**Created it, 05/10/15**

**Update it, 05/11/04**

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MATHEMATICAL FORMS 2 “2nd part”

**1. 2. -
ELECTRONICS **

**
FORMULATE 74
-**
(law of OHM) **:** Calculation of the resistance of a circuit knowing the tension
applied and the intensity of the current.

**Statement :**
The resistance, expressed in ohms, is obtained by dividing the tension expressed
into volts by the intensity of the current, expressed in amps (See theories ** 1**
and ** 1.2** in the electronic summary heading).

__Example__ :

** OBSERVATION :** Formula 74 can also apply by
expressing the tension and the intensity of the current respectively in
submultiples (or multiples) of the volt and the amp; in particular, one can
express the tension in millivolts

However, if the tension is expressed in volts and
the intensity of the current in milliamperes ** (mA ; 1 mA =
0,001 A)**, resistance will be expressed in
kilo-ohms **(k****W
; 1 kW =
1 000****
W****)**.
Sometimes one expresses the tension in volts and the intensity of the current in
microamperes **(µA ; 1 µA = 0,000 001 A) ;** in this
case, by applying formula 74, one obtains the resistance expressed in megohms **(M****W
;
1 MW =
1 000 000**** W).**

**
FORMULATE 75
-**
(law of OHM)** :** Calculation of the tension applied to a circuit knowing the
resistance and the intensity of the current.

**V = R x I**

(This formula is drawn from formula 74).

** OBSERVATION :** Formula 75 can also apply by
expressing the resistance and the intensity of the current respectively with the
submultiples of the ohm and the amp; one can express in particular resistance in
kilo-ohms (

**
FORMULATE 76
-**
** **(law of OHM)** :** Calculation of the intensity of the current knowing resistance and
the tension applied to the circuit.

(This formula is drawn from formula 74).

** OBSERVATION :** formula 76 can also apply by
expressing the tension and resistance respectively with the submultiples and
multiples of the volt and the ohm. For example, the tension can be expressed in
millivolts (

**
FORMULATE 77 -**
Calculation of the total resistance of a formed circuit of several resistances
connected in series knowing their values.

**Statement :** The value
of resistance equivalent to several resistances connected in series is obtained
by making the sum of the values of each resistance.

The values all of resistance must be expressed in
the same measuring unit (**W,
kW,
MW**).

__Example__:

**a) Data: R1 = 300
W (ohm) ;
R2 = 1 kW (kilo-ohm) =
1000 W**

**Total resistance : Rt = 300 + 1000 = 1300 W**

**b) Data : R1 = 2 kW
;
R2 = 0,5 kW ; R3 = 1500 W =
1,5 kW**

**Total resistance: Rt = 2 + 0,5 + 1,5 = 4 kW**

**c) Data : R1 = 0,5 MW
(megohm ;
1 MW =
1 000 000 W) ; R2 = 2,7 MW
; R3 =
300 kW = 0,3
MW ; R4 = 1
MW**

**Total resistance: Rt = 0,5 + 2,7 + 0,3 + 1 = 4,5 MW**

**FORMULATE 78** ** - ** Calculation
of the total conductance of a formed circuit of several resistances connected in
parallel knowing their values.

**Statement :** The
conductance resulting from the association of several resistances connected in
parallel is obtained by making the sum of the conductances of each resistance, (see
theory 2 in the electronic summary heading).

The values all of the conductances must be
expressed in the same measuring unit (**S** or** mS**
or **µS**).

__Example__:

**Data : G1 = 10 000 S ; G2 =
15 000 S ; G3 = 5 000 S ; G4 = 60 000 S**

**Total conductance : Gt = 10
000 S
+ 15 000 S + 5 000 S + 60 000 S = 90 000 S**

**
FORMULATE 79 -**
Calculation of resistance equivalent to several resistances connected in
parallel knowing their values.

Statement: Resistance equivalent to several
resistances connected in parallel is obtained by carrying out calculations in
three times** : ** initially the conductance of each resistance (formula 70) is
calculated; then one calculates the total conductance of resistances in parallel
(formula 78) **;** finally, one calculates equivalent resistance, i.e. the resistance
which corresponds to the total conductance (formula 71).

All the calculations indicated in the preceding statement can be represented by the following formula:

The values all of resistance must be expressed in
the same measuring unit **(W,
k****W,
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 resistances
connected in parallel** :**

**
FORMULATE 80 -
**
Calculation of the equivalent resistance of two resistances connected in
parallel knowing their values.

Statement: The equivalent resistance of two resistances connected in parallel is obtained by multiplying the values of two resistances and by dividing the whole by the sum of these two values.

The values all of resistance must be expressed in
the same measuring unit (**W,
k****W,
MW**).

__Example__:

**Data : R1 = 2 kW (kilo-ohm) = 2000
W
; R2 = 800 W**

Equivalent resistance of the two resistances connected in parallel:

**Re = (2 000 x 800) / (2 000 + 800)
= 1 600 000 / 2 800
571,4 W (value approached by defect).**

**OBSERVATION :** To calculate the value of the
equivalent resistance of two resistances, one can also be used for oneself of
formula 79.

**
FORMULATE 81**
** - ** Calculation of the value of resistance to be put
in parallel with another resistance of value known to obtain a given equivalent
resistance.

**Statement :**
The value of resistance to be put in parallel with another resistance of value
known to obtain a given equivalent resistance is calculated by multiplying the
value of the resistance known by equivalent resistance, the whole divided by the
difference in these two values.

The values all of resistance must be expressed in
the same measuring unit (**W,
k****W,
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 (value
approached by defect)**

**
FORMULATE
82 -**
** **
Calculation of resistance equivalent to two or several of the same resistances
value connected in parallel.

**Statement :**
** **
The equivalent resistance of two or several resistances of equal values
connected in parallel is obtained by dividing the value by the number of
resistances.

**Re = R / n**

Equivalent resistance will be expressed in the same measuring unit as that used to express the value of resistances.

__Example__:

**a) Data : R = 1 200 W
(ohm) ; n = 2**

**Equivalent resistance: Re = 1.200 / 2 = 600 W**

**b) Data : R = 150 kW
(kilo-ohm) ; n = 3 **

**Equivalent resistance: Re = 150 / 3 = 50 kW**

**c) Data : R = 2 MW
(megohm) ; n = 4**

**Equivalent resistance: Re = 2 / 4 = 0,5 MW = 500 kW**

**FORMULATE 83** **
-**
Calculation of the electromotive force (**f.e.m.**)
obtained while connecting in series two or several piles, knowing the
electromotive force of each pile.

**Statement :**
While putting in series two or several piles, one obtains an electromotive force
equal to the sum of the electromotive forces of each pile.

**E2 = electromotive force
of the second pile**

The electromotive forces all must be expressed in the same measuring unit.

__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.**

**
FORMULATE 84**
** -** Calculation of resistance internal of pile
knowing its ** f.e.m. ** (no-load voltage when it does
not provide any current) and the charging voltage when it provides a given
current.

**Statement :**
** **
The internal resistance of a pile is given by the difference between the** f.e.m.
**
and the charging voltage, the whole divided by the provided current.

__Example__:

**
FORMULATE 85
- ** Calculation of the
electric output of an apparatus knowing the tension applied and the intensity of
the absorptive current.

**Statement :** The
electric output, expressed in Watts, is obtained by multiplying the tension,
expressed in volts, by the intensity of the current expressed in amps.

__Example__:

**
FORMULATE 86**
** -** Calculation of the intensity of the current
absorptive by an apparatus knowing the tension applied and its electric output.

(This formula is drawn from formula 85)

__Example__:

**Data : P = 300 W ; V = 220 V**

**Intensity of the current : I
= 300 / 220 = 1,36 A ** (value approached by
defect)**.**

**
FORMULATE 87 -**
Calculation of the tension applied to an apparatus knowing the intensity of the
absorptive current and the electric output of this one.

(This formula is drawn from formula 85).

__Example__:

**
FORMULATE 88
- ** Calculation
of the electric power consumed by an apparatus knowing its electric output and
its operation life.

**Statement :**
** **
The power consumption by an apparatus, expressed in Watt-seconds, is obtained by
multiplying the power of the apparatus, expressed in Watts, by the operating
time expressed in seconds.

- Example :

**Data : P = 600
W ; t = 5 mn (minutes) = 300 s**

**Power
consumption : W =
600 x 300 = 180 000 W.s**

** OBSERVATION :** The Watt-second (

In practice, to indicate the domestic and
industrial consumption of the electric power, one uses a multiple of the
Watt-second, i.e. the kilowatt-hour (**kW.h**).

The result of the preceding example can be
expressed in **kW.h** by means of equivalence:

If in formula 88, the power is expressed in
kilowatts (**kW ; 1 kW = 1000 W**) and time in hours (**h,
1 h = 60 mn = 3 600 s**), the power consumption will be expressed in
kilowatt-hours.

For example, if **P = 800 W =
0,8 kW** and **“t” = 30 mn = 0,5 h**, the
power consumption will be of **:**

**
FORMULATE 89
-**
** **
Calculation of the electric output dissipated by Joule effect in a resistance
knowing the intensity of the current and the value of this resistance.

**Statement :** The
electric output, 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 amps.

__Example__:

**
FORMULATE 90
-**
Calculation of the value of a resistance knowing the dissipated electric output
and the intensity of the current which crosses it.

(This formula is drawn from formula 89)

__Example__:

**
FORMULATE 91 -**
Calculation of the intensity of the current which traverses a resistance knowing
the dissipated electric output and the value of this resistance.

**Statement :**
The intensity of the current, expressed in amps, which traverses a resistance,
is obtained by dividing the dissipated power expressed into Watts by the
resistance expressed in ohms, and extracting the square root of the quotient
obtained.

**
FORMULATE 92 -**
Calculation of the electric output dissipated by Joule effect in a resistance
knowing the tension applied and the value of this resistance.

**Statement :**
The electric output, expressed in Watts, dissipated in a resistance is obtained
by dividing the square of the tension, expressed into volts, by the resistance
expressed in ohms.

**P = V ^{2}
/ R**

**P = electric output out
of W (watt)**

**V = tension out of V
(volt)**

**R = resistance in W
(ohm)**

__Example__:

**
FORMULATE 93**
** -** Calculation of the value of a resistance knowing
the dissipated electric output and the tension applied.

(This formula is drawn from formula 92).

__Example__:

**
FORMULATE 94
- ** Calculation
of the tension applied to a resistance knowing the dissipated electric output
and the value of this resistance.

**Statement :**
The tension applied to a resistance, expressed in volts, is obtained by
multiplying the dissipated power expressed in Watts, by the resistance expressed
in ohms and extracting the square root of the product obtained.

**
FORMULATE 95**
** -** Calculation of the quantity of heat obtained by
transforming by Joule effect an electric quantity of power given (for the
calculation of the electric power, to see formula 88).

**Statement :**
The quantity of heat expressed in kilogram calories produced in a resistance by
Joule effect is obtained by multiplying the electric power expressed in ** Watt-seconds**
(**Joule**) dissipated in this resistance by number **0,000238.**

__Example__:

**Data :** Electric
power dissipated by a resistance ** W = 0,5 kW.h (kilowatt-hour)
= 3 600 000 x 0,5 = 1 800 000 W.s ** (for the
equivalence between the kilowatt-hour and it Watt-second, to see the observation
which follows formula 88).

Quantity of heat produced by resistance: **Qc
= 0,000 238 x 1 800 000 = 428,4 kcal**.

**
FORMULATE 96
-**
Calculation of the hot resistance of a driver knowing the increase in
temperature of material and the resistance of the driver to the ambient
temperature (**20° C**).

**Statement :** By
increasing the temperature of a driver, one increases his electric resistance.
For the calculation of the hot resistance of a driver, it is necessary to
supplement the statement in the following way: the hot resistance, expressed in
ohms, is obtained by adding the strength with the driver to the ambient
temperature (**20° C**) with the product of the
temperature coefficient of material, the strength to the ambient temperature and
the increase in temperature expressed in degrees Celsius.

The temperature coefficients of principal
conducting materials are deferred in the last column of right-hand side of table
III (**we defer same figure 1 below**).

**Rt **
= hot resistance (at the temperature** t**) in **W**
(ohm)

**R _{20}** = cold strength
(to the temperature of

= temperature coefficient of material

**t **
= temperature of the hot driver in **°C** (degrees
Celsius)

**t - 20** = increase in
temperature in **°C** (degrees Celsius)

__Example__:

**Data relating to a tungsten
driver : R20 = 30 W (cold resistance of the driver) ;
= 0,0045 (temperature coefficient of tungsten) ; t = 320 °C (temperature
of the driver).**

**Increase in temperature of the
driver : t - 20 = 320 - 20 = 300° C**

**Hot resistance of the driver : Rt
= 30 + 0,0045 x 30 x 300 = 30 + 40,5 = 70,5 W**

**FORMULATE 97****
-** Calculation of resistance per cold
meter (to **20° C**) of a driver knowing its section
and the resistivity of material.

* This formula is drawn from formula 64 (“to see form mathematics 2 - 1st part”) by giving to the length of the driver the value of 1 meter *.

__Example__:

Data relating to a driver of nickel-chromium **:**
** p
= 0,9 µ****W****.m ** (resistivity of nickel-chromium at temperature of **20°
C ****;** table III, figure 1) ; ** S = 0,007854
mm ^{2}
**
(section of the driver).

**Resistance per meter (to 20°
C) : R / m = 0,9 / 0,007854 = 115 W
/ m (approximate
value)**

** OBSERVATION :** In table IV (“to see the
form maths 2, 1st part, figure 2”), one deferred the values of resistance per
meter of the drivers of nickel-chromium, constantan and manganin for the
sections of the most frequent use in the electric applications.

**
FORMULATE 98 -**
Calculation of resistance per hot meter of a wire of heating knowing resistance
per cold meter (formula 97), the temperature coefficient of material (table III,
figure 1) and the operating temperature of the wire of heating.

**(R / m) t = resistance per meter to
the operating temperature of the wire of heating in W
/ m
(ohm per meter).**

**(R / m) _{20} = resistance
per meter to the ambient temperature in W
/ m (ohm per meter)**

**
= temperature coefficient of material**

**t = operating temperature of the
wire of heating in ° C (degrees Celsius)**

**t - 20 = increase in temperature
in ° C (degrees Celsius)**

(This formula is drawn from formula 96 in
substituent the symbol of resistance (**R**) to that of
resistance per meter, **R / m**).

__Example__:

**Data :** **(R
/ m) 20 = 5,65**
**
W /
m**
(resistance per cold meter of a wire of nickel-chromium having a diameter of **0,45
mm** **;** table IV, figure 2)** ;**
**= 0,00011** (temperature coefficient of
nickel-chromium; table III, figure 1)** ;** ** t
=
1020° C ** (operating temperature of the wire of heating).

**Increase in temperature
during the passage of the ambient temperature at the operating temperature of
the wire :** **t - 20 = 1020 - 20 = 1000° C**

**Resistance per meter to
1020° C :** ** (R / m) t = 5,65 + 0,00011 x 5,65
x 1000 =
5,65 + 0,6215 = 6,2715
6,28** ** W ****/
m
** **(value approached by excess).**

** OBSERVATION : ** The calculation of resistance
per hot meter of a wire of heating is necessary to determine the length of it
when one knows the value of the resistance which it must have during operation,
i.e. its hot resistance. For better explaining with a practical example, let us
see how one must proceed in the calculation of resistance to nickel-chromium of
an electric heating appliance.

Calculation of the resistance of a heating appliance (electric furnace, electric drying oven, etc…).

**Data :** The furnace is
supplied with the tension of **220 V** and must
dissipate a power of **600 W** at the temperature of
approximately **1020° C** (operating temperature of
the wire of heating).

**Process :**

**1)** One calculates the intensity of the current
which supplies the furnace under the normal conditions of operation **;** to this
end, formula 86 is used **:**

**2) ** While basing oneself on the value of the
calculated current, one chooses the section of the wire; for the current of **2,73
A**, it will be necessary to choose a wire of nickel-chromium with a
diameter **0,45 mm** corresponding to the section of
** 0,159043 mm ^{2} **(see table 4 of figure 2, 1st theory). Generally, to make this choice,
it is necessary to know the density of current necessary to maintain the wire of
nickel-chromium at the operating temperature.

Thereafter, in form 4, we will see how to proceed
to calculate the section of a discussion thread knowing the allowed density of
current by material and the intensity of the current which must pass in the wire**
;**
it is enough for you to now know the diameters of the wire which are indicated
in the fifth column on the basis of the left on table V (figure 3 below) in
correspondence with the various values of the current deferred in the third
column.

**3) ** One calculates the value of hot resistance
that the wire must present to dissipate the power of **600 W**,
the tension of **220 V** being applied; formula 93 to
this end is used:

**R = V ^{2} / P = 220^{2}
/ 600 = 48 400 / 600
= 80,666
80,67 W ** (value
approached by excess)

**4)** By consulting table IV (figure 2, form maths2,
1st part), one determines resistance per cold meter of the wire of
nickel-chromium having a diameter of **0,45 mm**
(section of ** 0,159043 mm ^{2}**)

**5)** Knowing the value of resistance per cold meter
of the wire nickel-chromium chosen, one calculates the value of resistance per
meter of this same wire to the operating temperature (**1020°
C**) **;** to this end, 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 (value approached by
excess)**

One can avoid this calculation by taking again
the value of 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 resistance per hot meter (**6,28 W**),
one calculates the length of the wire by dividing the first value by the second:

To build an electric furnace functioning with the
tension of **220 V**, at the temperature of
approximately ** 1000° C ** (**1020°
C ** in calculations) in order to dissipate a power of **600
W**, one can thus use a wire of nickel-chromium having the resistivity of
** 0,9 µ****W****.m ** (table III, figure 1).

The value of the resistivity of nickel-chromium was not mentioned in calculation, but was introduced previously into the formula 97 to calculate the resistances per meter indicated in table IV (figure 2, “1st part of the form mathematics N° 2”), i.e. the values of the resistances per meter cold used successively in formula 98 to calculate the resistances per meter hot indicated in table V (above).

We thus finish this 2nd part of the form of this lesson and finally, the third and the fourth will be examined a little later