The conductance of a conductor is defined as the facility it has to pass the electric current. It depends not only on the material used for its manufacture, but also on its geometry: length and cross-sectional area.
The symbol used for conductance is G, and it is the inverse of electrical resistance R, a slightly more familiar quantity. The SI International System unit for conductance is the inverse of the ohm, denoted as Ω-1 and receives the name of siemens (S).
Other terms used in electricity that sound similar to conductance and are related are conductivity and the driving, but they should not be confused. The first of these terms is an intrinsic property of the substance with which the conductor is made and the second describes the flow of electric charge through it..
For an electrical conductor with constant cross-sectional area TO, length L and conductivity σ, the conductance is given by:
G = σ.A / L
The higher the conductivity, the higher the conductance. Also, the greater the cross-sectional area, the easier it is for the conductor to pass current. On the contrary, the greater the length L, the lower the conductance, since the current carriers lose more energy on longer paths..
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The conductance G for a conductor with constant cross-sectional area is calculated according to the equation given above. This is important, because if the cross section is not constant, the integral calculus must be used to find both the resistance and the conductance.
Since it is the inverse of the resistance, the conductance G can be calculated knowing that:
G = 1 / R
In fact, the electrical resistance of a conductor can be measured directly with a multimeter, a device that also measures current and voltage..
As said at the beginning, the unit of conductance in the international system is the Siemens (S). A conductor is said to have a conductance of 1 S if the current through it increases by 1 ampere for each volt of potential difference.
Let's see how that is possible through Ohm's law, if it is written in terms of conductance:
V = I.R = I / G
Where V is the voltage or potential difference between the ends of the conductor e I current intensity. In terms of these magnitudes, the formula looks like this:
G = I / V
Formerly the unit for conductance was the mho (ohm written backwards) denoted as Ʊ, which is an inverted capital omega. This notation was deprecated and was replaced by the siemens in honor of the German engineer and inventor Ernst Von Siemens (1816-1892), pioneer of telecommunications, but both are totally equivalent.
1 mho = 1 siemens = 1 A / V (ampere / volt)
In other measurement systems the statsiemens (statS) (in the cgs or centimeter-gram-second system) and the absiemens (abS) (electromagnetic cgs system) with the "s" at the end, without indicating singular or plural, since they come from a proper name.
Some equivalences
1 statS = 1.11265 x 10 -12 siemens
1 abS = 1 x 109 siemens
As mentioned before, having the resistance, the conductance is immediately known when determining the inverse or reciprocal value. In this way an electrical resistance of 100 ohm is equivalent to 0.01 siemens, for example.
Here are two more examples of the use of conductance:
They are different terms, as already indicated. Conductivity is a property of the substance with which the conductor is made, while conductance is proper to the conductor.
Conductivity can be expressed in terms of G as:
σ = G. (L / A)
Here is a table with the conductivities of frequently used conductive materials:
Table 1. Conductivities, resistivities and thermal coefficient of some conductors. Reference temperature: 20 ºC.
| Metal | σ x 106 (Ye) | ρ x 10-8 (Ω.m) | α ºC-1 |
|---|---|---|---|
| Silver | 62.9 | 1.59 | 0.0058 |
| Copper | 56.5 | 1.77 | 0.0038 |
| Gold | 41.0 | 2.44 | 0.0034 |
| Aluminum | 35.4 | 2.82 | 0.0039 |
| Tungsten | 18.0 | 5.60 | 0.0045 |
| Iron | 10.0 | 10.0 | 0.0050 |
When you have circuits with resistors in parallel, it is sometimes necessary to obtain the equivalent resistance. Knowing the value of the equivalent resistance allows substituting a single value for the set of resistors.
For this resistor configuration, the equivalent resistance is given by:
Geq = G1 + Gtwo + G3 +… Gn
That is, the equivalent conductance is the sum of the conductances. If you want to know the equivalent resistance, simply invert the result.
a) Write Ohm's law in terms of conductance.
b) Find the conductance of a tungsten wire 5.4 cm long and 0.15 mm in diameter.
c) Now a current of 1.5 A is passed through the wire. What is the potential difference between the ends of this conductor?
From the preceding sections you have to:
V = I / G
G = σ.A / L
Substituting the latter in the first, it looks like this:
V = I /(σ.A/L) = I.L / σ.A
Where:
-I is the intensity of the current.
-L is the length of the conductor.
-σ is the conductivity.
-A is the cross-sectional area.
To calculate the conductance of this tungsten wire, its conductivity is required, which is found in Table 1:
σ = 18 x106 Ye
L = 5.4 cm = 5.4 x 10-two m
D = 0. 15 mm = 0.15 x 10-3 m
A = π.Dtwo / 4 = π. (0.15 x 10-3 m)two / 4 = 1.77 x 10-8 mtwo
Substituting in the equation we have:
G = σ.A / L = 18 x106 Ye . 1.77 x 10-8 mtwo / 0.15 x 10-3 m = 2120.6 S.
V = I / G = 1.5 A / 2120.6 S = 0.71 mV.
Find the equivalent resistance in the following circuit and knowing that ior = 2 A, calculate ix and the power dissipated by the circuit:
Resistors are listed: R1= 2 Ω; Rtwo= 4 Ω; R3= 8 Ω; R4= 16 Ω
Then the conductance is calculated in each case: G1 = 0.5 Ʊ; Gtwo = 0.25 Ʊ; G3 = 0.125 Ʊ; G4 = 0.0625 Ʊ
And finally they are added as indicated before, to find the equivalent conductance:
Geq = G1 + Gtwo + G3 +… Gn = 0.5 Ʊ + 0.25 Ʊ + 0.125 Ʊ + 0.0625 Ʊ = 0.9375 Ʊ
Therefore Req = 1.07 Ω.
The voltage across R4 is V4 = ior. R4 = 2 A. 16 Ω = 32 V, and it is the same for all resistors, since they are connected in parallel. Then it is possible to find the currents that flow through each resistance:
-i1 = V1 / R1 = 32 V / 2 Ω = 16 A
-itwo = Vtwo / Rtwo = 32 V / 4 Ω = 8 A
-i3 = V3 / R3 = 32 V / 8 Ω = 4 A
-ix = i1 + itwo + i3 + ior = 16 + 8 + 4 + 2 A = 30 A
Finally, the dissipated power P is:
P = (ix)two. Req = 30 A x 1.07 Ω = 32.1 W
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