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Form 1. 4. - Electromagnetism :
In this last form, you will find the following formulas concerning the electromagnetism and the various quantities relating to the periodic signals most commonly used in electronics.
FORMULA 124 Calculation of the magnetomotive force produced by a coil traversed by a current, knowing the number of turns of the winding and the intensity of the current.
Calculation of the magnetomotive force produced by a coil carried by a current
Enunciated : The magnetomotive force, expressed in ampere-turns, is obtained by multiplying the number of turns by the intensity of the current, expressed in amperes.
F = NI
F = magnetomotive force in A.t (ampere-turn)
N = number of turns
I = intensity of the current in A (ampere)
Example :
Data : N = 1 600 (number of turns of a coil) ; I = 0,05 A (intensity of the current flowing through the winding of the coil).
Magnetomotive force produced by the coil : F = 1 600 x 0,05 = 80 A.t.
FORMULA 125 - Calculation of the number of turns of a coil, knowing the magnetomotive force which it must produce and the intensity of the current which traverses the winding.
Calculation of the number of turns of a coil
N = F / I
N = number of turns of the inductor
F = magnetomotive force in A.t (ampere-turn))
I = intensity of the current in A (ampere)
(This formula is from formula 124).
Example :
Data : F = 100 A.t ; I = 0,025 A
Number of turns of the winding : N = 100 / 0,025 = 4 000.
FORMULA 126 - Calculation of the intensity of the current which must traverse the winding of a coil, knowing the magnetomotive force of the coil and the number of turns of the winding.
Calculation of the intensity of the current that must traverse the winding of a coil
I = F / N
I = intensity of the current in A (ampere)
F = magnetomotive force in A.t (ampere-turn)
N = number of turns of the coil
(This formula is from formula 124).
Example :
Data : F = 50 A.t. ; N = 800
Intensity of the current flowing through the winding of the coil :
I = 50 / 800 = 0,0625 A
FORMULA 127 - Calculation of the absolute magnetic permeability of a material, knowing the absolute permeability of the vacuum and the relative permeability of the material.
Calculation of the absolute magnetic permeability of a material
Enunciated : The absolute magnetic permeability of a material, expressed in microhenrys per meter, is obtained by multiplying the absolute magnetic permeability of the void, expressed in microhenrys per meter, by the relative permeability of the material.
µ = µo x µr
µ = absolute magnetic permeability of the material in µH / m (micro-henry per meter)
µo = absolute magnetic permeability of the vacuum equal to 1,256 µH / m (micro-henry per meter)
µr = relative magnetic permeability of the material (Table VII, Figure 1).
Example :
Data : µo = 1,256 µH / m ; µr = 2 000 (maximum value of the relative permeability of the magnetic alloy perminvar 45 - 25 (maximum value) :
µ = 1,256 x 2 000 = 2 512 µH / m.
In Table VII of Figure 1, the values of the relative magnetic permeability of some diamagnetic, paramagnetic and ferromagnetic materials have been given. It should be noted that in general, the diamagnetic materials have a relative permeability value slightly less than 1 and that the paramagnetic materials have relative permeability values slightly greater than 1 ; these values can be considered to be practically equal to 1 in all technical applications.
The absolute permeability of the 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 the ferromagnetic materials is not constant but changes according to the variation of the magnetization intensity ; for this reason, the maximum values are shown in Table VII. These values, however, are not sufficient to characterize the behavior of ferromagnetic materials ; for this purpose particular graphs, called magnetization curves, should be provided ; by these graphs (which are of little interest to the electronics engineer and therefore will not be taken into account in the form), it is possible to establish the value of the absolute permeability of the ferromagnetic materials in relation to the various values of the intensity. magnetization of these same materials.
FORMULA 128 - Calculation of the relative magnetic permeability of a material, knowing the absolute permeability of the vacuum and the material.
Calculation of the relative magnetic permeability of a material
µr = µ / µo
µr = relative magnetic permeability
µ = absolute permeability of the material in µH / m (microhenry per meter)
µo = absolute permeability of the vacuum = 1,256 µH / m (microhenry per meter)
(This formula is from Formula 127).
Example :
Data : µ = 8 792 µH / m (absolute magnetic permeability of silicon iron) ; µo = 1,256 µH / m
Relative permeability of silicon iron : µr = 8 792 / 1,256 = 7 000
(See Table VII, Figure 1).
FORMULA 129 - Calculation of the inductance of a coil with a single layer, without core, knowing the absolute permeability of the air, the section of the turns, the number of turns and the length of the coil.
Calculation of the inductance of a coil with a single layer
Enunciated : The inductance of a single-core, non-core coil, expressed in microhenrys, is obtained by multiplying the absolute air permeability, expressed in microhenrys per meter, by the section of the turns, expressed in square centimeters, by the square of the number of turns and dividing the product obtained by the length of the coil, expressed in centimeters and multiplied by 100.
L = µSN^{2} / 100 l
L = The inductance in µH (microhenry)
µ = absolute permeability of the air in µH / m (microhenry per meter)
S = section of turns in cm^{2}
N = number of turns
l = length of the coil in cm.
Example :
Data : µ 1,256 µH / m (absolute air permeability value, air is a paramagnetic substance ; see Table VII, Figure 1 and the observation following formula 127) ; S = 7,068 cm^{2} (section of a cylindrical coil with contiguous turns of 3 cm in diameter, for the calculation of the section, knowing the diameter, see Formula 19 of the form) ; N = 120 spires ; l = 3,6 cm (length of the coil).
Inductance of the coil :
L = (1,256 x 7,068 x 120^{2})^{ }/ (100 x 3,6) = (1,256 x 7,068 x 14 400) / (100 x 3,6)
L 127 834 / 360 355 µH
OBSERVATION :
The formula 129 for the calculation of the inductance is valid in theory, when it is admitted that all the magnetic flux produced by the current is embraced by the turns of the winding ; in practice, however, part of the magnetic flux produced is dispersed ; to account for the dispersed flow, empirical formulas are used for project calculations.
FORMULA 130 - Calculation of the flux embraced by the turns of a winding, knowing the inductance and the intensity of the current.
Calculation of the flux embraced by the turns of a winding
Enunciated : The flux embraced by the windings of a winding, expressed in webers, is obtained by multiplying the inductance, expressed in henries, by the intensity of the current flowing through the winding, expressed in amperes.
Fc = LI
Fc = flux embraced by the turns of a winding in Wb (weber)
L = The inductance in H (henry)
I = intensity of the current in A (ampere)
Example :
Data : L = 2,5 H ; I = 0,03 A.
Flux embraced by the turns of the winding : Fc = 2,5 x 0,03 = 0,075 Wb.
OBSERVATION :
Formula 130 refers to an ideal inductor, that is, to a coil in which all the magnetic flux produced by the current is embraced by the coils of the coil.
FORMULA 131 - Calculation of the inductance of a coil, knowing the flux embraced by the turns of the winding and the intensity of the current which runs through it.
Calculation of the inductance of a coil
L = Fc / I
L = The inductance in H (henry)
Fc = flux embraced in Wb (weber)
I = intensity of the current in A (ampere)
(This formula is from formula 130).
Example :
Data : Fc = 0,002 / 0,05 = 0,04 H
FORMULA 132 - Calculation of the intensity of the current which goes through a coil, knowing the inductance and the flux embraced by the turns of the winding.
Calculating the intensity of the current flowing through a coil
I = Fc / L
(This formula is from formula 130).
Example :
Data : Fc = 0,6 Wb ; L = 250 mH (millihenry) = 0,25 H.
Current intensity : I = 0,6 / 0,25 = 2,4 A.
FORMULA 133 - (NEUMANN's law). Calculation of the electromotive force induced in a turn, knowing the variation of the magnetic flux embraced by the turn and the time during which this variation is accomplished.
Calculation of the induced electromotive force in a coil
Enunciated : The induced electromotive force, expressed in volts, is obtained by dividing the variation of the kissed flux expressed in webers, by the time during which this variation occurs, expressed in seconds.
E = (F"c - F'c) / (t" - t')
E = induced electromotive force in 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 the flux in Wb (weber) which is often written Fc
t" = final moment
t' = initial moment
t" - t' = time interval in seconds that 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
FORMULA 134 - Calculation of the electromotive force of self-induction, knowing the inductance of the winding, the variation of the intensity of the current which traverses it and the time during which this variation occurs.
Calculation of the electromotive force of self-induction
Enunciated : The electromotive force of self-induction, expressed in volts, is obtained by multiplying the inductance, expressed in henries, by the variation of the intensity of the current, expressed in amperes and dividing the product obtained by the time during which this variation occurs in seconds.
E = L x (I" - I') / (t" - t') = LI / t
E = electromotive force of self-induction in V (volt)
L = The inductance in 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 (ampere) which is often written I
t" = final moment
t' = initial moment
t" - t' = time interval in seconds that is often written t
Example :
Data : L = 2,5 H (inductance of a coil 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
FORMULA 135 - Calculation of the inductance of a winding, knowing the electromotive force of self-induction, the variation of the intensity of the current and the time during which this variation occurs.
Calculation of the inductance of a winding
L = E x (t" - t') / (I" - I') = Et / I
L = The inductance in H (henry)
E = electromotive force of self-induction in V (volt)
t" = final moment
t' = initial moment
t" - t' = t = time interval 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 (ampere)
(This formula is from formula 134)
Example :
Data : E = 120 V ; t = 0,01 s ; I = 0,8 A.
Inductance of winding : L = (120 x 0,01) / 0,8 = 1,2 / 0,8 = 1,5 H
FORMULA 136 - Calculation of the total inductance presented by two or more coils connected in series and not coupled together, knowing the inductance of each of the coils.
Enunciated : The inductance presented at the same time by two or more coils connected in series is obtained by adding the inductances of the coils.
Lt = L1 + L2 + L3 + ... + Ln
Lt = total inductance
L1 = inductance of the first coil
L2 = inductance of the second coil
L3 = inductance of the third coil
Ln = inductance of the last coil
The various inductances must all be expressed in the same unit of measurement.
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) ; L2 = 1 µH (0,000001 H) ; L3 = 50 µH ; L4 = 150 µH
Total inductance : Lt = 300 + 1 + 50 + 150 = 501 µH.
FORMULA 137 - Calculation of the equivalent inductance presented by two or more coils of equal value, connected in parallel and not coupled together, knowing their inductance.
Enunciated : The inductance presented by two or more equal coils connected in parallel, is obtained by dividing their inductance by the number of coils.
Leq = L / n
Leq = equivalent inductance
L = The inductance of each coil
n = number of coils connected in parallel
The equivalent inductance shall be expressed in the same unit of measurement as that used to indicate the inductance of the coils.
Example :
a) Data : L = 2 H (henry) ; number (n) of coil = 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.
FORMULA 138 - Calculation of the equivalent inductance of two coils of different value, connected in parallel and not coupled together, knowing their inductance.
Calculation of the equivalent inductance of two coils of different value
Enunciated : The sum of the inductances presented by two coils of different inductance, connected in parallel, is obtained by multiplying the two values and dividing the product obtained by the sum of these same values.
Leq = (L1 x L2) / (L1 + L2)
Leq = equivalent inductance
L1 = inductance of a coil
L2 = inductance of the other coil.
The inductance values must all be expressed in the same unit of measurement.
Example :
Data : L1 = 12 mH (millihenry) ; L2 = 6 mH.
Equivalent inductance : Leq = (12 x 6) / (12 + 6) = 72 / 18 = 4 mH.
FORMULA 139 - Calculation of the inductance of a coil to be connected in parallel with another coil of known value, to obtain a known equivalent inductance (the two coils must not be coupled together).
Calculation of the inductance of a coil to be connected in parallel with another coil of known value
Enunciated : The inductance of a coil to be connected in parallel with another coil, to obtain a given equivalent inductance, is calculated by multiplying the value of the known coil by the equivalent inductance and dividing the product by the difference of these. values.
Li = (L x Leq) / (L - Leq)
Li = unknown inductance
L = value of available coil
Leq = equivalent inductance that we want to obtain.
The inductance values must all be expressed in the same unit of measurement.
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
FORMULA 140 - Calculation of the equivalent inductance of several coils connected in parallel and not coupled together, knowing their inductance.
Enunciated : The equivalent inductance of several coils connected in parallel, is obtained by performing the calculations in three stages : first, we calculate the inverse of the inductance of each coil, which amounts to dividing the number 1 by the value of the coil ; then, we add the values of the inverses ; finally, the equivalent inductance is calculated by dividing the number 1 by the sum of the inverses.
The inductance values must all be expressed in the same unit of measurement.
Example :
Data : L1 = 2 mH (millihenry) ; L2 = 4 mH ; L3 = 4 mH.
OBSERVATION :
If one must calculate the equivalent inductance of two coils connected in parallel, one can use the Formula 140, but it is simpler to use The Formula 138 ; in addition, in the case where the inductances of the coils connected in parallel are equal to one another, the Formula 137 should be used.
FORMULA 141 - Calculation of the energy stored by a coil, traversed by a current, knowing the inductance and the intensity of the current.
Calculation of the energy stored by a coil
Enunciated : The energy stored by a coil, expressed in joules, is obtained by multiplying the inductance, expressed in henries, by the square of the intensity of the current, expressed in amperes and dividing by 2 the product obtained.
W = (L x I^{2}) / 2
W = electrical energy in J (joule)
L = The inductance in H (henry)
I = intensity of the current in A (ampere)
Example :
Data : L = 50 mH (millihenry) = 0,05 H ; I = 100 mA (milliampere) = 0,1 A
Stored electric energy :
W = (0,05 x 0,1^{2}) / 2 = 0,05 x 0,01 / 2 = 0,00025 J.
FORMULA 142 - Calculation of the frequency of a periodic quantity (for example, of a current with sinusoidal pace), knowing the period, that is to say the duration of each cycle.
Calculation of the frequency of a periodic quantity, for example, of a sinusoidal current
Enunciated : The frequency, expressed in hertz, is obtained by dividing the number 1 by the period expressed in 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 shall be expressed in kilohertz, (kHz, 1 kHz = 1000 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).
FORMULA 143 - Calculation of the period of a periodic quantity (for example, of an alternating current with sinusoidal pace) knowing the frequency, that is to say the number of cycles accomplished during the unit of time .
Enunciated : The period, expressed in seconds, is obtained by dividing the number 1 by the frequency, expressed in hertz.
T = 1 / f
T = period in s (seconde)
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 kilohertz (kHz, 1 kHz = 1000 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).
FORMULA 144 - Calculation of the effective value of an alternating and sinusoidal current (or voltage), knowing the maximum value.
Calculation of the effective value of a current or voltage of the alternating and sinusoidal type
Enunciated : The effective value of a sinusoidal alternating current (or voltage) is obtained by dividing the maximum value by the square root of 2.
REMARKES :
Multiplying the numerator and the denominator of this formula by 1.41, we obtain the following simplified formula :
likewise, we will have :
Ieff 0,707 x Imax
Ieff = Effective value of current in A (ampere)
Imax = maximum value of the current in A (ampere).
Example :
a) Data : maximum value of the alternating current, Imax = 0,8 A (ampere).
Valeur Effective value of alternating current : Ieff = 0,707 x 0,8 = 0,5656 A.
b) Data : maximum value of AC voltage : Vmax = 311 V (volt).
Effective Value of Alternative Voltage : Veff 0,707 x 311 = 220 V.
OBSERVATION :
The effective value of an alternating current (or voltage) also depends on the waveform. Indeed, if the wave is sinusoidal, (Figure 2-a), the efficient value is equal to the product of the maximum value by the 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 the number 0.577 (Table VIII, Figure 3). In general, a precise number corresponds to each waveform, number between zero and 1, which multiplied by the maximum value, makes it possible to obtain the efficient value.
FORMULA 145 - Calculation of the maximum value of a sinusoidal alternating current (or voltage), knowing the Efficient value..
Calculation of the maximum value of a sinusoidal alternating current or voltage
Vmax = Veff x 1,41
Vmax = maximum value of the voltage in V (volt)
Veff = effective value of the voltage in V (volt)
Square root of 2 1,41
Imax = Ieff x 1,41
Imax = maximum value of current in A (ampere)
Ieff = effective value of the current in A (ampere)
(The formulas above are taken from formulas 144).
Example :
a) Data : Effective value of the alternating current, Ieff = 2,5 A (ampere)
2,5 A. Maximum value of the alternating current : Imax = 1,41 x 2,5 = 3,5
b) Data : Effective value of the alternative voltage, Veff = 160 V.
Maximum value of the alternative voltage, Vmax = 1,41 x 160 = 225,6 V.
OBSERVATION :
The number that multiplies the Effective value depends on the waveform. Indeed, if the wave is sinusoidal (Figure 2-a), we will use the factor 1,414 ; but if the wave is rectangular (Figure 2-b), we will use the factor 1 and the maximum value will be equal to the effective value ; in addition, if the wave is triangular (Figure 2-c), we will use the factor 1.73 (Table VIII, Figure 3).
FORMULA 146 - Calculation of the peak-to-peak value of a sinusoidal alternating voltage, knowing its effective value.
Calculation of the peak-to-peak value of a sinusoidal AC voltage
Enunciated : The peak-to-peak value of a sinusoidal AC voltage is obtained by multiplying the effective value by the fixed number 2,82.
Vcc = 2,82 x Veff
Vcc = peak to peak value of the voltage
Veff = Effective value of the voltage
Icc = 2,82 x Ieff
Icc = current peak to peak value
Ieff = effective value of the current
Example :
Data : Effective value of Alternative voltage, Veff = 220 V (volt)
Peak to peak value of the voltage, Vcc = 220 x 2,82 = 620,4 V.
OBSERVATION :
The fixed number that multiplies the effective value depends on the waveform. Indeed, if the wave is sinusoidal (Figure 2-a), one will use the factor 2,82 ; but if the wave is rectangular (Figure 2-b), use the factor 2 (Table VIII, Figure 3) ; if on the contrary, the wave is triangular (Figure 2-c), we will use the factor 3.46 (Table VIII, Figure 3).
FORMULA 147 - Calculation of the average value of a reciprocating sinusoidal current (or voltage), knowing the effective value.
Calculation of the average value of a sinusoidal alternating current or voltage
Enunciated : The average value of a sinusoidal alternating current (or voltage) is obtained by multiplying the effective value by the fixed number 0,9.
Vm = 0,9 x Veff
Vm = average value of the voltage in V (volt)
Veff = effective value of the voltage in V (volt)
Im = 0,9 x Ieff
Im = average value of current in A (ampere)
Ieff = effective value of the current in A (ampere)
Example :
a) Data : Effective value of alternating current, Ieff = 0,35 A (ampere)
Mean value of the alternating 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 fixed number that multiplies the Effective value depends on the waveform. Indeed, if the wave is sinusoidal (Figure 2-a), one will use the factor 0.9 but if the wave is rectangular (Figure 2-b), one will use the factor 1 and because, the average value will be equal to the effective value ; finally, if the wave is triangular (Figure 2-c), we will use the factor 0.866 (table VIII, figure 3).
FORMULA 148 - Calculation of the pulsation of a periodic quantity, knowing its frequency.
Calculation of the pulsation of a periodic quantity
Enunciated : The pulse, expressed in radians per second, is given twice the product of the number P by the frequency, expressed in hertz.
w = 2 n f 6,28 f
w = pulse in rd / s (radian per second)
P = symbol of the number 3,14...
f = frequency in Hz (hertz)
Example :
Data : f = 400 Hz
Pulsation : w 6,28 x 400 = 2 512 rd / s.
FORMULA 149 - Calculation of the frequency of a periodic quantity, knowing the pulsation.
Calculation of the frequency of a periodic quantity
f = w / 2P 0,159 w
f = frequency in Hz (hertz)
w = pulse in rd / s (radian per second)
P = fixed number symbol 3,14...
(This formula is from formula 148).
Example :
Data : w = 2 000 rd / s.
Frequency : f 0,159 x 2 000 = 318 Hz.
FORMULA 150 - Calculation of the pulsation of a periodic quantity, knowing its period.
Calculation of the pulsation of a periodic quantity
w = 2P / T 6,28 / T
w = pulse in rd / s (radian per second)
P = fixed number symbol 3,14...
T = period in s (second)
The present formula is obtained by replacing in formula 148, the frequency f, by the second member of the Formula 142, namely : 1 / T.
Example :
Data : T = 0,02 s (alternating current period at 50 Hz).
Pulsation : w = 6,28 / 0,02 = 314 rd / s
FORMULA 151 - Calculation of the period of a periodic quantity, knowing its pulsation.
Calculation of the period of a periodic quantity
T = 2P / w 6,28 / w
T = period in s (second)
P = fixed number symbol 3,14...
w = pulse in rd / s (radian per second)
(This formula is from formula 150).
Example :
Data : w = 628 rd / s (pulsation of an alternating current at 100 Hz)
Period : T = 6,28 / 628 = 0,01 s.
FORMULA 152 - Calculation of the capacitive reactance of a capacitor, knowing the capacity of the capacitor and the frequency of the alternating current which crosses it.
Calculation of the capacitive reactance of a capacitor
Enunciated : The capacitive reactance, expressed in ohms, is obtained by dividing the number 1 by 6,28 (2P) by the frequency, expressed in hertz and by the capacity, expressed in farads.
Xc = 1 / (2PfC) 1 / (6,28 fC)
Xc = capacitive reactance in W (ohm)
P = fixed number symbol 3,14...
f = frequency in Hz (hertz)
C = capacity in F (farad)
This formula is also written :
Xc = 1 / Cw
since according the 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 units of measurement of frequency and capacity with the corresponding multiples and submultiples ; in particular, it is possible to find the frequency expressed in kilohertz (kHz) or in megahertz (MHz) and the capacity in nanofarads (nF), in picofarads (pF), or in microfarads (µF). In all these cases, formula 152 can be used, taking into account the following :
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 kilohertz and the capacity in nanofarads, the reactance will be expressed in megohms,
If the frequency is expressed in kilohertz 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,
Finally, if the frequency is expressed in megahertz and the capacity in picofarads, the reactance will be expressed in megohms.
FORMULA 153 - Calculation of the capacity of a capacitor, knowing its capacitive reactance for a given frequency.
Calculation of the capacitance of a capacitor
C = 1 / (2 P f Xc) 1 / (6,28 f Xc)
C = capacity in F (farad)
P = fixed number symbol 3,14...
f = frequency in Hz (hertz)
Xc = capacitive reactance in W (ohm)
(This formula is 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).
FORMULA 154 - Calculation of the inductive reactance of a coil, knowing its inductance and the frequency of the alternating current which crosses it.
Calculation of the inductive reactance of a coil
Enunciated : The inductive reactance, expressed in ohms, is obtained by multiplying the fixed number 6,28 (2P) by the frequency, expressed in hertz and by the inductance, expressed in henries.
XL = 2 P f L 6,28 f L
XL = inductive reactance in W (ohm)
P = fixed number symbol 3,14...
f = frequency in Hz (hertz)
L = frequency in 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 :
The formula 154 can also be used by expressing the frequency in kilohertz (kHz) and the inductance in millihenrys (mH), or the frequency in megahertz (MHz) and the inductance in microhenrys (µH) ; in either case, the inductive reactance will be expressed in ohms.
FORMULA 155 - Calculation of the inductance of a coil, knowing its inductive reactance for a given frequency.
Calculation of the inductance of a coil
L = XL / (2 P f) XL / 6,28 f
L = The inductance in H (henry)
XL = inductive reactance in W (ohm)
P = fixed number symbol 3,14...
f = frequency in Hz (hertz)
(This formula is from formula 154).
Example :
Data : XL = 20 000 = 20 000 W ; f = 700 kHz (kilohertz) = 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 finish our fourth mathematical forms and reminders. ...).
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