Calculation of the magnetomotive force Calculation of the number of whorls of a reel Calculation of the intensity of the current of a reel Calculation of the absolute magnetic permeability Calculation of the relative magnetic permeability Calculation of the inductance of a reel Calculation of the flow embraced by the whorls Calculation of the inductance of a reel Intensity of the current which traverses a reel Electromotive force induced in a whorl Calculation of the electromotive force of self-induction Calculation of the inductance of a rolling up Calculation of the total inductance of 2 reels in series Calculation of inductance Leq connected in parallel Calculation of Leq of 2 reels connected in parallels Calculation of the inductance of an unknown reel Calculation of Leq of several reels connected in parallel Calculation of the energy stored by a reel Calculation of the frequency of a periodic size Calculation of the period of a periodic size Calculation of the effective value of a current Calculation of the maximum value of a current Peak value with peak of a sinusoidal alternating voltage Calculation of the average value of a current Calculation of the pulsation of a periodic size Calculation of the frequency of a periodic size Calculation of the pulsation of a periodic size Calculation of the period of a periodic size Calculation of the capacitive reactance of a condenser Calculation of the capacitance of a capacitor Calculation of the inductive reactance of a reel Calculation of the inductance of a reel Return to the synopsis To contact the author Low of page

Created it, 05/10/15

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MATHEMATICAL FORMS 4    “6th Part”

In this last form, you will find the continuation of the formulas concerning electromagnetism and the various sizes relating to the periodic signals most usually used in electronics.

FORMULATE 124 - Calculation of the magnetomotive force produced by a reel traversed by a current, knowing the number of whorls of rolling up and the intensity of the current.

Statement : The magnetomotive force, expressed in ampere turns, is obtained by multiplying the number of whorls by the intensity of the current, expressed in amps.

F = NI

F = magnetomotive force in A.t (ampere turn)

N = number of whorls

I = intensity of the current in A (amp)

• Example :

Data :  N = 1 600 (a number of whorls of a reel) I = 0,05 A (intensity of the current which traverses the rolling up of the reel).

Magnetomotive force produced by the reel : F = 1 600 x 0,05 = 80 A.t.

FORMULATE 125 - Calculation of the number of whorls of a reel, knowing the magnetomotive force that it must produce and the intensity of the current which traverses rolling up.

N = F / I

N = number of whorls of the inductor

F = magnetomotive force in A.t (ampere turn)

I = intensity of the current in A (amp)

(This formula is drawn from formula 124).

• Example :

Data : F = 100 A.t ;  I = 0,025 A

Number of whorls of rolling up : N = 100 / 0,025 = 4 000.

FORMULATE 126 - Calculation of the intensity of the current which must traverse the rolling up of a reel, knowing the magnetomotive force of the reel and the number of whorls of rolling up.

I = F / N

I = intensity of the current in A (amp)

F = magnetomotive force in A.t (ampere turn)

N = number of whorls of the reel

(This formula is drawn from formula 124).

Example :

Data :  F = 50 A.t.  ;  N = 800

Intensity of the current which traverses the rolling up of the reel :

I = 50 / 800 = 0,0625 A

FORMULATE 127 - Calculation of the absolute magnetic permeability of a material, knowing the absolute permeability of the vacuum and the relative permeability of material.

Statement : The absolute magnetic permeability of a material, expressed out of microphone-henrys per meter, is obtained by multiplying the absolute magnetic permeability of the vacuum, expressed out of microphone-henrys per meter, by the relative permeability of material.

µ = µo x µr

µ = absolute magnetic permeability of material in µH / m (microphone-Henry per meter)

µo = absolute magnetic permeability of the vacuum equalizes with 1,256 µH / m (micro-henry per meter)

µr = relative magnetic permeability of material (table VII, figure 1).

• Example :

Data: µo = 1,256 µH / m ;  µr = 2 000 (maximum value of the relative permeability of magnetic alloy perminvar 45 - 25 (maximum value) :

µ = 1,256 x 2 000 = 2 512 µH / m.

OBSERVATION :

To table VII of figure 1, one gave the values of the magnetic permeability relating to some diamagnetic, paramagnetic and ferromagnetic materials. It will be noted that in general, the diamagnetic materials have a value of relative permeability slightly lower than 1 and that the paramagnetic materials have values of relative permeability slightly higher than 1 ; these values can be regarded as practically equal to 1 in all the technical applications.

The absolute permeability of diamagnetic and paramagnetic materials is thus practically equal to that of the vacuum.

µ = µo x µr = 1,256 x 1 = 1,256 µH / m

The permeability of ferromagnetic materials is not constant but changes according to the variation of the intensity of magnetizing; for this reason, one indicated to table VII the maximum values. These values are not however enough to characterize the behavior of ferromagnetic materials; to this end, it would be necessary to provide particular graphs, called curves of magnetizing; by these graphs (which present little interest for the electronics specialist and which will thus not be taken into account in the form), one can establish the value of the absolute permeability of ferromagnetic materials in relation to the various values of the intensity of magnetizing of these same materials.

FORMULATE 128 - Calculation of the relative magnetic permeability of a material, knowing the absolute permeability of the vacuum and material.

µr = µ / µo

µr = relative magnetic permeability

µ = absolute permeability of material in µH / m (microhenry per meter)

µo = absolute permeability of the vacuum = 1,256 µH / m (microhenry per meter)

(This formula is drawn from formula 127).

• Example :

Data :  µ = 8,792 µH / m (absolute magnetic permeability of iron silicon)µo = 1,256 µH / m

Relative permeability of iron silicon : µr = 8 792 / 1,256 = 7 000

(See table VII, figure 1).

FORMULATE 129 - Calculation of the inductance of a reel with only one layer, without core, knowing the absolute permeability of the air, the section of the whorls, the number of whorls and the length of the reel.

Statement : The inductance of a reel with only one layer, without core, expressed in microhenrys, is obtained by multiplying the absolute permeability of the air, expressed in microhenrys per meter, by the section of the whorls, expressed in square centimetres, by the square of the number of whorls and by dividing the product obtained by the length of the reel, expressed in centimetres and multiplied by 100.

L = µSN2 / 100 I

L = inductance in µH (microhenry)

µ = absolute permeability of the air in µH / m (microhenry per meter)

S = section of the whorls in cm2

N = number of whorls

I = length of the reel in cm.

• Example :

Data : µ 1,256 µH / m (value of the absolute permeability of the air ; the air is a paramagnetic substance; to on this subject see table VII, figure 1 and the observation which follows formula 127) ; S = 7,068 cm2 (section of a cylindrical reel with jointed whorls 3 cm in diameter; for the calculation of the section, knowing the diameter, to see formula 19 of the form) ; N = 120 whorls ; I = 3,6 cm (length of the reel).

Inductance of the reel :

L = (1,256 x 7,068 x 1202) / (100 x 3,6) = (1,256 x 7,068 x 14 400) / (100 x 3,6)

L 127 834 / 360 355 µH

OBSERVATION :

Formula 129 for the calculation of inductance is valid in theory, when it is admitted that all the magnetic flux produced by the current is embraced by the whorls of rolling up; in practice, it however happens that part of the magnetic flux product is dispersed; to take account of dispersed flow, one has recourse to empirical formulas for the design calculations.

FORMULATE 130 - Calculation of the flow embraced by the whorls of a rolling up, knowing the inductance and the intensity of the current.

Statement: The flow embraced by the whorls of a rolling up, expressed in webers, is obtained by multiplying the inductance, expressed in henrys, by the intensity of the current which traverses rolling up, expressed in amps.

Fc = LI

Fc = flow embraced by the whorls of a rolling up in Wb (weber)

L = inductance out of H (Henry)

I = intensity of the current in A (amp)

• Example :

Data :  L = 2,5 H ;  I = 0,03 A.

Flow embraced by the whorls of rolling up :  Fc = 2,5 x 0,03 = 0,075 Wb.

OBSERVATION :

Formula 130 refers to an ideal inductor, i.e. with a reel in which all the magnetic flux produced by the current is embraced by the whorls of rolling up.

FORMULATE 131 - Calculation of the inductance of a reel, knowing the flow embraced by the whorls of rolling up and the intensity of the current which traverses it.

L = Fc / I

L = inductance out of H (Henry)

Fc = flow embraced in Wb (weber)

I = intensity of the current in A (amp)

(This formula is drawn from formula 130).

• Example :

Data :  Fc = 0,002 / 0,05 = 0,04 H

FORMULATE 132 - Calculation of the intensity of the current which traverses a reel, knowing inductance and the flow embraced by the whorls of rolling up.

I = Fc / L

(This formula is drawn from formula 130).

• Example :

Data : Fc = 0,6 Wb ;  L = 250 mH (millihenry) = 0,25 H.

Intensity of the current : I = 0,6 / 0,25 = 2,4 A.

FORMULATE 133 - (law of NEUMANN). Calculation of the electromotive force induced in a whorl, knowing the variation of the magnetic flux embraced by the whorl and the time during which achieves this variation.

Statement : The induced electromotive force, expressed in volts, is obtained by dividing the variation of embraced flow expressed into webers, by the time during which this variation occurs, expressed in seconds.

E = (F"c - F'c) / (t" - t')

E = induced electromotive force out of V (volt)

F"c = value of the flow embraced at the end of the interval considered

F'c = value of the flow embraced at the beginning of the interval considered

F"c - F'c = variation of flow in Wb (weber) which is often written Fc

t" = urgent final

t' = urgent initial

t"- t' = interval of time in seconds which is often written t

The formula is also written :

E = Fc /  t

• Example :

Data : F"c - F'c = Fc = 2,2 Wb ;  t"- t' = t = 0,02 s.

Induced electromotive force:  E = 2,2 / 0,02 = 110 V

FORMULATE 134 - Calculation of the electromotive force of self-induction, knowing the inductance of rolling up, the variation of the intensity of the current which traverses it and the time during which occurs this variation.

Statement : The electromotive force of self-induction, expressed in volts, is obtained by multiplying the inductance, expressed in henrys, by the variation of the intensity of the current, expressed in amps and dividing the product obtained by the lasting time which occurs this variation expressed in seconds.

E = L x (I" - I') / (t" - t') = LI / t

E = electromotive force of self-induction out of V (volt)

L = inductance out of H (henry)

I" = final intensity of the current

I' = initial intensity of the current

I" - I' = variation of the intensity of the current in A (amp) which is often written I

t" = urgent final

t' = urgent initial

t" - t' = interval of time in seconds which is often written t

• Example:

Data :  L = 2,5 H (inductance of a rolling up with core)I = 0,6 A;  t = 0,01 s.

Electromotive force of self-induction:

E = (2,5 x 0,6) / 0,01 = 1,5 / 0,01 = 150 V

FORMULATE 135 - Calculation of the inductance of a rolling up, knowing the electromotive force of self-induction, the variation of the intensity of the current and the time during which occurs this variation.

L = E x (t" - t' / (I" - I') = Et /  I

L = inductance out of H (henry)

E = electromotive force of self-induction out of V (volt)

t" = urgent final

t' = urgent initial

t" - t' = t = interval of time in seconds

I" = final intensity of the current

I' = initial intensity of the current

I" - I' = I = variation of the intensity of the current in A (amp)

(This formula is drawn from formula 134)

• Example :

Data: E = 120 V ;  t = 0,01 s ;  I = 0,8 A.

Inductance of rolling up : L = (120 x 0,01) / 0,8 = 1,2 / 0,8 = 1,5 H

FORMULATE 136 - Calculation of the total inductance presented by two or several reels connected in series and not coupled between them, knowing the inductance of each reel.

Statement : The inductance presented at the same time by two or several reels connected in series is obtained by adding inductances with the reels.

Lt = L1 + L2 + L3 +… + Ln

Lt = total inductance

L1 = inductance of the first reel

L2 = inductance of the second reel

L3 = inductance of the third reel

Ln = inductance of the last reel

Various inductances all must be expressed in the same measuring unit.

• Examples :

a) Data : L1 = 0,5 H (henry) ;  L2 = 0,5 H ;  L3 = 1,5 H ;  Ln = 2 H.

Total inductance : Lt = 0,5 + 0,5 + 1,5 + 2 = 4,5 H.

b) Data : L1 = 20 mH (millihenry ; 1 mH = 0,001 H) ;  L2 = 5 mH.

Total inductance : Lt = 20 + 5 = 25 mH.

c) Data : L1 = 300 µH (microhenry ; 1 µH = 0,000001 H) ; L2 = 50 µH ;  L3 = 150 µH

Total inductance : Lt = 300 + 50 + 150 = 500 µH.

FORMULATE 137 - Calculation of the equivalent inductance presented by two or several reels of value equal, connected in parallel and not coupled between them, knowing their inductance.

Statement : The inductance presented by two or several equal reels connected in parallel, is obtained by dividing their inductance by the number of reels.

Leq = L / n

Leq = equivalent inductance

L = inductance of each reel

n = number of reels connected in parallel

Equivalent inductance will be expressed in the same measuring unit as that used to indicate the inductance of the reels.

• Example :

a) Data : L = 2 H (Henry) ;  numbers (n) of reel = 4

Equivalent inductance : Leq = 2 / 4 = 0,5 H.

b) Data : L = 50 mH (millihenry ; 1 mH = 0,001 H) ;  n = 2

Equivalent inductance : Leq = 50 / 2 = 25 mH.

c) Data : L = 600 µH (microhenry ; 1 µH = 0,000001 H) ;  n = 3

Equivalent inductance : Leq = 600 / 3 = 200 µH.

FORMULATE 138 - Calculation of the equivalent inductance of two reels of value different, connected in parallel and not coupled between them, knowing their inductance.

Statement : The sum of the inductances presented by two reels of different inductance, connected in parallel, is obtained by multiplying the two values and by dividing the product obtained by the sum of these same values.

Leq = (L1 x L2) / (L1 + L2)

Leq = equivalent inductance

L1 = inductance of a reel

L2 = inductance of the other winds.

The values all of inductance must be expressed in the same measuring unit.

• Example :

Data: L1 = 12 mH (millihenry) ;  L2 = 6 mH.

Equivalent inductance : Leq = (12 x 6) / (12 + 6) = 72 / 18 = 4 mH.

FORMULATE 139 - Calculation of the inductance of a reel to be connected in parallel to another reel of known value, to obtain a known equivalent inductance (the two reels should not be coupled between them).

Statement : The inductance of a reel to be connected in parallel to another reel, to obtain a given equivalent inductance, is calculated by multiplying the value of the reel known by equivalent inductance and by dividing the product by the difference in these values.

Li = (L x Leq) /( L - Leq)

Li = unknown inductance

L = value of the reel available

Leq = equivalent inductance which one wants to obtain.

The values all of inductance must be expressed in the same measuring unit.

• Example :

Data : L = 800 µH = 800 µH (microhenry)Leq = 600 µH

Unknown inductance : Li = (800 x 600) / (800 - 600) = 480 000 / 200 = 2 400 µH

FORMULATE 140 - Calculation of equivalent inductance several reels connected in parallel and not coupled between them, knowing their inductance.

Statement : Equivalent inductance several reels connected in parallel, is obtained by carrying out calculations in three times: initially, one calculates the reverse of the inductance of each reel, which amounts dividing number 1 by the value of the reel; then, one adds the values with the opposite ; finally, one calculates equivalent inductance by dividing number 1 by the sum of the opposite.

The values all of inductance must be expressed in the same measuring unit.

• Example :

Data : L1 = 2 mH (millihenry) ;  L2 = 4 mH ;  L3 = 4 mH.

OBSERVATION :

If one must calculate the equivalent inductance of two reels connected in parallel, one can use formula 140, but it is simpler to resort to formula 138 ; moreover, if inductances of the reels connected in parallel are equal between them, it is advisable to resort to formula 137.

FORMULATE 141 - Calculation of the energy stored by a reel, traversed by a current, knowing the inductance and the intensity of the current.

Statement : The energy stored by a reel, expressed in joules, is obtained by multiplying the inductance, expressed in henrys by the square of the intensity of the current, expressed in amps and dividing by 2 the product obtained.

W = (L x I2) / 2

W = power electric in J (joule)

L = inductance out of H (henry)

I = intensity of the current in A (amp)

• Example :

Data : L = 50 mH (millihenry) = 0,05 H   ;   I = 100 mA (milliampere) = 0,1 A

Stored electric power :

W = (0,05 x 0,12) / 2 = 0,05 x 0,01 / 2 = 0,00025 J.

FORMULATE 142 - Calculation of the frequency of a periodic size (for example, of a current with sinusoidal pace), knowing the period, i.e. duration of each cycle.

Statement : The frequency, expressed in hertz, is obtained by dividing number 1 by the period expressed into seconds.

f = 1 / T

f = frequency in Hz (hertz)

T = period in s (second)

• Example :

Data : T = 0,02 s.

Frequency : f = 1 / 0,02 = 50 Hz.

OBSERVATION :

If, in formula 142, the period is expressed in milliseconds (ms ; 1 ms = 0,001 s), the frequency will be expressed in kilocycle, (kHz; 1 kHz = 1 000 Hz) ; if on the contrary, the period is expressed in microseconds (µs ; 1 µs = 0,000001 s), the frequency will be expressed in megahertz (MHz ; 1 MHz = 1 000 000 Hz).

FORMULATE 143 - Calculation of the period of a periodic size (for example, of a AC current with sinusoidal pace) knowing the frequency, i.e. the number of cycles achieved during the unit of time.

Statement : The period, expressed in seconds, is obtained by dividing number 1 by the frequency, expressed in hertz.

T = 1 / f

T = period in s (second)

f = frequency in Hz (hertz)

• Example :

Data : f = 1 000 Hz.

Period : T = 1 / 1 000 = 0,001 s

OBSERVATION :

If, in formula 143, the frequency is expressed in kilocycle (kHz ; 1 kHz = 1 000 Hz), the period will be expressed in milliseconds (ms ; 1 ms = 0,001 s) ; if, on the contrary the frequency is expressed in megahertz (MHz ; 1 MHz = 1 000 000 Hz), the period will be expressed in microseconds (µs ; 1 µs = 0,000001 s).

FORMULATE 144 - Calculation of the effective value of a current (or tension) of type alternative and sinusoidal, knowing the maximum value.

Statement: The effective value of a current (or tension) of alternative type sinusoidal is obtained by dividing the maximum value by the square root of 2.

NOTE :

By multiplying the numerator and the denominator of this formula per 1,41, one also obtains the following simplified formula :

in the same way, one will have :

Ieff 0,707 x Imax

Ieff = Effective value of the current in A (amp)

Imax = maximum value of the current in A (amp).

• Example :

a) Data : maximum value of the AC current, Imax = 0,8 A (amp).

Effective value of the AC current : Ieff = 0,707 x 0,8 = 0,5656 A.

b) Data : maximum value of the alternating voltage : Vmax = 311 V (volt).

Effective value of the alternating voltage : Veff 0,707 x 311 = 220 V.

OBSERVATION :

The effective value of a current (or tension) alternating also depends on the form of wave. Indeed, if the wave is sinusoidal, (figure 2-a), the effective value is equal to the product of the maximum value by number 0,707 ; on the other hand, if the wave is rectangular (figure 2-b) the effective value is equal to the maximum value. If the wave is triangular (figure 2-c), the effective value is equal to the product of the maximum value by number 0,577 (table VIII, figure 3). In general, a precise number corresponds to each form of wave, numbers ranging between zero and 1, which multiplied by the maximum value, makes it possible to obtain the effective value.

FORMULATE 145 - Calculation of the maximum value of a current (or tension) alternate sinusoidal, knowing the effective value.

Vmax = Veff x 1,41

Vmax = maximum value of the tension out of V (volt)

Veff = Effective value of the tension out of V (volt)

square root 2 1,41

Imax = Ieff x 02

Imax = maximum value of the current in A (amp)

Ieff = effective value of the current in A (amp)

(The formulas above are drawn from formulas 144).

• Example :

a) Data : effective value of the AC current, Ieff = 2,5 A (amp)

2,5 A. maximum Valeur of the AC current : Imax = 1,41 x 2,5 = 3,5

b) Data : effective value of the alternating voltage, Veff = 160 V.

Maximum value of the alternating voltage, Vmax = 1,41 x 160 = 225,6 V.

OBSERVATION :

The number which multiplies the effective value depends on the form of wave. Indeed, if the wave is sinusoidal (figure 2-a), factor 1,414 will be used ; but if the wave is rectangular (figure 2-b), one will use factor 1 and the maximum value will be thus equal to the effective value ; moreover, if the wave is triangular (figure 2-c), one will use the factor 1,73 (table VIII, figure 3).

FORMULATE 146 - Calculation of the peak value to peak of a sinusoidal alternating voltage, knowing its effective value.

Statement: The peak value with peak of a sinusoidal alternating voltage, is obtained by multiplying the effective value by the number fixes 2,82.

Vcc = 2,82 x Veff

Vcc = peak value with tension peak

Veff = Effective value of the tension

Icc = 2,82 x Ieff

Icc = peak value with peak of the current

Ieff = effective value of the current

• Example :

Data : Effective value of the alternating voltage, Veff = 220 V (volt)

Peak value to tension peak, Vcc = 220 x 2,82 = 620,4 V.

OBSERVATION :

The number fixes which multiplies the effective value depends on the form of wave. Indeed, if the wave is sinusoidal (figure 2-a), factor 2,82 will be used; but if the wave is rectangular (figure 2-b), factor 2 will be used (table VIII, figure 3); if on the contrary, the wave is triangular (figure 2-c), one uses factor 3,46 (table VIII, figure 3).

FORMULATE 147 - Calculation of the average value of a current (or tension) alternate sinusoidal, knowing the effective value.

Statement: The average value of a current (or tension) alternate sinusoidal is obtained by multiplying the effective value by the number fixes 0,9.

Vm = 0,9 x Veff

Vm = average value of the tension out of V (volt)

Veff = Effective value of the tension out of V (volt)

Im = 0,9 x Ieff

Im = average value of the current in A (amp)

Ieff = effective value of the current in A (amp)

• Example :

a) Data : Effective value of the AC current, Ieff = 0,35 A (amp)

Average value of the AC current, Im = 0,9 x 0,35 = 0,315 A

b) Data : Effective value of the alternating voltage :  Veff = 220 V.

Average value of the alternating voltage : Vm = 0,9 x 220 = 198 V.

OBSERVATION:

The number fixes which multiplies the effective value depends on the form of wave. Indeed, if the wave is sinusoidal (figure 2-a), factor 0,9 will be used but if the wave is rectangular (figure 2-b), one will use factor 1 and so the average value will be equal to the effective value ; finally, if the wave is triangular (figure 2-c), factor 0,866 will be used (table VIII, figure 3).

FORMULATE 148 - Calculation of the pulsation of a periodic size, knowing its frequency.

Statement: The pulsation, expressed in radians a second, is given by twice the product of the P number by the frequency, expressed in hertz.

w = 2 Õ f 6,28 f

w = pulsation in rd / s (radian per second)

P = symbol of number 3,14 …

f = frequency in Hz (hertz)

• Example :

Data : f = 400 Hz

Pulsation: w 6,28 x 400 = 2 512 rd / s.

FORMULATE 149 - Calculation of the frequency of a periodic size, knowing the pulsation.

f = w / 2Õ 0,159 w

f = frequency in Hz (hertz)

w = pulsation in rd / s (radian per second)

Õ = symbol of a number fixes 3,14 …

(This formula is drawn from formula 148).

• Example :

Data : w = 2.000 rd / s.

Frequency : f 0,159 x 2 000 = 318 Hz.

FORMULATE 150 - Calculation of the pulsation of a periodic size, knowing its period.

w = 2Õ / T 6,28 / T

w = pulsation in rd / s (radian per second)

Õ = symbol of the number fixes 3,14 …

T = period in S (second)

The present formula is obtained while replacing in formula 148, the frequency f, by the second member of formula 142, that is to say1 / T.

• Example :

Data : T = 0,02 s (period of the AC current with 50 Hz).

Pulsation: w = 6,28 / 0,02 = 314 rd / s

FORMULATE 151 - Calculation of the period of a periodic size, knowing its pulsation.

T = 2Õ / w 6,28 / w

T = period in s (second)

Õ = symbol of the number fixes 3,14 …

w = pulsation in rd / s (radian per second)

(This formula is drawn from formula 150).

• Example:

data: w = 628 rd / s (pulsation of a AC current with 100 Hz)

Period : T = 6,28 / 628 = 0,01 s.

FORMULATE 152 - Calculation of the capacitive reactance of a condenser, knowing the capacity of the condenser and the frequency of the AC current which crosses it.

Statement: The capacitive reactance, expressed in ohms, is obtained by dividing number 1 by 6,28 (2Õ) by the frequency, expressed in hertz and by the capacity, expressed in farads.

Xc = 1 / (2Õ fC)  1 / (6,28 fC)

Xc = capacitive reactance in W (ohm)

Õ = symbol of the number fixes 3,14…

f = frequency in Hz (hertz)

C = capacity out of F (farad)

This formula is also written:

Xc = 1 / Cw

since according to formula 148, w = 6,28 f.

• Example :

Data : f = 3 000 Hz ;  C = 500 nF (nanofarad) = 0,0000005 F.

Capacitive reactance : Xc 1 / (6,28 x 3 000 x 0,0000005) = 1 / 0,00942 106,15 W.

OBSERVATION :

In calculations, it is often more convenient to replace the measuring units of the frequency and the capacity by the multiples and submultiples corresponding; in particular, it sometimes happens to find the frequency expressed in kilocycle (kHz) or in megahertz (MHz) and the capacity in nanofarads (nF), picofarads (pF), or in microfarads (µF). In all these cases, one can use formula 152, while holding account of what follows:

• If the frequency is expressed in hertz and the capacity in microfarads, the reactance will be expressed in megohms,

• If the frequency is expressed in kilocycle and the capacity in nanofarads, the reactance will be expressed in megohms,

• If the frequency is expressed in kilocycle and the capacity in microfarads, the reactance will be expressed in kilo-ohms,

•  If the frequency is expressed in megahertz and the capacity in nanofarads, the reactance will be expressed in kilo-ohms,

• Lastly, if the frequency is expressed in megahertz and the capacity in picofarads, the reactance will be expressed in megohms.

FORMULATE 153 - Calculation of the capacitance of a capacitor, knowing its capacitive reactance for a given frequency.

C = 1 / (2 Õ f Xc) 1 / (6,28 f Xc)

C = capacity out of F (farad)

Õ = symbol of the number fixes 3,14…

f = frequency in Hz (hertz)

Xc = capacitive reactance in W (ohm)

(This formula is drawn from formula 152).

• Example :

Data : F = 5 000 Hz ;  Xc = 400 W.

Capacity :  C 1 / (6,28 x 5 000 x 400) 1 / 12 560 000 = 0,000 000 079 617 F

C 79,61 nF (nanofarad).

FORMULATE 154 - Calculation of the inductive reactance of a reel, knowing its inductance and the frequency of the AC current which crosses it.

Statement : The inductive reactance, expressed in ohms, is obtained by multiplying the number fixes 6,28 (2Õ) by the frequency, expressed in hertz and by inductance, expressed in henrys.

XL = 2 Õ f L 6,28 f L

XL = inductive reactance in W (ohm)

Õ = symbol of the number fixes 3,14 …

f = frequency in Hz (hertz)

L = inductance out of H (henry)

or :  XL = Lw           Since          w = 6,28 f (formula 148).

• Example :

Data : f = 250 000 Hz ;  L = 0,006 H.

Inductive reactance : XL = 6,28 x 250 000 x 0,006 = 9 420 W.

OBSERVATION :

Formula 154 can be also used by expressing the frequency in kilocycle (kHz) and inductance in millihenrys (mH), or the frequency in megahertz (MHz) and inductance in microhenrys (µH); in one and the other case, the inductive reactance will be expressed in ohms.

FORMULATE 155 - Calculation of the inductance of a reel, knowing its inductive reactance for a given frequency.

L = XL / (2 Õ f) XL / 6,28 f

L = inductance out of H (henry)

XL = inductive reactance in W (ohm)

Õ = symbol of the number fixes 3,14 …

f = frequency in Hz (hertz)

(This formula is drawn from formula 154).

• Example:

Data : XL = 20 000 = 20 000 W ;  f = 700 kHz (kilocycles) = 700 000 Hz.

Inductance : L = 20 000 / (6,28 x 700 000) = 20 000 / 4 396 000 0,004 549 H

= 4,549 mH (millihenry).

(We thus finish our fourth mathematical forms and of recalls…).

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Daniel