Asnetcomp Tutorials: August 2016

Friday, August 26, 2016

XL, XC - Phase Shift by calculus (A part of Reciprocal- 2)

＜Electronics Tutorials＞some of which I recently read several times.They are good reading materials as tutorials as the name shows. At the beginning of their ＜Series Resonance Circuit＞ we can find the following mostly familiar formulas.

http://www.electronics-tutorials.ws/accircuits/series-resonance.html

series circuit characteristics

sptt additional explanation and comments

The first and the second equations were already introduced at the end of Post ＜Reciprocal＞.

"
Inductors act like resistors as XL (which is regarded as resistance) = 2 $π$ fL. On the other and Capacitors act as reciprocal of Resistors as XC (which is regarded as resistance) = 1/2 $π$ fC.

"
Please note that XC is a reciprocal form. So the chart becomes y = 1/x form like below.

The difference is x : 2 $π$ fC. <x> is a variable while f is variable but 2 $π$ is constant and C can be either a constant or the second variable.

＜f＞ has the unit ＜(Number) / sec＞. The unit of C is Farad but originally (from wiki "Farad")

"
A farad has the base SI representation of: s⁴ × A² × m⁻² × kg⁻¹

It can further be expressed as:

{\mbox{F}}=\,\mathrm {{\frac {A\cdot s}{V}}={\dfrac {\mbox{J}}{{\mbox{V}}^{2}}}={\dfrac {{\mbox{W}}\cdot {\mbox{s}}}{{\mbox{V}}^{2}}}={\dfrac {\mbox{C}}{\mbox{V}}}={\dfrac {{\mbox{C}}^{2}}{\mbox{J}}}={\dfrac {{\mbox{C}}^{2}}{{\mbox{N}}\cdot {\mbox{m}}}}={\dfrac {{\mbox{s}}^{2}\cdot {\mbox{C}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {{\mbox{s}}^{4}\cdot {\mbox{A}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {\mbox{s}}{\Omega }}} ={\dfrac {{\mbox{s}}^{2}}{\mbox{H}}}

where F=farad, A=ampere, V=volt, C=coulomb, J=joule, m=metre, N=newton, s=second, W=watt, kg=kilogram, Ω=ohm, H=henry.

"
If we simply take ＜s / ‎Ω＞, then 2 $π$ fC is in ＜/ ‎Ω＞(Reciprocal of ‎Ω). Therefore ＜1 / ‎2 $π$ fC＞ is in ‎Ω.

＜f＞can be several Hz to several Giga Hz but for AM ranging 500KHz - 1600KHz, for FM 80MHz (80,000KHz) to 120MHz (120,000KHz). C can be several pico Farad to several micoro Farad.

For AM Tuning (which uses LC Resonance and Oscillation) we can use ＜f＞as 1,000KHz and C as 100pF. If we use original units

1,000KHz ---- ＞ 1,000,000Hz

100pF ---- ＞100 / 1,000, 000, 000,000 F or 1 / 10,000,000.000 F. This is very small number as compared with Farad or even uF (1/1,000,000 F). These small numbers make differences is a remarkable thing.
We put these two numbers
1 / ‎2 $π$ fC = 1 / 2 x 3.14 x 1,000,000 (Hz in /s) x (1/ 10,000,000,000 (F in s / ‎Ω)
= 1 / 2 x 3.14 x 1 x (1 / 10,000 (Ω)
= 1 / 2 x 3.14 x 0.0001
= 1 / 0.000628 = 1,591 Ω (=XC)

We can set L as 100uH.

XL = 2 $π$ fL = 2 x 3.14 x 1,000,000 (Hz in /s) x 100 / 1,000,000 H = 6.28 x 100 (Hz in /s) H
= 628 (supposed to be in Ω).

The unit of H is Henry and originally (from wiki "Henry")

"

{\mbox{H}}={\dfrac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{s}}^{2}\cdot {\mbox{A}}^{2}}}={\dfrac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{C}}^{2}}}={\dfrac {\mbox{J}}{{\mbox{A}}^{2}}}={\dfrac {{\mbox{T}}\cdot {\mbox{m}}^{2}}{\mbox{A}}}={\dfrac {\mbox{Wb}}{\mbox{A}}}={\dfrac {{\mbox{V}}\cdot {\mbox{s}}}{\mbox{A}}}={\dfrac {{\mbox{s}}^{2}}{\mbox{F}}}={\dfrac {\mbox{1}}{{\mbox{F}}\cdot {\mbox{Hz}}^{2}}}=\Omega \cdot {\mbox{s}}

where

Wb = weber, T = tesla, J = joule, m = meter, s = second, A = ampere, V = volt, C = coulomb,
F = farad, Hz = hertz, Ω = ohm

"

We can take simply ＜s Ω ＞.

Next we must review the fith equation.

Total Circuit Reactance XT = XC - XL or XL - XC
Why these or and why not XT = XC + XL ?

We are not able to find it in this article but find it in the previous article ＜Series RLC Circuit Analysis
＞ (http://www.electronics-tutorials.ws/accircuits/series-circuit.htmlhttp://www.electronics-tutorials.ws/accircuits/series-circuit.htm), which I will not quote in this post as it is a bit long and I have already quoted the above formulas in this post. Too much copies. We must think independently here. But we need the following drawing.

series rlc circuit analysis

We discussed a much simpler case of Series of Resistors at the end of Post ＜Reciprocal＞.

"

R = R₁ + R₂ + R₃ + ...

Generally

VT = V1 + V2 +V3.

IR = V
Therefore
ITRT = I1R1 + I2R2 + I3R3
Since IT = I1 = I2 = I3

Therefore RT = R1 + R2 + R3
Inductors act like resistors as XL (which is regarded as resistance) = 2 $π$ fL. On the other and Capacitors act as reciprocal of Resistors as XC (which is regarded as resistance) = 1/2 $π$ fC.
"
These two schematics are similar in terms of the arrangement - in Series, but the difference is big. We can re-arrange R-L-C Series in terms of "resistance".

------- R --- XL(2 $π$ fL) --- Xc (1/2 $π$ fC) --------

Then use the same reasoning

Generally

VT = VR + VL +VC.

IR(X) = V
Therefore
ITR(X)T = IRR + ILXL(2 $π$ fL) + ICXc (1/2 $π$ fC)
Since IT = IR = IL = IC

Therefore R(X)T = R + XL(2 $π$ fL) + Xc (1/2 $π$ fC)
-------
So back to the same question. Why not XT = XC + XL ?
and why not R(X)T = R + XL(2 $π$ fL) + Xc (1/2 $π$ fC)

Look at the graph again.

Vs (Voltage Source) is AC not DC. And here I is also AC.

I(t) = Im sin (ωt), where ω (omega) = 2 $π$ f

The reactions of an inductor and a capacitor to AC are very interesting and they act oppositely or RECIPROCALLY or inversely in terms of frequency (cycle of up and down (voltage) or back and forth (current) as

XL = 2 $π$ fL.
XC = 1/2 $π$ fC

show.

Again back to
Total Circuit Reactance XT = XC - XL or XL - XC
Why or and why not XT = XC + XL ?

For simplicity we omit R here and use the same reasoning.

Generally

VT = VL +VC.

IX = V (please recall IR=V (Ohm's Law))

Therefore

IXT = IXL + IXc ( IT = IL = IC)
Therefore XT = XL + Xc

What is wrong ? Well very wrong. An inductor and a capacitor do not act in the same way as a resistor against or in AC (current and voltage). AC is not always sinusoidal but quite often so (like Radio signals). Some basic equations should be introduced here and used to get correct (correctly work) equations. And Calculus is needed to get the points.

1) Calculus Stage One

Inductor

v = L ----
dt

Please note that when there is no change of current (DC Current) then v = 0.

Capacitor

i = C ----
dt

Please note that when there is no change of voltage (DC voltage) then i = 0.

Above we considered

VT = VL +VC.

Unlike in the case of resistors VT, VL, VC are not constant. (should be re-write as vT = vL +vC )
di

VL = v = L ----
dt

VL (v) changes with time as i (I(t) = Im sin (ωt)) changes. The faster the change of ＜i＞becomes the higher VL (v) becomes. The faster means: the frequency becomes higher, say 50Hz to 1000Khz (1KHz) or to 1,000,000Hz (1MHz). AM Radio frequency 500KHz - 1,600KHz. FM frequency 76 - 108MHz. These are quite faster than 50Hz. When considering even at 50Hz ＜i＞ changes (up from the original current level to the maximum current level and down to the minimum current level and up again to the original current level, called one cycle) 50 times changes (cycles) per one second, 1KHz and 1MHz make extremely fast change of up and down. It is not easy to imagine these changes of the current as compared with changes of the voltage (up and down). Actually we cannot see the current flow and the change of the current flow. Just imagine. At least we can imagine the change of the direction of the current (back and forth). VL (v) changes with time though very regularly and periodically as i (I(t) = Im sin (ωt)) changes. Therefore VL (v) at 1MHz is much higher than the VL (v) at 1Hz. This is the point.

Next VC. The relation between ＜i＞ and ＜v＞ under AC or sinusoidal change is

i = C ----
dt

This is not a reciprocal of

v = L ----
dt

but looks like it. C replaces L and ＜i＞ and ＜v＞ are exchanged. Easy to remember. We want to have the value of Vc, not Ic. To get Vc we need an integration procedure.
＜Electronics Tutorials＞uses Q (electric charge on the plate on a capacitor) in ＜Series RLC Circuit Analysis＞, but does not elaborate it much.

Q
---- = v
C

Q comes from
dQ
i = -------
dt

＜What "electric current" is ? ＞ is a big issue. Quoting wiki "electric current"

"
For a steady flow of charge through a surface, the current I (in amperes) can be calculated with the following equation:

I={Q \over t}\,,

where Q is the electric charge transferred through the surface over a time t. If Q and t are measured in coulombs and seconds respectively, I is in amperes.
More generally, electric current can be represented as the rate at which charge flows through a given surface as:

I={\frac {\mathrm {d} Q}{\mathrm {d} t}}\,.

"
This explanation is very much like those found in many common text books. Please see my tutorials ＜What is Electric Current?＞and ＜What is Electric Current? - 2＞.

Q dQ dq
What is the difference between I = ------ and I = ------ (or i = ------ ) ?
t dt dt

The definition of the unit of current, Ampere is ( from wiki ＜Ampere＞)

"
. . . . . .

Conversely, a current of one ampere is one coulomb of charge going past a given point per second:

{\rm {1\ A=1{\tfrac {C}{s}}.}}

In general, charge Q is determined by steady current I flowing for a time t as Q = It.

"
Then there is s distinctive difference. In this explanation " In general" is not a generalization. " steady current I flow" is a very special case. We can directly (not through differentiation) count Q even under the condition of "going past a given point per second", which can be imagined by an analogy of water flow through a point (or water flow through a pipe (or a hose) / its cross sectional Area.

                                      dQ                                    dq
Real generalization is I = ------ or more correctly i = ------
                                       dt                                       dt

where ＜i＞and ＜q＞ are not fixed or constant values but those changing with time. In this case when ＜q＞ is constant (not changing with time) ＜i＞ is zero (0), no flow of current ?

                 Q                   dq
Again, I = -----   and i = ------ are different things.
                  t                    dt

The latter is not a generalization of the former.

Then how to get Q to get Vc ?
This Q (which is generally defined as integration of ＜i＞), the Q related with a capacitor comprises a equation with ＜v＞. Look at the Unit ＜F＞ again.

{\mbox{F}}=\,\mathrm {{\frac {A\cdot s}{V}}={\dfrac {\mbox{J}}{{\mbox{V}}^{2}}}={\dfrac {{\mbox{W}}\cdot {\mbox{s}}}{{\mbox{V}}^{2}}}={\dfrac {\mbox{C}}{\mbox{V}}}={\dfrac {{\mbox{C}}^{2}}{\mbox{J}}}={\dfrac {{\mbox{C}}^{2}}{{\mbox{N}}\cdot {\mbox{m}}}}={\dfrac {{\mbox{s}}^{2}\cdot {\mbox{C}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {{\mbox{s}}^{4}\cdot {\mbox{A}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {\mbox{s}}{\Omega }}} ={\dfrac {{\mbox{s}}^{2}}{\mbox{H}}}

Here ＜C＞ is Coulomb (the unit of charge) and ＜F＞ is Farad.
                         As              C
We can see F = ----- and -----.
                          V               V

Unit wise,

FV
----   = A
s

or

CV
---- = i
t

This relates with the basic differential equation.

i = C ----
dt

Still we are not able to get Vc ?

Vc seems simply an accumulation of the charge(s) on a capacitor. And this Vc changes with Ic (= I(t) = Im sin (ωt)). The amount and direction of Ic change with time. Wait a minute. What is "the amount of Ic" ? i = dQ / dt. So the greater change of the amount of Q, not the greater the Q is, the greater i becomes. Please recall that the greater the change of of velocity (speed), not the greater the velocity (speed) is, the bigger the force is created.

Q
---- = v, or Q = Cv
C

When C is fixed, Q changes simply proportionally with v while i changes proportionally with derivative of v in a capacitor. The faster v changes the more current flows (or more correctly the faster Q changes).

This explains more correctly

First: the more current flows the faster Q changes.

The more current flows in one direction the faster Q increases.
The more current flows in the opposite direction the faster Q decreases.

This is

I={\frac {\mathrm {d} Q}{\mathrm {d} t}}\,.

And it can be applied in a capacitor circuit. In a circuit with no capacitor this cannot work. Just simply

I={Q \over t}\,,

where I and there for Q are constant.

2) Calculus Stage Two

To answer to the question. Why not XT = XC + XL but

XT = XC - XL or XL - XC

We need Trigonometric Calculus. ＜Electronics Tutorials＞ does not use Trigonometric Calculus and simply say

"
We recall from the previous tutorial about series RLC circuits that the voltage across a series combination is the phasor sum of V_R, V_L and V_C. Then if at resonance the two reactances are equal and cancelling, the two voltages representing V_L and V_C must also be opposite and equal in value thereby cancelling each other out because with pure components the phasor voltages are drawn at +90^o and -90^o respectively.
Then in a series resonance circuit as V_L = -V_C the resulting reactive voltages are zero and all the supply voltage is dropped across the resistor. Therefore, V_R = V_supply and it is for this reason that series resonance circuits are known as voltage resonance circuits, (as opposed to parallel resonance circuits which are current resonance circuits).

" We need only two Trigonometric Calculus to explain XT = XC - XL or XL - XC

d
---- sin (x) = cos (x)
dx

and
d
∫ sin (x) dx = - cos (x), this comes from more basic ---- cos (x) = - sin (x).
dx

To be convinced you read and understand some text book. But I like the following intuitive ways. From also ＜Electronics Tutorials＞

Inductor

v = L ----
dt

When ＜i＞ is sinusoidal like I(t) = Im sin (ωt) by using

d
---- sin (x) = cos (x)
dx

＜v＞becomes like V(t) = (Vm) cos (ωt), which is shown in the above nice drawing VL, exactly.

Next

Capacitor

i = C ----
dt

To integrate the both sides we use

∫ sin (x) dx = - cos (x)

∫ sin (ωt) dx = - C Vm cos (ωt), which is also shown in the above nice drawing VC, exactly.

d
This process looks like a math magic. ----- disappeared and cos (ωt) came out.
dt

But isn't it beautiful operation ?

sptt

Saturday, August 20, 2016

Reciprocal - 2 (LC Resonant Circuits)

At the end of Post ＜Reciprocal＞I wrote

"
More interesting reciprocal nature will be found in the combination of LC Resonance Circuits, which I will review later.

"

LC Resonance Circuits are in a wonder land. We can see them from many different points of views. Let's look at them in the reciprocal view. I mostly quote from ＜Electronics Tutorials＞ which I recently read several times.This is a good reading material as a tutorial as the name shows.

http://www.electronics-tutorials.ws/accircuits/series-resonance.html

< Skipped>

The first and the second equations were already introduced at the end of Post ＜Reciprocal＞

"
Inductors act like resistors as XL (which is regarded as resistance) = 2 $π$ fL. On the other hand Capacitors act as reciprocal of Resistors as XC (which is regarded as resistance) = 1/2 $π$ fC.

"
Please note that XC is a reciprocal form. So the chart becomes y = 1/x form like below.

The difference is x : 2 $π$ fC. <x> is a variable while f is variable but 2 $π$ is constant and C can be either a constant or the second variable.

＜f＞ has unit ＜(Number) / sec＞. The unit of C is Farad but originally (from wiki "Farad")

"
A farad has the base SI representation of: s⁴ × A² × m⁻² × kg⁻¹
It can further be expressed as:

{\displaystyle {\mbox{F}}=\,\mathrm {{\frac {A\cdot s}{V}}={\dfrac {\mbox{J}}{{\mbox{V}}^{2}}}={\dfrac {{\mbox{W}}\cdot {\mbox{s}}}{{\mbox{V}}^{2}}}={\dfrac {\mbox{C}}{\mbox{V}}}={\dfrac {{\mbox{C}}^{2}}{\mbox{J}}}={\dfrac {{\mbox{C}}^{2}}{{\mbox{N}}\cdot {\mbox{m}}}}={\dfrac {{\mbox{s}}^{2}\cdot {\mbox{C}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {{\mbox{s}}^{4}\cdot {\mbox{A}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {\mbox{s}}{\Omega }}} ={\dfrac {{\mbox{s}}^{2}}{\mbox{H}}}}

{\mbox{F}}=\,\mathrm {{\frac {A\cdot s}{V}}={\dfrac {\mbox{J}}{{\mbox{V}}^{2}}}={\dfrac {{\mbox{W}}\cdot {\mbox{s}}}{{\mbox{V}}^{2}}}={\dfrac {\mbox{C}}{\mbox{V}}}={\dfrac {{\mbox{C}}^{2}}{\mbox{J}}}={\dfrac {{\mbox{C}}^{2}}{{\mbox{N}}\cdot {\mbox{m}}}}={\dfrac {{\mbox{s}}^{2}\cdot {\mbox{C}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {{\mbox{s}}^{4}\cdot {\mbox{A}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {\mbox{s}}{\Omega }}} ={\dfrac {{\mbox{s}}^{2}}{\mbox{H}}}

where F=farad, A=ampere, V=volt, C=coulomb, J=joule, m=metre, N=newton, s=second, W=watt, kg=kilogram, Ω=ohm, H=henry.

"
If we simply take ＜s / ‎Ω＞, 2 $π$ fC is in ＜/ ‎Ω＞(Reciprocal of ‎Ω). Therefore ＜1 / ‎2 $π$ fC＞ is in ‎Ω.

＜f＞can be several Hz to several Giga Hz but for AM ranging 500KHz - 1600KHz, for FM 80MHz (80,000KHz) to 120MHz (120,000KHz). C can be several pico Farad to several micoro Farad.

For AM Tuning (which uses LC Resonance and Oscillation) we can use
＜f＞as 1,000KHz and C as 100pF. If we use original units

1,000KHz ---- ＞ 1,000,000Hz

100pF ---- ＞100 / 1,000, 000, 000,000 F or 1 / 10,000,000.000 F. This is very small number as compared with Farad or even uF (1/1,000,000 F). These small numbers make differences is a remarkable thing.

We put these two numbers

1 / ‎2 $π$ fC = 1 / 2 x 3.14 x 1,000,000 (Hz in /s) x (1/ 10,000,000,000 (F in s / ‎Ω)

= 1 / 2 x 3.14 x 1 x (1 / 10,000 (Ω)

= 1 / 2 x 3.14 x 0.0001

= 1 / 0.000628 = 1,591 Ω (=XC)

We can set L as 100uH.

XL = 2 $π$ fL = 2 x 3.14 x 1,000,000 (Hz in /s) x 100 / 1,000,000 H = 6.28 x 100 (Hz in /s) H

= 628 (supposed to be in Ω)

The unit of H is Henry and originally (from wiki "Henry")

{\mbox{H}}={\dfrac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{s}}^{2}\cdot {\mbox{A}}^{2}}}={\dfrac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{C}}^{2}}}={\dfrac {\mbox{J}}{{\mbox{A}}^{2}}}={\dfrac {{\mbox{T}}\cdot {\mbox{m}}^{2}}{\mbox{A}}}={\dfrac {\mbox{Wb}}{\mbox{A}}}={\dfrac {{\mbox{V}}\cdot {\mbox{s}}}{\mbox{A}}}={\dfrac {{\mbox{s}}^{2}}{\mbox{F}}}={\dfrac {\mbox{1}}{{\mbox{F}}\cdot {\mbox{Hz}}^{2}}}=\Omega \cdot {\mbox{s}}

where

Wb = weber, T = tesla, J = joule, m = meter, s = second, A = ampere, V = volt, C = coulomb,
F = farad, Hz = hertz, Ω = ohm

We can take ＜s Ω ＞

Next we must review the firth equation.

Total Circuit Reactance XT = XC - XL or XL - XC

Why or and why not XT = XC + XL ?

We are not able to find it in this article but find it in the previous article ＜Series RLC Circuit Analysis＞ (http://www.electronics-tutorials.ws/accircuits/series-circuit.htmlhttp://www.electronics-tutorials.ws/accircuits/series-circuit.htm), which I do not quote this article as it is a bit long and I have already quoted this article. Too much copies. We must think independently here.

We discussed a much simpler case of Series of Resistors at the end of Post ＜Reciprocal＞.

"

R = R₁ + R₂ + R₃ + ...

Generally

VT = V1 + V2 +V3.

IR = V

Therefore

ITRT = I1R1 + I2R2 + I3R3

Since IT = I1 = I2 = I3

Therefore RT = R1 + R2 + R3

As already mentioned Inductors act like resistors as XL (which is regarded as resistance) = 2 $π$ fL while Capacitors act as reciprocal of Resistors as XC (which is regarded as resistance) = 1/2 $π$ fC.

"

These two schematics are similar in terms of the arrangement - in Series, but the difference is big. We can re-arrange R-L-C Series in terms of "resistance".

------- R --- XL(2 $π$ fL) --- Xc (1/2 $π$ fC) --------

Then use the same reasoning

Generally

VT = VR + VL +VC.

IR(X) = V

Therefore

ITR(X)T = IRR + ILXL(2 $π$ fL) + ICXc (1/2 $π$ fC)

Since IT = IR = IL = IC

Therefore R(X)T = R + XL(2 $π$ fL) + Xc (1/2 $π$ fC)

-------

So back to the same question. Why not XT = XC + XL ?

and why not R(X)T = R + XL(2 $π$ fL) + Xc (1/2 $π$ fC)

Vs (Voltage Source) is AC not DC. And here I is also AC.

I(t) = Im sin (ωt), where ω (omega) = 2 $π$ f

The reactions of an inductor and a capacitor to AC are very interesting and they act oppositely or RECIPROCALLY or inversely in terms of frequency (cycle of up and down (voltage) or back and forth (current) as

XL = 2 $π$ fL.
XC = 1/2 $π$ fC

show.

Again back to

Total Circuit Reactance XT = XC - XL or XL - XC

Why or and why not XT = XC + XL ?

For simplicity we omit R here and use the same reasoning.

Generally

VT = VL +VC.

IX = V (please recall IR=V (Ohm's Law))

Therefore

IXT = IXL + IXc ( IT = IL = IC)

Therefore XT = XL + Xc

What is wrong ? Well very wrong. An inductor and a capacitor do not act in the same way as a resistor against or in AC (current and voltage). AC is not always sinusoidal but quite often so (like Radio signals). Some basic equations should be introduced here and used to get correct (correctly work) equations. And Calculus is needed to get the points.

1) Calculus Stage One

Inductor

v = L ----
dt

Please note that when there is no change of current (DC Current) then v = 0.

Capacitor

i = C ----
dt

Please note that when there is no change of voltage (DC voltage) then i = 0.

Above we considered

VT = VL +VC.

Unlike in the case of resistors VT, VL, VC are not constant. (should be re-write as vT = vL +vC )

di

VL = v = L ----
dt

i = C ----
dt

This is not a reciprocal of

v = L ----
dt

but looks like it. C replaces L and ＜i＞ and ＜v＞ are exchanged. Easy to remember. We want to have the value of Vc, not Ic. To get Vc we need an integration procedure.

＜Electronics Tutorials＞uses Q (electric charge on the plate on a capacitor) in ＜Series RLC Circuit Analysis＞, but does not elaborate it much.

Q
---- = v
C

Q comes from

dQ
i = -------
dt

＜What "electric current" is ? ＞ is a big issue. Quoting wiki "electric current"

"
For a steady flow of charge through a surface, the current I (in amperes) can be calculated with the following equation:

I={Q \over t}\,,

I={\frac {\mathrm {d} Q}{\mathrm {d} t}}\,.

"

This explanation is very much like those found in common text books. Please see my tutorials ＜What is Electric Current?＞and ＜What is Electric Current? - 2＞.

Q dQ dq
What is the difference between I = ------ and I = ------ (or i = ------ ) ?
t dt dt

The definition of the unit of current, Ampere is ( from wiki ＜Ampere＞)

"
. . . . . .

Conversely, a current of one ampere is one coulomb of charge going past a given point per second:

{\rm {1\ A=1{\tfrac {C}{s}}.}}

In general, charge Q is determined by steady current I flowing for a time t as Q = It.

"

Then there is s distinctive difference. In this explanation " In general" is not a generalization. " steady current I flow" is a very special case. We can directly (not through differentiation) count Q even under the condition of "going past a given point per second", which can be imagined by an analogy of water flow through a point (or water flow through a pipe (or a hose) / its cross sectional Area.

                                              dQ                                    dq
Real generalization is I = ------ or more correctly i = ------
                                               dt                                       dt

where ＜i＞and ＜q＞ are not fixed or constant values but those changing with time. In this case when ＜q＞ is constant (not changing with time) ＜i＞ is zero (0), no flow of current ?

                   Q                   dq
Again, I = -----   and i = ------ are different things. The latter is not a generalization of the former.
                    t                    dt

Then how to get Q to get Vc ?

This Q (which is generally defined as integration of ＜i＞), the Q related with a capacitor comprises a equation with ＜v＞. Look at the Unit ＜F＞ again.

{\mbox{F}}=\,\mathrm {{\frac {A\cdot s}{V}}={\dfrac {\mbox{J}}{{\mbox{V}}^{2}}}={\dfrac {{\mbox{W}}\cdot {\mbox{s}}}{{\mbox{V}}^{2}}}={\dfrac {\mbox{C}}{\mbox{V}}}={\dfrac {{\mbox{C}}^{2}}{\mbox{J}}}={\dfrac {{\mbox{C}}^{2}}{{\mbox{N}}\cdot {\mbox{m}}}}={\dfrac {{\mbox{s}}^{2}\cdot {\mbox{C}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {{\mbox{s}}^{4}\cdot {\mbox{A}}^{2}}{{\mbox{m}}^{2}\cdot {\mbox{kg}}}}={\dfrac {\mbox{s}}{\Omega }}} ={\dfrac {{\mbox{s}}^{2}}{\mbox{H}}}

Here ＜C＞ is Coulomb (the unit of charge) and ＜F＞ is Farad.

                          As              C
We can see F = ----- and ----
                            V             V

Unit wise,

FV
----   = A
s

or

CV
---- = i
t

This relates with the basic differential equation.

i = C ----
dt

I={\frac {\mathrm {d} Q}{\mathrm {d} t}}\,.

And it can be applied in a capacitor circuit. In a circuit with no capacitor this cannot work. Just simply

I={Q \over t}\,,

where I and there for Q are constant.

2) Calculus Stage Two

To answer to the question. Why not XT = XC + XL but

XT = XC - XL or XL - XC

We need Trigonometric Calculus. ＜Electronics Tutorials＞ does not use Trigonometric Calculus and simply say

"
We recall from the previous tutorial about series RLC circuits that the voltage across a series combination is the phasor sum of V_R, V_L and V_C. Then if at resonance the two reactances are equal and cancelling, the two voltages representing V_L and V_C must also be opposite and equal in value thereby cancelling each other out because with pure components the phasor voltages are drawn at +90^o and -90^o respectively.
Then in a series resonance circuit as V_L = -V_C the resulting reactive voltages are zero and all the supply voltage is dropped across the resistor. Therefore, V_R = V_supply and it is for this reason that series resonance circuits are known as voltage resonance circuits, (as opposed to parallel resonance circuits which are current resonance circuits).

"

We need only two Trigonometric Calculus to explain XT = XC - XL or XL - XC.

d
---- sin (x) = cos (x)
dx

and

d
∫ sin(x)dx = - cos (x), this comes from more basic ---- cos(x) = - sin (x).
dx

To be convinced you read and understand some text book. But I like the following intuitive ways. From also ＜Electronics Tutorials＞

＜skipped＞

Inductor

i = C ----
dt

To integrate the both side we use

sptt