Created it, 05/10/15
Update it, 05/11/05
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FOOT- NOTE :
As in the preceding forms, the units used in calculations are borrowed from system S.I. The formulas contained in this recall relate to electrical engineering and the condensers.
FORMULATE 99 -
Calculation
of the electric output of an apparatus connected to the sector knowing the
operating time, the corresponding number of turns of the disc visible with the
meter and the number of revolutions per kilowatt-hour indicated for each type of
meter.
Statement : The electric output, expressed in Watts, is obtained while multiplying by 3 600 000 the number of revolutions of the turning disc (figure 1) and by dividing the product obtained by the number of revolutions per kilowatt-hour (2 160 turns / kW.h in the case of the meter of figure 1) and by the operating time expressed into seconds.
P = (3 600 000 x n) / (N x t)
P = electric output out of W (watt)
n = number of revolutions of the disc of the meter
N = number of revolutions per kilowatt-hour (turns / kW.h) of the meter
t = time in seconds put by the disc to achieve 'n' turns
Example :
Data : n = 18 turns (counted when the apparatus which one wants to determine the electric output is connected with the sector) ; N = 2.160 / kW.h (indicated on the dial of the meter of figure 1) ; t = 30 s (time put to achieve 18 turns, measured with a stop watch).
Electric output of the apparatus
:P = (3 600
000 x 18) / (2 160 x 30) = 64 800 000 / 64 800 = 1 000 WOBSERVATION :
To determine the number of turns (n) achieved by the turning disc of the meter, one hopes how much time, during time considered, the small red feature or black (figure 1) passes in front of the small window through which one sees this disc
On the meters, the indication of the number of “turns to the kilowatt-hour” is almost always replaced by the equivalent indication “turn per kilowatt-hour” or by the symbol turns / kW.h.
FORMULATE 100 - Calculation
of the cost of the electric power consumed by an apparatus knowing the quantity
of power consumption and the unit price, expressed into frank per kilowatt-hour.
C = pW
C = cost of the electric power out of F (frank)
p = price unit of the electric power out of F / kW.h (frank by kilowatt-hour)
W = power electric in kW.h (kilowatt-hour)
Example :
Data : p = 0,40 F kW.h ; W = 97,8 kW.h (see the indications carried on figure 1).
Cost : C = 0,40 x 97,8 = 39,12 F
FORMULATE 101 - Calculation
of the cost per hour of the power consumption by an electrical appliance knowing
the power of the apparatus and the unit price of energy.
Statement : The cost per hour, expressed into frank, from the power consumption by an electrical appliance is obtained while dividing by 1.000 the product between the power, expressed in Watts, and the unit price, expressed into frank per kilowatt-hour.
C / h = P x p / 1 000
C / h = cost per hour out of F (frank)
P = electric output out of W (watt)
p = price unit of the electric power out of F / kW.h (frank by 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 is 4 centimes
FORMULATE 102 -
Calculation
of the capacity of a body, knowing the quantity of electricity and the potential
of this body.
Statement : The electric capacity, expressed in farads, is obtained by dividing the quantity of electricity, expressed into coulombs, by the potential expressed in volts.
C = Q / V
C = capacity electric out of F (farad)
Q = quantity of electricity out of C (coulomb)
V = potential (volt)
Example :
Data : Q = 0,00002 C ; V = 10 V
Electric capacity : C = 0,00002 / 10 = 0,000002 = 2 x 10-6 F
OBSERVATION :
Since the calculated electric capacity is relatively small, one can express the result in microfarads (µF), 1 µF = 10-6 F.
From where C = 2 µF
FORMULATE 103 - Calculation
of the capacitance of a capacitor, knowing the quantity of electricity presents
on one of the reinforcements and the potential difference between the
reinforcements.
Statement : The capacitance of a capacitor, expressed in farads is obtained by dividing the quantity of electricity expressed into coulombs, presents on one or the other braces, by the potential difference between these reinforcements, expressed in volts.
C = Q / V
C = capacity of the condenser out of F (farad)
Q = quantity of electricity out of C (coulomb)
V = potential difference out of V (volt)
The present formula is similar to formula 102 and the sizes which it contains are expressed in the same measuring units; this is why the processes of calculation are identical in both cases.
Example :
Data : Q = 0,000008 C ; V = 200 V
Capacity of the condenser : C = 0,000008 / 200 = 4 x 10-8 F
OBSERVATION :
Generally, the capacities of the condensers are very small compared to the measuring unit, which is the farad, and for this reason, one usually uses the submultiples of the farad, that is to say the microfarad (µF ; 1 µF = 10-6 F), is the nanofarad (nF ; 1 nF = 10-9 F) and picofarad (pF ; 1 pF = 10-12). By expressing the result of the preceding example, by using the various submultiples of the farad, one obtains:
C = 4 x 10-8 F = 0,04 µF = 40 nF = 40 000 pF
FORMULATE 104 - Calculation
of the quantity of electricity present on one or the other plate of a capacitor,
knowing the potential difference between the reinforcements and the capacity of
the condenser.
Q = CV
Q = quantity of electricity out of C (coulomb)
C = capacity out of F (farad)
V = potential difference out of V (volt)
(This formula is drawn from formula 103).
Example :
Data : C = 20 nF = 2 x 10-8 F (for equivalence between the nanofarad and the unit farad, to see the observation which follows formula 103) ; V = 500 V.
Quantity of electricity present on a reinforcement of the condenser :
Q = 2 x 10-8 x 500 = 10-5 C
FORMULATE 105 - Calculation
of the potential difference (tension) existing between the plates of a capacitor,
knowing the quantity of electricity present on a reinforcement and the capacity
of the condenser.
V = Q / C
V = potential difference out of V (volt)
Q = quantity of electricity out of C (coulomb)
C = capacity out of F (farad)
(This formula is drawn from formula 103).
Example :
Data : Q = 0,0006 C ; C = 5 µF = 5 x 10-6 F (for equivalence between the microfarad and the measuring unit farad, to see the observation which follows formula 103).
Potential difference existing between the reinforcements of the condenser :
V = 0,0006 / (5 x 10-6) = 120 V
FORMULATE 106 -
Calculation
of the absolute permittivity of a material, knowing the absolute permittivity of
the vacuum (or the air) and the permittivity relating to the vacuum (or the air)
of this same material.
Statement : The absolute permittivity of a material, expressed in picofarad per meter, is obtained by multiplying the permittivity relating to the vacuum (or the air) of material by the absolute permittivity of the vacuum (or the air), expressed in picofarad per meter.
e = eo x er
e = constant dielectric absolute of a material in pF / m (picofarad per meter).
eo = constant dielectric absolute of vacuum (8,86 pF / m (picofarad per meter).
er = constant dielectric relative of material (see table VI, figure 2).
Example :
Data : eo = 8,86 pF / m ; er = 5 (permittivity relating to the bakelized paperboard, to see table VI, figure 2).
Absolute permittivity of the bakelized paperboard :
e = 8,86 x 5 = 44,3 pF / m
OBSERVATION :
Sometimes, in the books of physics and the technical handbooks, this same absolute permittivity is expressed in farad per meter (F/m). The farad per meter is the measuring unit adopted in the international system, the picofarad per centimetre and the picofarad per meter is 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
On table VI figure 2, one indicated the permittivities relating to the vacuum, or the dry air, of some insulating materials which can be of an interest for the electronics specialist.
FORMULATE 107 - Calculation
of the relative permittivity of a material, knowing the absolute permittivities
of the vacuum (or the air) and of material.
er = e / eo
er = constant dielectric relative of material
e = constant dielectric absolute of material in pF / m (picofarad per meter)
eo = constant dielectric absolute of the vacuum 8,86 pF / m (picofarad / m)
(This formula is drawn from formula 106).
Example :
Data : e = 16,834 pF / m (absolute permittivity of beeswax).
eo = 8,86 pF / m (absolute permittivity of the vacuum ; to see the observation which follows formula 106.
Permittivity relating to beeswax : er = 16,834 / 8,86 = 1,9
(Value indicated to table VI, figure 2).
FORMULATE 108 - Calculation
of the capacitance of a capacitor to air, formed by two equal, plane and
parallel plates, knowing their surface, the distance enters the involved
surfaces and the absolute permittivity of the air.
Statement : The capacity of the condenser with air (figure 3-a), expressed in picofarad, is obtained by multiplying the absolute permittivity of air (8,86 pF / m ; formulate 106) by the surface of a plate, expressed in square centimetres and dividing the product obtained by the distance between the plates, expressed in millimetres and multiplied by 10.
C = (eo x S) / (10 x d)
C = capacity of the condenser in pF (picofarad)
eo = constant dielectric absolute of the air = 8,86 pF / m (picofarad per meter)
S = surface of a plate in cm2
d = distance enters the plates in mm.
Example : (figure 3-a)
Data: eo 8,86 pF / m ; S = 64 cm2; d = 1,5 mm
Capacity of the condenser : C = (8,86 x 64) / (10 x 1,5) = 567,04 / 15 = 37,8 pF
FORMULATE 109
- Calculation
of the capacitance of a capacitor to air formed by three or several equal, plane
and parallel plates, connected as shown figure 3-b, knowing the surface of the
plates, the distance which separates them, the number of plates and the absolute
permittivity of the air.
Statement : The capacitance of a capacitor to air (figure 3-b) is obtained by multiplying the capacity 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 condenser in pF (picofarad)
Eo
= constant dielectric absolute of the air = 8,86
pF / m (picofarad per meter)
S = surface of a plate in cm2
d = distance enters 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
Capacity of the condenser :
C = (8,86 x 12) / (10 x 0,8) x (11 - 1) = (106,32 / 8) x 10 = 13,29 x 10 = 132,9 pF
FORMULATE 110 - Calculation
of the electric power stored by a condenser, knowing the quantity of electricity
present on one or the other braces and the existing tension between them.
Statement : The energy stored by a condenser, expressed in joules, is obtained by multiplying the quantity of electricity presents on a reinforcement, expressed in coulombs, by the tension between the reinforcements, expressed in volts and dividing by 2 the product obtained.
W = QV / 2
W = power electric in J (joule)
Q = quantity of electricity out of C (coulomb)
V = tension out of V (volt)
Example :
Data : Q = 0,0004 C (quantity of electricity present on a reinforcement of the condenser);
V = 200 V (tension enters the reinforcements of the condenser).
Electric power stored by the condenser :
W = (0,0004 x 200) / 2 = 0,08 / 2 = 0,04 J
FORMULATE 111 -
Calculation
of the quantity of electricity present on the plate of a capacitor, knowing the
value of the energy stored by the condenser and the tension existing between its
reinforcements.
Q = 2W / V
Q = quantity of electricity out of C (coulomb)
W = power electric in J (joule)
V = tension out of V (volt)
(This formula is drawn from formula 110)
Example :
Data : W = 0,08 J = 0,08 J (energy stored by the condenser).
V = 160 V (tension enters the reinforcements of the condenser).
Quantity of electricity present on a reinforcement of the condenser :
Q = (2 x 0,08) / 160 = 0,16 / 160 = 0,001 C
FORMULATE 112 - Calculation
of the tension existing between the plates of a capacitor, knowing the value of
the stored energy and the quantity of electricity presents on a reinforcement.
V = 2W / Q
V = tension out of V (volt)
W = power electric in J (joule)
Q = quantity of electricity out of C (coulomb)
(This formula is drawn from formula 110)
Example:
Data : W = 1,5 J (energy stored by the condenser).
Q = 0,06 C (quantity of electricity present on a reinforcement of the condenser).
Tension enters the reinforcements of the condenser :
V = (2 x 1,5) / 0,06 = 3 / 0,06 = 50 V
FORMULATE 113 - Calculation
of the electric power stored by a condenser, knowing the capacity and the
tension existing between the reinforcements.
Statement : The energy stored by a condenser, expressed in joules, is obtained by multiplying the capacity expressed in farads by the square of the tension, expressed in volt and dividing by two the product obtained.
W = CV2 / 2
W = power electric in J (joule)
C = capacity out of F (farad)
V = tension out of V (volt)
Example :
Data : C = 400 µF = 0,0004 F ; V = 50 V
Stored electric power :
W = (0,0004 x 502) / 2 = (0,0004 x 2 500) / 2 = 1 / 2 = 0,5 J
FORMULATE
114 -
Calculation of the capacitance of a capacitor, knowing the value of the stored
energy and the tension existing between the reinforcements.
C = 2W / V2
C = capacity out of F (farad)
W = power electric in J (joule)
V = tension out of V (volt)
(This formula is drawn from formula 113).
Example :
Data : W = 0,05 J (energy stored by the condenser).
V = 500 V (tension enters the reinforcements of the condensers).
Capacity : C = (2 x 0,05) / 5002 = 0,1 / 250 000 = 4 x 10-7 F = 0,4 µF (microfarad)
FORMULATE 115 - Calculation
of the tension existing between the plates of a capacitor, knowing the value of
the stored energy and the capacity of the condenser.
Statement : The tension existing between the plates of a capacitor, expressed in volts, is obtained by dividing the double of the stored energy, expressed into joules, by the capacity expressed in farads and extracting the square root of the quotient obtained.
Example :
Data : W = 4,9 J (electric power stored by the condenser).
C = 80 µF = 8 x 10-5 F (capacity of the condenser)
Existing tension enters the reinforcements of the condenser :
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FORMULATE 116 - Calculation
of the intensity of the existing electric field in the dielectric one of a
condenser, knowing the tension and the distance enters the reinforcements.
Statement : The intensity of the existing electric field in the dielectric one of a condenser, expressed in kilovolts per meter is obtained by dividing the tension existing between its reinforcements, expressed in volts, by the distance which separates them expressed in millimetres.
E = V / d
E = intensity of the electric field in kV / m (kilovolt per meter)
V = tension out of V (volt)
d = distance enters the reinforcements 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 dielectric is called dielectric rigidity.
In table VI figure 2, beside the values of the permittivity relating to some insulating materials (dielectric), the values of dielectric rigidity are carried. One will observe that dielectric rigidity, in this table, is expressed in kilovolts per centimetre (kV / cm) instead of kilovolts per meter (kV / m), as indicated in formula 116. The kilovolt per centimetre and the kilovolt per meter are multiples of the volt per meter (V / m), measuring unit of 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
FORMULATE 117 - Calculation
thickness of an insulating layer, knowing the dielectric rigidity of material
and the electric tension that it will have to support between one and the other
surface thickness.
Statement: The thickness of the insulating layer, expressed in millimetres, is obtained by dividing the tension, expressed into volts, by the dielectric rigidity of material, expressed in kilovolts per centimetre (table VI, figure 2) and multiplied by 100.
d = V / (100 x Rd)
d = thickness of insulator in mm
V = tension out of V (volt)
Rd = Dielectric rigidity in kV / cm (kilovolt per centimetre)
Example:
Data : V = 12 000 Rd (tension to be applied between opposite surfaces of dielectric).
Rd = 600 kV / cm (minimal dielectric rigidity of the mica; to see table VI, figure 2).
Thickness of the layer of mica necessary to obtain insulation with the tension indicated :
d = 12 000 / (100 x 600) = 12 000 / 60 000 = 0,2 mm
FORMULATE 118 - Calculation
of the tension which a thickness of insulating material will be able to support,
knowing the thickness and the dielectric rigidity of material.
V = 100 x Rd x d
V = tension out of V (volt)
Rd = Dielectric rigidity in kV / cm (kilovolt per centimetre)
d = thickness in mm
(This formula is drawn from formula 117)
Example :
Data : Rd = 400 kV / cm (minimal dielectric rigidity of paraffined paper, to see table VI, figure 2) ; d = 0,025 mm (thickness of a paraffined paper sheet).
Tension which one wants to apply between the two pages of a paraffined paper sheet :
V = 100 x 400 x 0,025 = 1 000 V
FORMULATE 119 - Calculation
of the total capacity of two or several condensers connected in parallel,
knowing the capacity of each condenser.
Statement : The total capacity of two or several condensers connected in parallel is obtained by adding their capacities.
Ct = C1 + C2 + C3 +… + Cn
Ct = total capacity
C1 = capacity of the first condenser
C2 = capacity of the second condenser
C3 = capacity of the third condenser
Cn = capacity of the last condenser
The values all of capacity must be expressed with the same measuring unit.
Examples :
a) Data : C1 = 10 pF (picofarad) ; C2 = 18 pF ; C3 = 8 pF ; Cn = 24 pF.
Total capacity : Ct = 10 + 18 + 8 + 24 = 60 pF
b) Data : C1 = 50 nF (nanofarad) ; C2 = 30 nF ; C3 = 15 nF ; C4 = 60 nF ; Cn = 5 nF.
Total capacity : Ct = 50 + 30 + 15 + 60 + 5 = 160 nF
c) Data : C1 = 2 µF (microfarad) ; C2 = 5 µF ; C3 = 5 µF
Total capacity : Ct = 2 + 5 + 5 = 12 µF
FORMULATE 120
- Calculation
of the equivalent capacity of two or several condensers of equal value,
connected in series, knowing their capacity.
Statement : The equivalent capacity of two or several condensers having the same capacity, connected in series, is obtained by dividing the capacitance of a capacitor by the number of condensers.
Ceq = C / n
Ceq = equivalent capacity
C = capacity of each condenser
n = number of condensers.
The equivalent capacity is obtained with the same measuring unit as that used to indicate the capacity of the condensers.
Examples :
a) Data : C = 420 pF (picofarad) ; n = 2.
Equivalent capacity : Ceq = 420 / 2 = 210 pF.
b) Data : C = 40 nF (nanofarad) ; n = 4
Equivalent capacity : Ceq = 40 / 4 = 10 nF.
c) Data : C = 2 µF (microfarad) ; n = 3
Equivalent capacity : Ceq = 2 / 3
0,666 µF = 666 nF.
FORMULATE 121 - Calculation
of the equivalent capacity of two condensers of different value, connected in
series, knowing their respective capacity.
Statement : The equivalent capacity of two condensers connected in series is obtained by multiplying the capacity of the two condensers and by dividing the product obtained by the sum of these capacities.
Ceq = (C1 x C2) / (C1 + C2)
Ceq = equivalent capacity
C1 = capacitance of a capacitor
C2 = capacity of the other condenser.
The values all of capacity must be expressed with the same measuring unit.
Example :
Data : C1 = 40 nF (nanofarad) ; C2 = 60 nF.
Equivalent capacity of the two condensers connected in series :
Ceq = (40 x 60) / (40 + 60) = 2 400 / 100 = 24 nF
FORMULATE 122 - Calculation
of the capacitance of a capacitor to be connected in series to another condenser
of known capacity, to obtain a given equivalent capacity.
Statement : The capacitance of a capacitor to be connected in series to another condenser, to obtain a given equivalent capacity, is calculated by multiplying the capacity of the condenser known by the equivalent capacity and by dividing the product by the difference in these same values.
Ci = (C x Ceq) / (C - Ceq)
Ci = unknown capacity
C = capacity of the condenser available
Ceq = equivalent capacity which one wants to obtain
The values all of capacity must be expressed with the same measuring unit.
Example :
Data : C = 500 pF (picofarad) ; Ceq = 400 pF.
Unknown capacity : Ci = (500 x 400) / (500 - 400) = 200 000 / 100 = 2 000 pF
FORMULATE 123 - Calculation
of the equivalent capacity several condensers connected in series, knowing their
capacity.
Statement : The equivalent capacity several condensers connected in series, is obtained by carrying out calculations in three times: one calculates initially the reverse of the capacity of each condenser, which amounts dividing number 1 by the value of the capacity. Then, one adds the values with the opposite. Lastly, one calculates the equivalent capacity by dividing number 1 by the sum of the opposite.
Ceq = equivalent capacity
C1 = capacity of the first condenser
C2 = capacity of the second condenser
C3 = capacity of the third condenser
Cn = capacity of the last condenser
The values all of capacity must be expressed with the same measuring unit.
Example:
Data : C1 = 500 pF (picofarad) ; C2 = 2 000 pF ; C3 = 400 pF ; Cn = 200 pF.
Equivalent capacity :
OBSERVATION :
If one must calculate the equivalent capacity of two condensers connected in series, one can use formula 123, but it is simpler to resort to formula 121 ; Moreover, if the capacities of the condensers connected in series are equal, one can then use formula 120.
End of this form and we will finish the last heading “form mathematics 4” concerning electromagnetism.
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