Calculation of the resistance of a circuit Calculation of the tension Calculation of the intensity of the current Total resistance of a circuit Total conductance of a circuit Equivalent resistance Equivalent resistance of two resistances Value of unknown resistance Resistance equivalent to two Calculation of the electromotive force Resistance interns of a pile Electric output of an apparatus Calculation of the intensity of the absorptive current Tension applied to an apparatus Consumed electric power Calculation of the electric output Calculation of the value of a resistance Calculation of the intensity of the current Dissipated electric output Calculation of the value of a resistance Tension applied to a resistance Calculation of the quantity of heat Hot resistance of a driver Resistance per cold meter Resistance per hot meter Wire of heating of Nickel-chromium Return to the synopsis To contact the author Low of page

Created it, 05/10/15

Update it, 05/11/04

N° Visitors

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).

R = V / I

R = resistance in W (ohm)

V = tension out of V (volt)

I = intensity of the current in A (amp)

Example :

Data : V = 150 V ;  I = 0,2 A

Resistance of the circuit : R = 150 / 0,2 = 750 W

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 (mV ; 1 mV = 0,001 V) and intensity of the current in milliamperes (mA ; 1 mA = 0,001 A) and thus resistance will be expressed in ohms.

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 (kW ; 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 (MW ; 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

V = tension out of V (volt)

R = resistance in W (ohm)

I = intensity of the current in A (amp)

(This formula is drawn from formula 74).

Example :

Data : R = 1500 W ;  I = 0,1 A

Tension applied : V = 1500 x 0,1 = 150 V

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 (kW) and intensity of the current in milliamperes (mA) and thus the tension will be expressed in volts.

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

I = V / R

I = intensity of the current in A (amp)

V = tension out of V (volt)

R = resistance in W (ohm)

(This formula is drawn from formula 74).

Example :
Data: V = 90 V ;  R = 180 W

Intensity of the current :  I = 90 / 180 = 0,5 A

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 (mV) and resistance in ohm (W) : in this case, the intensity of the current will be expressed in milliamperes (mA) ; in the same way, the tension can be expressed in volts and resistance in kilo-ohms (kW), the intensity of the current will be expressed in milliamperes.

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.

Rt = R1 + R2 + Rn, etc…

Rt = total equivalent resistance

R1 = value of the first resistance

R2 = value of the second resistance

Rn = value of the last 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).

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 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:

Re = equivalent resistance

R1 = value of the first resistance

R2 = value of the second resistance

Rn = value of the last resistance

The values all of resistance must be expressed in the same measuring unit (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 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.

Re = (R1 x R2) / (R1 + R2)

Re = equivalent resistance

R1 = value of a resistance

R2 = value of other resistance

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

Ri = (R x Re) / (R - Re)

Ri = unknown resistance

R = value of resistance available

Re = equivalent resistance which one wants to obtain

The values all of resistance must be expressed in the same measuring unit (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 (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

Re = equivalent resistance

R = value of resistances

n = number of resistances

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.

Et = E1 + E2 +… + En

Et = total electromotive force

E1 = electromotive force of the first pile

E2 = electromotive force of the second pile

En = electromotive force of the last 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.

Ri = (E - V) / I

Ri = resistance interns in W (ohm)

E = f.e.m. “no-load voltage” out of V (volt)

V = charging voltage out of V (volt)

I = current provides in A (amp).
• 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

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.

P = VI

P = electric output out of W (watt)

V = tension out of V (volt)

I = intensity of the current in A (amp)
• Example :

Data : V = 200 V ;  I = 1,5 A

Power : P = 200 x 1,5 = 300 W

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

I = P / V

I = intensity of the current absorptive in A (amp)

P = electric output out of W (watt)

V = tension applied out of V (volt)

(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.

V = P / I

V = tension applied out of V (volt)

P = electric output out of W (watt)

I = intensity of the current absorptive in A (amp).

(This formula is drawn from formula 85).

• Example :

Data : P = 1 200 W ;  I = 5,455 A

Tension applied : V = 1 200 / 5,455 = 220 V (approximate value)

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.

W = Pt

W = power consumption in W.s (watt-second)

P = electric output out of W (watt)

t = time in s (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 (W.s), measuring unit used to express the electric quantity of consumed power, is equivalent to 1 joule (J), measuring unit of energy and mechanical work.

1 W.s = 1 J

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).

1 kW.h = 3 600 000 W.s ;  1 W.s = 1 / 3 600 000 kW.h

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

180 000 W.s = 180 000 / 3 600 000 kW.h = 0,05 kW.h

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 :

W = 0,8 x 0,5 = 0,4 kW.h

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.

P = RI2

P = electric output out of W (watt)

R = resistance in W (ohm)

I = intensity of the current in A (amp)
• Example :

Data :  R = 125 W ;  I = 0,3 A

Dissipated electric output : P = 125 x 0,32 = 125 x 0,09 = 11,25 W

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

R = P / I2

R = resistance in W (ohm)

P = power dissipated out of W (watt)

I = intensity of the current in A (amp)

(This formula is drawn from formula 89)

• Example :

Data : P = 100 W ;  I = 0,5 A

Resistance : R = 100 / 0,52 = 100 / 0,25 = 400 W

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 = V2 / R

P = electric output out of W (watt)

V = tension out of V (volt)

R = resistance in W (ohm)

• Example :

Data : V = 220 V ;  R = 242 W

Dissipated electric output : P = 2202 / 242 = 48 400 / 242 = 200 W

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

R = V2 / P

R = resistance in W (ohm)

V = tension applied out of V (volt)

P = electric output dissipated out of W (watt)

(This formula is drawn from formula 92).

• Example :

Data: 220 V ;  P = 400 W

Resistance :  R = 2202 / 400 = 48 400 / 400 = 121 W

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.

Qc = 0,000238 W (Value approached by defect)

Qc = quantity of heat in kilogram calories

W = power electric in W.s (Watt-second) or in J (Joule)
• 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)

R20 = cold strength (to the temperature of 20° C) in W (ohm)

= 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.

(R / m) = (p / S)

R / m = resistance per meter in W / m (ohm per meter)

p = resistivity in µW.m (microhmmeter)

S = section in mm2

* 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 µW.m (resistivity of nickel-chromium at temperature of 20° C ; table III, figure 1) ; S = 0,007854 mm2 (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 = (R / m) 20 + (R / m) 20 (t - 20)

(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 :

I = P / V = 600 / 220 = 2,73 A (value approached by excess)

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 mm2 (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 = V2 / P = 2202 / 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 mm2) the value indicated on the table is of 5,65 W / m.

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:

80,67 / 6,28 = 12,84 W

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 µW.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

 Click here for the following lesson or in the synopsis envisaged to this end. High of page Preceding page Following page

Daniel