Enunciated :
The resistance, expressed in ohms, is obtained by dividing the voltage expressed in volts by the intensity of the current, expressed in amperes (see theories 1 and 1.2 in the electronic summary section).
R = V / I
R = resistance in W (ohm)
V = voltage in V (volt)
I = intensity of the current in A (ampere)
Example :
Data : V = 150 V ; I = 0,2 A
Circuit resistance : R = 150 / 0,2 = 750 W
OBSERVATION : Formula 74 can also be applied by expressing the voltage and current respectively in submultiples (or multiples) of the volt and the ampere ; in particular, one can express the voltage in millivolts (mV, 1 mV = 0.001 V) and the current intensity in milliamperes (mA, 1 mA = 0.001 A) and so the resistance will be expressed in ohms.
However, if the voltage is expressed in volts and the current in milliamperes (mA, 1 mA = 0.001 A), the resistance will be expressed in kilo-ohms (kΩ, 1 kΩ = 1000 Ω). The voltage is sometimes expressed in volts and the intensity of the current in microamperes (µA, 1 µA = 0.000 001 A) ; in this case, applying formula 74, the resistance expressed in megohms (MΩ, 1 MΩ = 1 000 000 Ω) is obtained.
FORMULA 75 - (OHM law) : Calculation of the voltage applied to a circuit knowing the resistance and the intensity of the current.
OBSERVATION : The formula 75 can also be applied by expressing the resistance and the intensity of the current respectively with the sub-multiples of the ohm and the ampere ; in particular, the resistance in kilo-ohms (kΩ) and the current intensity in milliamps (mA) can be expressed and the voltage will be expressed in volts.
FORMULA 76 - (OHM law) : Calculation of the current intensity knowing the resistance and the voltage applied to the circuit.
OBSERVATION : Formula 76 can also be applied by expressing voltage and resistance respectively with submultiples and multiples of volts and Ohms. For example, the voltage can be expressed in millivolts (mV) and the resistance in ohm (Ω) : in this case, the intensity of the current will be expressed in milliamperes (mA) ; similarly, the voltage can be expressed in volts and the resistance in kilo-ohms (kΩ), the intensity of the current will be expressed in milliamperes.
FORMULA 77 - Calculation of the total resistance of a circuit formed of several connected resistors in series knowing their values.
Enunciated :
The resulting conductance of the association of several parallel connected resistors is obtained by summing the conductances of each resistance, (see theory 2 in the electronic summary section).
Gt = G1 + G2 + ... + Gn
Gt = total conductance
G1 = conductance of the first resistance
G2 = conductance of the second resistance
Gn = conductance of the last resistance
The values of the conductances must all be expressed in the same unit of measure (S
ou mS or µS).
Example :
Data : G1 = 10 000 S ;
G2 = 15 000 S ; G3 = 5 000 S ; G4 = 60 000 S
Total conductivity : Gt = 10 000
S + 15 000 S + 5 000 S + 60 000 S = 90 000 S
FORMULA 79 - Computation of the resistance equivalent to several resistances connected in parallel knowing their values.
Enunciated : The resistance equivalent to several resistances connected in parallel is obtained by performing the calculations in three stages: first, the conductance of each resistor (formula 70) is calculated ; then the total conductance of the resistances in parallel is calculated (formula 78) ; finally, we calculate the equivalent resistance, that is to say the resistance which corresponds to the total conductance (formula 71).
All the calculations indicated in the preceding statement can be translated by the following formula :
Re = equivalent resistance
R1 = value of the first resistance
R2 = value of the second resistance
Rn = value of the last resistance
Resistance values must all be expressed in the same unit of measure (W,
kW, MW).
Example :
Data : R1 = 200 W
(ohm) ; R2 = 1 kW (kilo-ohm) = 1000 W
; R3 = 20 W ; R4 = 500 W
; R5 = 100 W
Resistance equivalent to the five resistors connected in parallel :
FORMULA 80 - Calculation of the equivalent resistance of two connected resistors in parallel knowing their values.
Enunciated : The equivalent resistance of two resistors connected in parallel is obtained by multiplying the values of the two resistances and dividing the whole by the sum of these two values.
Re = (R1 x R2) / ( R1 + R2)
Re = equivalent resistance
R1 = value of a resistance
R2 = value of the other resistance
The resistance values must all be expressed in the same unit of measure (W,
kW, MW).
Example :
Data : R1 = 2 kW
(kilo-ohm) = 2000 W ; R2 = 800 W
Equivalent resistance of the two resistors connected in parallel :
Re = (2 000 x 800) / (2 000 + 800) = 1 600 000 / 2 800
571,4 W (approximate value by default).
OBSERVATION : To calculate the value of the equivalent resistance of two resistances, one can also use Formula 79.
FORMULA 81 -
Calculation of the value of the resistance to be paralleled with another resistance of known value to obtain a given equivalent resistance.
Enunciated : The value of the resistance to be paralleled with another resistance of known value to obtain a given equivalent resistance is calculated by multiplying the value of the known resistance by the equivalent resistance, all divided by the difference of these two values.
Ri = (R x Re) / (R - Re)
Ri = unknown resistance
R = value of available resistance
Re = equivalent resistance that we want to obtain
The resistance values must all be expressed in the same unit of measure (W,
kW, MW).
Example :
Data : R = 2000 W
(ohm) ; Re = 600 W
Unknown resistance : Ri = (2
000 x 600) / (2 000 - 600) = 1 200 000 / 1 400 = 857 W (approximate value by default)
FORMULA 82 -
Calculation of the resistance equivalent to two or more resistors of the same value connected in parallel.
Enunciated :
The equivalent resistance of two or more resistors of equal values connected in parallel is obtained by dividing the value by the number of resistors.
Re = R / n
Re = equivalent resistance
R = resistance value
n = number of resistances
The equivalent resistance shall be expressed in the same unit of measure as that used to express the value of the resistors.
FORMULA 83 -
Computation of the electromotive force (f.e.m.) obtained by connecting in series two or more piles, knowing the electromotive force of each pile.
Enunciated : By putting two or more cells in series, an electromotive force equal to the sum of the electromotive forces of each stack is obtained.
Et = E1 + E2 + ... + In
Et = total electromotive force
E1 = electromotive force of the first battery
E2 = electromotive force of the second battery
En = electromotive force of the last battery.
The electromotive forces must all be expressed in the same unit of measure.
Example :
Data : E1 = 4,5 V (volt)
; E2 = 4,5 V ; E3 = 9 V ; E4 = En = 9 V
Total electromotive force :
Et = 4,5 + 4,5 + 9 + 9 = 27 V.
FORMULA 84 -
Calculation of the internal resistance of a battery knowing its f.e.m. (no-load voltage when no current is supplied) and the load voltage when supplying a given current.
Enunciated :
The internal resistance of a stack is given by the difference between the f.e.m. and the load voltage, all divided by the current supplied.
Ri = (E - V) / I
Ri = internal resistance in W (ohm)
E = f.e.m. "Vacuum voltage" in V (volt)
V = load voltage in (volt)
I = current supplied in A (ampere).
Example :
Data : E = 4,5 V
; V = 4,2 V ; I = 0,3 A
Internal resistance
: Ri = (4,5 - 4,2) / 0,3 = 0,3 / 0,3 = 1 W
FORMULA 85 - Calculation of the electrical power of an apparatus knowing the applied voltage and the intensity of the current absorbed.
Enunciated : The electric power, expressed in watts, is obtained by multiplying the voltage, expressed in volts, by the intensity of the current expressed in amperes.
P = VI
P = electric power in W (watt)
V = voltage in V (volt)
I = intensity of the current in A (ampere)
Example :
Data : V = 200 V
; I = 1,5 A
Power : P = 200 x 1,5 = 300 W
FORMULA 86 -
Calculation of the intensity of the current absorbed by an apparatus knowing the tension applied and its electrical power.
Enunciated : The energy consumed by a device, expressed in watts-seconds, is obtained by multiplying the power of the device, expressed in watts, by the operating time expressed in seconds.
W = Pt
W = energy consumed in W.s (watt-seconde)
P = electric power in W (watt)
t = time in s (seconds)
Example
:
Data : P = 600 W ; t = 5 mn (minutes) = 300 s
Consumed energy : W = 600 x 300 = 180 000 W.s
OBSERVATION : The watt-seconds (W.s), unit of measurement used to express the amount of electrical energy consumed, is equivalent to 1 joule (J), a unit of measurement of energy and mechanical work.
1 W.s = 1 J
In practice, to indicate the domestic and industrial consumption of electrical energy, a multiple of the watt-second, that is to say the kilowatt-hour (kW.h) is used.
If in the Formula 88 the power is expressed in kilowatts (kW, 1 kW = 1000 W) and the time in hours (h, 1 h = 60 mn = 3 600 s), the energy consumed will be expressed in kilowatt hours.
For example, if P = 800
W = 0,8 kW et "t"
= 30 mn = 0,5 h, the energy consumed will be :
W = 0,8 x 0,5 = 0,4 kW.h
FORMULA 89 -
Calculation of the electric power dissipated by Joule effect in a resistance knowing the intensity of the current and the value of this resistance.
Enunciated :
The electric power, expressed in watts, dissipated in a resistance is obtained by multiplying the resistance, expressed in ohms, by the square of the current which crosses it, expressed in amperes.
P = RI2
P = electric power in W (watt)
R = resistance in W (ohm)
I = intensity of the current in A (ampere)
Example :
Data : R = 125 W ; I = 0,3 A
Electrical power dissipated : P = 125 x 0,32 = 125 x 0,09 = 11,25 W
FORMULA 90 - Calculation of the value of a resistor knowing the electric power dissipated and the intensity of the current which crosses it.
FORMULA 91- Calculation of the intensity of the current which flows through a resistor knowing the dissipated electrical power and the value of this resistance.
Enunciated : The amperage current, which flows through a resistor, is obtained by dividing the dissipated power expressed in watts by the resistance expressed in ohms, and extracting the square root of the quotient obtained.
FORMULA 92 - Calculation of the electrical power dissipated by Joule effect in a resistor knowing the applied voltage and the value of this resistance.
Enunciated : The electrical power, expressed in watts, dissipated in a resistor is obtained by dividing the square of the voltage, expressed in volts, by the resistance expressed in ohms.
P = V2 / R
P = electric power in W (watt)
V = voltage in V (volt)
R = resistance in W (ohm)
Example :
Data : V = 220 V ; R = 242 W
Electrical power dissipated : P = 2202 / 242 = 48 400 / 242 = 200 W
FORMULA 93 -
Calculation of the value of a resistor knowing the electric power dissipated and the tension applied.
Enunciated :
The voltage applied to a resistor, expressed in volts, is obtained by multiplying the dissipated power expressed in watts, the resistance expressed in ohms and extracting the square root of the product obtained.
FORMULA 95 - Calculation of the amount of heat obtained by Joule transforming a given amount of electrical energy (for the calculation of electrical energy, see formula 88).
Enunciated : The amount of heat expressed in kilo-calories produced in a Joule resistance is obtained by multiplying the electrical energy expressed in watts-seconds (Joule) dissipated in this resistance by the number 0.000238.
Qc = 0,000238 W (approximate value by default)
Qc = amount of heat in kilo-calories
W = electrical energy in W.s (watt-seconde) or in J (Joule)
Example :
Data : Electrical energy dissipated by a resistance W = 0.5 kW.h (kilowatt-hours) = 3 600 000 x 0.5 = 1 800 000 Ws (for the equivalence between the kilowatt-hour and the watt-second, see observation following Formula 88).
Quantity of heat produced by the resistance :Qc = 0,000 238 x 1 800 000 = 428,4 kcal.
FORMULA 96-
Calculation of the hot resistance of a driver knowing the increase of temperature of the material and the resistance of the conductor to the ambient temperature (20° C).
Enunciated : Increasing the temperature of a conductor increases its electrical resistance. For the calculation of the hot resistance of a conductor, the statement must be completed as follows : The resistance in hot, expressed in ohms, is obtained by adding the resistance of the conductor to the ambient temperature (20° C. ) with the product of the temperature coefficient of the material, the resistance at ambient temperature and the temperature increase expressed in degrees Celsius.
The temperature coefficients of the main conducting materials are reported in the last right column of Table III (we report the same Figure 1 below).
Rt
= hot resistance (at temperature t) in W (ohm)
R20
= resistance to cold (at a temperature of 20° C)
in W (ohm)
=
temperature coefficient of the material
t
= hot conductor temperature in °C (degrees Celsius)
t - 20
= temperature increase in °C (degrees Celsius)
Example :
Data relating to a tungsten conductor : R20 = 30 W (cold resistance of the conductor) ;
= 0,0045 (temperature coefficient of tungsten) ; t = 320° C (conductor temperature).
Increase in temperature of the conductor : t - 20 = 320 - 20 = 300° C
Conductor hot resistance : Rt = 30 + 0,0045 x 30 x 300 = 30 + 40,5 = 70,5 W
FORMULA 97-Calculation of the resistance per cold meter (at 20° C) of a driver knowing its section and the resistivity of the material.
R / m = resistance per meter in Ω / m (ohm per meter)
p = resistivity in µW.m (microhm-meter)
S = section in mm2
* This formula is taken from formula 64 ("see mathematical form 2 - 1st part") giving the length of the driver the value of 1 meter *.
Example :
Data relating to a nickel-chromium conductor :p = 0,9 µW.m
(resistivity of nickel-chromium at a temperature of 20°
C ; table III, Figure 1) ;S = 0,007854 mm2(conductor section).
Resistance per meter (at 20° C) : R / m = 0,9 / 0,007854 = 115 W / m
(approximate value)
OBSERVATION : In Table IV ("see mathematical form 2, 1st part, Figure 2"), the resistance values per meter of the nickel-chromium, constantan and manganin conductors were reported for the most frequent use sections. in electrical applications.
FORMULA 98 - Calculation of the resistance per meter when hot of a heating wire knowing the resistance per cold meter (formula 97), the coefficient of temperature of the material (table III, Figure 1) and the operating temperature heating wire.
(R / m) t = resistance per meter at the operating temperature of the heating wire in W / m (ohm per meter).
(R / m)20 = resistance per meter at ambient temperature in W / m (ohm per meter)
= temperature coefficient of the material
t = operating temperature of heating wire in ° C (degrees Celsius)
t - 20 = temperature increase in ° C (degrees Celsius)
(This formula is taken from formula 96 by substituting the symbol of resistance (R)
for that of resistance by meter, R / m).
Example :
Data :
(R / m) 20 = 5,65
W
/ m (resistance per cold meter of a nickel-chromium wire having a diameter of 0,45
mm ; table IV, Figure 2) ;
= 0,00011 (temperature coefficient of nickel-chromium ;
tableau III, figure 1) ; t = 1020° C (operating temperature of the heating wire).
Increase in temperature during the transition from ambient temperature to the operating temperature of the wire : t - 20 = 1020 - 20 = 1000° C
Resistance per meter at 1020° C : (R / m) t = 5,65 + 0,00011 x 5,65 x 1000 = 5,65 + 0,6215 = 6,2715
6,28 W / m (approximate value excess).
OBSERVATION : The calculation of the resistance per meter when heating a heating wire is necessary to determine its length when one knows the value of the resistance which it must have during the operation, that is to say its resistance hot. To better explain with a practical example, let's see how one should proceed in the calculation of the nickel-chromium resistance of an electric heater.
Calculation of the resistance of a heating device (electric oven, electric oven, etc ...).
Data : The oven is powered with 220 V and must dissipate a power of 600 W at a temperature of about 1020° C (operating temperature of the heating wire).
Process :
1) The intensity of the current supplied to the furnace under normal operating conditions is calculated ; for this purpose, Formula 86 is used :
I = P / V = 600 / 220 =
2,73 A (approximated value by excess)
2) Based on the value of the calculated current, the section of the wire is selected ; for the current of 2.73 A, it will be necessary to choose a nickel-chromium wire of diameter 0.45 mm corresponding to the section of 0.159043 mm2 (see Table 4 of Figure 2, 1st theory). In general, to make this choice, it is necessary to know the current density necessary to maintain the nickel-chromium wire at the operating temperature.
Subsequently, in the form 4, we will see how to calculate the section of a wire knowing the current density allowed by the material and the intensity of the current that must pass through the wire ; All you need now is to know the diameters of the wires which are indicated in the fifth column starting from the left in Table V (Figure 3 below) in correspondence with the various values of the current reported in the third column.
3) One calculates the value of the hot resistance that the wire must present to dissipate the power of 600 W, the voltage of 220 V being applied; the Formula 93 is used for this purpose :
R = V2 / P = 2202
/ 600 = 48 400 / 600 = 80,666
80,67 W
(approximate value over)
4) By consulting Table IV (Figure 2, Mathematical Form 2, Part 1), the resistance per cold meter of the nickel-chromium wire having a diameter of 0.45 mm (section 0.159043 mm2) is determined ; the value shown in the table is 5.65 W / m.
5) Knowing the value of the resistance per cold meter of the chosen nickel-chromium wire, the value of the resistance per meter of this same wire is calculated at the operating temperature (1020°
C) ; for this purpose, Formula 98 is used :
(R / m) t = (R / m)20 + (R
/ m)20 (t - 20)
= 5,65 + 0,00011 x 5,65 x 1000
= 5,65 + 0,6215 = 6,2715
6,28 W / m (approximated value by excess)
This calculation can be avoided by taking the value of the resistance per hot meter in table V (Figure 3) in correspondence with the values of 600 W and 220 V.
6) Knowing the value of the hot resistance of the wire (80,67 W)
and the value of the resistance per meter when hot (6,28
W), the length of the wire is calculated by dividing the first value by the second :
80,67 / 6,28 = 12,84 meters
To build an electric furnace operating with the voltage of 220 V, at a temperature of about 1000° C (1020° C in calculations) so as to dissipate a power of 600 W, we can use a nickel-chromium wire having the resistivity of 0,9 µW.m
(table III, Figure 1).
The value of the resistivity of the nickel-chromium was not mentioned in the calculation, but was introduced previously in the formula 97 to calculate the resistances per meter indicated in the table IV (Figure 2, "1st part of the mathematical form N° 2"), that is to say the values of the resistances per cold meter used successively in the Formula 98 to calculate the resistance per meter when hot indicated in Table V (above).
Calculation of the resistance of a circuit
571,4 W (approximate value by default).
=
temperature coefficient of the material
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