As in the previous forms, the units used in the calculations are borrowed from the system S.I. ... The formulas contained in this reminder concern electrotechnics and capacitors.
FORMULA 99 -
Calculation of the electrical power of a device connected to the sector knowing the operating time, the corresponding number of revolutions of the disk visible to the meter and the number of revolutions per kilowatt-hour indicated for each type of meter.
Enunciated:
The electric power, expressed in watts, is obtained by multiplying the number of revolutions of the rotating disk by 3 600 000 (Figure 1) and dividing the product obtained by the number of revolutions per kilowatt-hour (2 160 revolutions / kW.h in the case of the counter of Figure 1) and the operating time expressed in seconds.
P = (3 600 000 x n) / (N x t)
P = electric power in W (watt)
n = number of revolutions of the meter disk
N = number of revolutions per kilowatt-hour (revolutions / kW.h) of the meter
t = time in seconds set by the disk to complete 'n' laps
Example
:
Data :n = 18 turns (counted when the device whose power is to be determined is connected to the mains) ; N = 2 160 / kW.h (indicated on the meter dial of Figure 1) ;t = 30 s (time taken to complete 18 laps, measured with a stopwatch).
Electrical power of the device :
P = (3 600 000 x 18) / (2 160 x 30) = 64 800 000 / 64 800 = 1 000 W
OBSERVATION :
To determine the number of revolutions (n) accomplished by the rotating disk of the counter, we count how many times, during the time considered, the small red or black line (Figure 1) passes in front of the small window through which we see this disk
On the meters, the number of "turns per kilowatt-hour" is almost always replaced by the equivalent indication "turn per kilowatt-hour" or by the symbol turns / kW.h.
FORMULA 100 - Calculation of the cost of the electrical energy consumed by a device knowing the quantity of energy consumed and the unit price, expressed in francs per kilowatt-hour.
C = cost of electric energy in F (franc) "but now it is in euro" !
p = unit price of electric energy in F / kW.h (franc per kilowatt hour)
W = electrical energy in kW.h (kilowatt hour)
Example
:
Data :
p = 0,40 F kW.h ;
W = 97,8 kW.h (see the indications shown in Figure 1).
Cost : C = 0,40 x 97,8 = 39,12 F
FORMULA 101 -
Calculation of the cost per hour of the energy consumed by an electrical apparatus knowing the power of the apparatus and the unit price of the energy.
Enunciated :
The cost per hour, expressed in francs, of the energy consumed by an electrical appliance is obtained by dividing by 1 000 the product between the power, expressed in watts, and the unit price, expressed in francs per kilowatt -hour.
C / h = P p / 1 000
C / h = cost per hour in F (franc)
P = electric power in W (watt)
p = unit price of electric energy in F / kW.h (franc per kilowatt hour)
Example
:
Data :
P = 100 W (power of a small lamp) ;
p = 0,40 F / kW.h
Cost per hour : C / h = (100 x 0,40) / 1 000 = 0,04 F or 4 cents
FORMULA 102 -
Calculation of the capacity of a body, knowing the quantity of electricity and the potential of this body.
Enunciated :
The electrical capacity, expressed in farads, is obtained by dividing the quantity of electricity, expressed in coulombs, by the potential expressed in volts.
C = Q / V
C = electrical capacitance in F (farad)
Q = quantity of electricity in C (coulomb)
V = potential (volt)
Example
:
Data : Q = 0,00002 C ; V = 10 V
Electrical capacity : C = 0,00002 / 10 = 0,000002 = 2 x 10-6 F
OBSERVATION :
Since the calculated electrical capacitance is relatively small, the result can be expressed in microfarads
(µF), 1 µF = 10-6 F.
Where
C = 2 µF
FORMULA 103 - Calculation of the capacity of a capacitor, knowing the quantity of electricity present on one of the reinforcements and the potential difference between the reinforcements.
Enunciated :
The capacity of a capacitor, expressed in farads is obtained by dividing the amount of electricity expressed in coulombs, present on one or the other reinforcement, by the potential difference between these frames, expressed in volts.
C = Q / V
C = capacity of the capacitor in F (farad)
Q = quantity of electricity in C (coulomb)
V = potential difference in V (volt)
This formula is similar to Formula 102 and the quantities contained therein are expressed in the same units of measurement ; this is why the calculation methods are identical in both cases.
Example
:
Data : Q =
0,000008 C ; V = 200 V
Capacitor capacity : C = 0,000008 / 200 = 4 x 10-8F
OBSERVATION :
In general, the capacities of the capacitors are very small compared to the unit of measurement, which is farad, and for this reason, Farad submultiples are commonly used, the microfarad (µF
; 1 µF = 10-6
F),
see the nanofarad (nF
; 1 nF = 10-9
F) and the picofarad
(pF
; 1 pF = 10-12).
By expressing the result of the preceding example, using the various sub-multiples of the farad, we obtain :
C = 4 x 10-8 F = 0,04 µF = 40 nF = 40 000 pF
FORMULA 104 -
Calculation of the quantity of electricity present on one or the other reinforcement of a capacitor, knowing the potential difference between the armatures and the capacity of the capacitor.
Data : C = 20
nF = 2 x 10-8 F (for equivalence between nanofarad and farad unit, see observation following Formula 103)
;V
= 500 V.
Quantity of electricity present on a capacitor frame :
Q = 2 x 10-8 x 500 = 10-5 C
FORMULA 105 -Calculation of the difference of potential (tension) existing between the armatures of a capacitor, knowing the quantity of electricity present on a frame and capacitance of the capacitor.
Data : Q = 0,0006 C ; C = 5 µF = 5 x 10-6 F
(for the equivalence between the microfarad and the Farad measurement unit, see the observation following the Formula 103).
Potential difference existing between the capacitor plates :
V = 0,0006 / (5 x 10-6) = 120 V
FORMULA 106 -
Calculation of the absolute dielectric constant of a material, knowing the absolute dielectric constant of the vacuum (or air) and the dielectric constant relative to the vacuum (or to the air) of this same material.
Enunciated :
The absolute dielectric constant of a material, expressed in picofarad per meter, is obtained by multiplying the dielectric constant relative to the vacuum (or air) of the material by the absolute dielectric constant of the vacuum (or air), expressed in picofarad per meter.
e
= eo
x er
e =
absolute dielectric constant of a material in pF / m (picofarad per meter).
eo = absolute dielectric constant of the vacuum (8.86 pF / m (picofarad per meter).
er = relative dielectric constant of the material (see Table VI, Figure 2).
Example :
Data : eo
= 8,86 pF / m ; er
= 5 (dielectric constant relative to the bakelized cardboard, see table VI, Figure 2).
Absolute dielectric constant of the bakelized cardboard :
e = 8,86 x 5 = 44,3 pF / m
OBSERVATION :
Sometimes, in physics books and technical manuals, this same absolute dielectric constant is expressed in farad per meter (F / m). Farad per meter is the unit of measurement adopted in the international system, picofarad per centimeter and picofarad per meter are submultiples of the same unit :
1 pF / cm = 10-10
F / m ; 1 F / m = 1010 pF / cm
1 pF / m = 10-12
F / m ; 1 F / m = 1012 pF / m
1 pF / cm = 100 pF / m ; 1 pF / m = 0,01 pF / cm
In Table VI, Figure 2, the dielectric constants relating to the vacuum, or dry air, of some insulating materials which may be of interest to the electronics engineer are indicated.
FORMULA 107 -
Calculation of the relative dielectric constant of a material, knowing the absolute dielectric constants of the vacuum (or air) and the material.
e
= absolute dielectric constant of the material in pF / m (picofarad per meter)
eo
= absolute dielectric constant of the vacuum 8,86 pF / m (picofarad / m)
(This formula is from formula 106).
Example :
Data : e
= 16,834 pF / m
(absolute dielectric constant of beeswax).
eo
= 8,86 pF / m (absolute dielectric constant of vacuum, see observation following formula 106).
Dielectric constant for beeswax : er
= 16,834 / 8,86 = 1,9
(Value shown in Table VI, Figure 2).
FORMULA 108 -
Calculation of the capacitance of an air capacitor, formed by two equal plates, flat and parallel, knowing their surface, the distance between the surfaces in presence and the absolute dielectric constant of the air.
Enunciated : The capacitance of the air capacitor (Figure 3-a), expressed in picofarad, is obtained by multiplying the absolute dielectric constant of air (8.86 pF / m, Formula 106) by the surface of a plate, expressed in square centimeters and dividing the product obtained by the distance between plates, expressed in millimeters and multiplied by 10.
C = (eo
x S) / (10 x d)
C = capacity of the capacitor in pF (picofarad)
eo
= absolute dielectric constant of the air = 8,86 pF / m (picofarad per meter)
S = surface of a plate in cm2
d = distance between the plates in mm.
Example :
(Figure 3-a)
Data :
eo
8,86 pF / m ; S = 64 cm2 ; d = 1,5 mm
Capacitor capacity : C = (8,86 x 64) / (10 x 1,5) = 567,04 / 15 = 37,8 pF
FORMULA 109 -
Calculation of the capacity of an air capacitor formed by three or more equal, flat and parallel plates, connected as shown in Figure 3-b, knowing the surface of the plates, the distance which separates them, the number of plates and the absolute dielectric constant of the air.
Enunciated :
The capacitance of an air capacitor (Figure 3-b) is obtained by multiplying the capacitance of two adjacent plates, calculated with Formula 108, by the number of plates minus one.
C = (Eo x S) / (10 x d) x (n - 1)
C = capacity of the capacitor in pF (picofarad)
Eo
= absolute dielectric constant of the air =
8,86 pF / m (picofarad per meter)
S = surface of a plate in cm2
d = distance between the plates in mm
n = number of plates
Example :
Data :Eo = 8,86 pF / m ; S = 12 cm2 ; d = 0,8 mm
; n = 11 plates
Capacitor capacity :
C = (8,86 x 12) / (10 x 0,8) x (11 - 1) = (106,32 / 8) x 10 = 13,29 x 10 = 132,9 pF
FORMULA 110 - Calculation of the electrical energy stored by a capacitor, knowing the quantity of electricity present on one or the other armature and the existing tension between them.
Enunciated :
The energy stored by a capacitor, expressed in joules, is obtained by multiplying the quantity of electricity present on a frame, expressed in coulombs, by the tension between the armatures, expressed in volts and by dividing by 2 the product obtained.
W = QV / 2
W = electrical energy in J (joule)
Q = quantity of electricity in C (coulomb)
V = voltage in V (volt)
Example :
Data :Q = 0,0004 C
((amount of electricity present on a Condenser Armature) ;
V = 200 V
(voltage between the capacitor plates).
Electrical energy stored by the capacitor :
W = (0,0004 x 200) / 2 = 0,08 / 2 = 0,04 J
FORMULA 111 -
Calculation of the quantity of electricity present on the reinforcement of a capacitor, knowing the value of the energy stored by the capacitor and the tension existing between its reinforcements.
Data :
W =
0,08J =0,08 J
(energy stored by the capacitor).
V = 160 V
(voltage between the capacitor plates).
Quantity of electricity present on a capacitor frame :
Q = (2 x 0,08) / 160 = 0,16 / 160 = 0,001 C
FORMULA 112 - Calculation of the existing tension between the armatures of a capacitor, knowing the value of the energy stored and the quantity of electricity present on a reinforcement.
Enunciated :
The energy stored by a capacitor, expressed in joules, is obtained by multiplying the capacity expressed in farads by the square of the voltage, expressed in volts and dividing by two the product obtained.
W = CV2 / 2
W = electrical energy in J (joule)
C = capacity in F (farad)
V = voltage in V (volt)
Example :
Data :C = 400 µF = 0,0004 F ; V = 50 V
Stored electric energy :
W = (0,0004 x 502) / 2 = (0,0004 x 2 500) / 2 = 1 / 2 = 0,5 J
FORMULA 114 - Calculation of the capacitance of a capacitor, knowing the value of the energy stored and the tension existing between the armatures.
Data :W = 0,05 J (energy stored by the capacitor).
V = 500 V
(voltage between the capacitor plates).
Capacity : C = (2 x 0,05) / 5002 = 0,1 / 250 000 = 4 x 10-7 F = 0,4 µF (microfarad)
FORMULA 115 - Calculation of the existing tension between the armatures of a capacitor, knowing the value of the stored energy and capacitance of the capacitor.
Enunciated :
The voltage between the armatures of a capacitor, expressed in volts, is obtained by dividing twice the stored energy, expressed in joules, by the capacity expressed in farads and extracting the square root of the quotient obtained.
Example :
Data : W = 4,9 J (electrical energy stored by the capacitor).
C = 80 µF = 8 x 10-5 F (capacitor capacitance)
Existing voltage between the capacitor plates :
FORMULA 116 - Calculation of the intensity of the electric field existing in the dielectric of a capacitor, knowing the tension and the distance between the armatures.
Enunciated :
The intensity of the electric field existing in the dielectric of a capacitor, expressed in kilovolts per meter is obtained by dividing the tension existing between its armatures, expressed in volts, by the distance which separates them expressed in millimeters.
E = V / d
E = electric field strength in kV / m (kilovolt per meter)
V = voltage in V (volt)
d = distance between the armatures in mm
Example :
Data :V = 350 V
; d = 0,7 mm.
Intensity of the electric field : E = 350 / 0,7 = 500 kV / m
OBSERVATION :
The maximum intensity of the electric field admitted by a dielectric is called dielectric strength.
In Table VI Figure 2, beside the values of the dielectric constant relative to some insulating materials (dielectric), are taken the values of the dielectric strength. It will be observed that the dielectric strength in this table is expressed in kilovolts per centimeter (kV / cm) instead of kilovolts per meter (kV / m), as indicated in formula 116. Kilovolt per centimeter and kilovolt per meter are multiples of volts per meter (V / m), the unit of measurement for the intensity of the electric field :
1 kV / m = 1 000 V / m ; 1 kV / cm = 100 000 V / m
1 kV / m = 0,01 kV / cm ; 1 kV / cm = 100 kV / m
FORMULA 117 -
Calculation of the thickness of an insulating layer, knowing the dielectric rigidity of the material and the electric tension which it will have to support between the one and the other surface of the thickness.
Enunciated :
The thickness of the insulating layer, expressed in millimeters, is obtained by dividing the voltage, expressed in volts, by the dielectric strength of the material, expressed in kilovolts per centimeter (Table VI, Figure 2) and multiplied by 100.
d = V / (100 x Rd)
d = thickness of the insulation in mm
V = voltage in V (volt)
Rd = dielectric strength in kV / cm (kilovolt per centimeter)
Example :
Data :V = 12 000 Rd
(voltage to be applied between the opposite surfaces of the dielectric).
Thickness of the layer of mica necessary to obtain the insulation at the indicated voltage :
d = 12 000 / (100 x 600) = 12 000 / 60 000 = 0,2 mm
FORMULA 118 -
Calculation of the voltage that a thickness of insulating material can withstand, knowing the thickness and the dielectric strength of the material.
Enunciated :
The equivalent capacitance of two or more capacitors with the same capacitance, connected in series, is obtained by dividing the capacity of a capacitor by the number of capacitors.
Ceq = C / n
Ceq = equivalent capacity
C = capacity of each capacitor
n = number of capacitors.
The equivalent capacitance is obtained with the same unit of measure as that used to indicate the capacity of the capacitors.
Enunciated :
The equivalent capacitance of two capacitors connected in series is obtained by multiplying the capacitance of the two capacitors and dividing the product obtained by the sum of these capacitors.
Ceq = (C1 x C2) / (C1 + C2)
Ceq = equivalent capacity
C1 = capacity of a capacitor
C2 = capacitance of the other capacitor.
The capacity values must all be expressed with the same unit of measurement.
Example :
Data :C1 = 40 nF (nanofarad) ; C2 = 60 nF.
Equivalent capacity of two capacitors connected in series: :
Ceq = (40 x 60) / (40 + 60) = 2 400 / 100 = 24 nF
FORMULA 122 - Calculation of the capacitance of a capacitor to be connected in series with another capacitor of known capacity, to obtain a given equivalent capacitance.
Enunciated :
The capacitance of a capacitor to be connected in series to another capacitor, to obtain a given equivalent capacitance, is calculated by multiplying the capacitance of the known capacitor by the equivalent capacitance and dividing the product by the difference of these same values.
Ci = (C x Ceq) / (C - Ceq)
Ci = unknown capacity
C = capacitance of capacitor available
Ceq = equivalent capacity that we want to obtain
The capacity values must all be expressed with the same unit of measurement.
Enunciated :
The equivalent capacity of several capacitors connected in series, is obtained by executing the computations in three stages : one calculates first the inverse of the capacity of each capacitor, which amounts to dividing the number 1 by the value of the capacity. Then we add the values of the inverses. Finally, we calculate the equivalent capacity by dividing the number 1 by the sum of the inverses.
Ceq = equivalent capacity
C1 = capacity of the first capacitor
C2 = capacitance of the second capacitor
C3 = capacity of the third capacitor
Cn = capacity of the last capacitor
The capacity values must all be expressed with the same unit of measurement.
If one must calculate the equivalent capacitance of two capacitors connected in series, one can use the Formula 123, but it is simpler to resort to The Formula 121; In addition, in the case where the capacitances of the capacitors connected in series are equal, then the formula 120 can be used.
Electrical power of a device
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0,666 µF = 666 nF.
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