ( = instantaneous voltage across the capacitor, divided by instantaneous current flowing through the capacitor ) The fact that this value ( I will call it Rc ) varies from +infinity to -infinity... twice during each cycle... is actually 100% correct.

The instantaneous resistance evaluates to a positive Rc quantity during the 1st and 3rd "1/4 cycle" intervals of a sine wave driving waveform, and it evaluates to a negative Rc quantity in the 2nd and 4th "1/4 cycle" intervals of the same driving waveform.

This simply indicates that energy is flowing IN TO the capacitor during the 1st and 3rd ( 1/4 cycle ) intervals, ( i.e the circuit is "charging" the cap = +Rc ) and energy is flowing OUT OF the capacitor during the 2nd and 4th ( 1/4 cycle ) intervals. ( i.e the cap is "discharging" energy back into the circuit = -Rc )

( repeating, this all assumes the driving waveform is a sine wave... and ONLY a sine wave )

The fact that the Rc value is totally "wild" and varies from +infinity to -infinity... twice... during each cycle... means the "resistance" of a capacitor driven with a sine wave DOES NOT HAVE A SPECIFIC VALUE THAT CAN BE IDENTIFIED OR USED IN ANY CIRCUIT CALCULATIONS. Stated differently, the concept of RESISTANCE as an expression of the voltage / current ratio in a capacitor ( driven with a sine wave ) IS USELESS.

That is why the voltage / current ratio of a capacitor is NEVER identified with the word RESISTANCE... instead, a NEW quantity is "invented" which is similar, and much more useful... called REACTANCE, which is also expressed in Ohms.

Reactance is defined as the RATIO of MAXIMUM VOLTAGE to MAXIMUM CURRENT, within each ( applied ) sine wave cycle... For a capacitor, maximum VOLTAGE occurs at w = +1/4 cycle, when SIN(w) = +1, and maximum current occurs at w = +0/4 cycle, when COS(w) = +1. Substituting these constants back into your equation will yield the well-known ( basic algebra ) equation for capacitive reactance...

Xc = 1 / ( 2Pi f C )

So... that equation is NOT TRUE at every instant in time... it expresses the ratio of MAX voltage to MAX current, but it ignores the fact that these two maxima do NOT occur simultaneously.... and there is nothing in the equation to even "hint" that this ( unethical ) "trick" is being done... that explains A LOT...

That's why summing together ( plain algebra ) values of R and X must be done with vector addition, instead of algebraic addition... the vectors take the "timing differences" into account when the sum is done... algebraic sums can't do that.

You will never, ever see an explanation like this anywhere in a text book because no-one wants to take the time to explain all this stuff... because it's basically a big mess that everyone wishes would just go away... and it does go away, if you use higher math... but "mere mortals" must muddle through life with plain old algebra, and it just won't do the job properly, in this case.

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  • \$\begingroup\$ Utter rubbish.. \$\endgroup\$
    – Andy aka
    Aug 9, 2020 at 12:10
  • \$\begingroup\$ Wow, I’d suggest not giving advice unless you triple check your math and assumptions. There are multiple statements that don’t follow logically, along with being invalid from a physics perspective. This reads like someone self taught, redefining terminology and making up mechanisms. Fine to be self taught, but you should validate your views and assumptions as well. \$\endgroup\$ Aug 9, 2020 at 13:24

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