CONDUCTANCE: FORMULAS, CALCULATION, EXAMPLES, EXERCISES - PHYSICAL - 2025

The conductance of a conductor is defined as how easy it is to let an electric current pass. 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 unit for conductance is the inverse of the ohm, denoted Ω ^-1 and is called siemens (S).

Figure 1. The conductance depends on the material and the geometry of the conductor. Source: Pixabay.

Other terms used in electricity that sound similar to conductance and are related are conductivity and conduction, but they should not be confused. The first of these terms is an intrinsic property of the substance from which the conductor is made, and the second describes the flow of electrical charge through it.

For an electric conductor with constant cross section of area A, length L and conductivity σ, the conductance is given by:

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.

How is conductance calculated?

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, you have to use integral calculus to find both resistance and conductance.

Since it is the inverse of the resistance, the conductance G can be calculated knowing that:

In fact, the electrical resistance of a conductor can be measured directly with a multimeter, a device that also measures current and voltage.

Units of conductance

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:

Where V is the voltage or potential difference between the ends of the conductor and I is the current intensity. In terms of these magnitudes, the formula looks like this:

Formerly the unit for conductance was the mho (ohm written backwards) denoted as Ʊ, which is an inverted capital omega. This notation fell into disuse and was replaced by Siemens in honor of the German engineer and inventor Ernst Von Siemens (1816-1892), pioneer of telecommunications, but both are totally equivalent.

Figure 2. Conductance versus resistance. Source: Wikimedia Commons. Think tank

In other measurement systems the statsiemens (statS) (in the cgs or centimeter-gram-second system) and the absiemens (abS) (electromagnetic cgs system) are used with the “s” at the end, without indicating singular or plural, and that come from a proper name.

Some equivalences

1 statS = 1.11265 x 10 ^-12 siemens

1 abS = 1 x 10 ⁹ siemens

Examples

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:

Conductivity and conductance

They are different terms, as already indicated. Conductivity is a property of the substance that the conductor is made of, 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 10 6 (S / m)	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.

Figure 3. Association of resistors in parallel. Source: Wikimedia Commons. No machine-readable author provided. Soteke assumed (based on copyright claims)..

For this resistor configuration, the equivalent resistance is given by:

G _eq = G ₁ + G ₂ + G ₃ +… G _n

That is, the equivalent conductance is the sum of the conductances. If you want to know the equivalent resistance, simply reverse the result.

Exercises

- Exercise 1

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?

Solution to

From the preceding sections you have to:

V = I / G

Substituting the latter in the first, it looks like this:

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.

Solution b

To calculate the conductance of this tungsten wire, its conductivity is required, which is found in Table 1:

σ = 18 x10 ⁶ S / m

L = 5.4 cm = 5.4 x 10 ^-2 m

D = 0. 15 mm = 0.15 x 10 ^-3 m

A = π.D ² /4 = π. (0.15 x 10 ^-3 m) ² /4 = 1.77 x 10 ^-8 m ²

Substituting in the equation we have:

G = σ. A / L = 18 x10 ⁶ S / m. 1.77 x 10 ^-8 m ² / 0.15 x 10 ^-3 m = 2120.6 S.

Solution c

V = I / G = 1.5 A / 2120.6 S = 0.71 mV.

- Exercise 2

Find the equivalent resistance in the following circuit and knowing that i _o = 2 A, calculate i _x and the power dissipated by the circuit:

Figure 4. Circuit with resistors in parallel. Source: Alexander, C. 2006. Fundamentals of electrical circuits. 3rd. Edition. McGraw Hill.

Solution

The resistances are listed: R ₁ = 2 Ω; R ₂ = 4 Ω; R ₃ = 8 Ω; R ₄ = 16 Ω

Then the conductance is calculated in each case: G ₁ = 0.5 Ʊ; G ₂ = 0.25 Ʊ; G ₃ = 0.125 Ʊ; G ₄ = 0.0625 Ʊ

And finally they are added as indicated before, to find the equivalent conductance:

G _eq = G ₁ + G ₂ + G ₃ +… G _n = 0.5 Ʊ + 0.25 Ʊ + 0.125 Ʊ + 0.0625 Ʊ = 0.9375 Ʊ

Therefore R _eq = 1.07 Ω.

The voltage across R ₄ is V ₄ = i _o. R ₄ = 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 resistor:

-i ₁ = V ₁ / R ₁ = 32 V / 2 Ω = 16 A

-i ₂ = V ₂ / R ₂ = 32 V / 4 Ω = 8 A

-i ₃ = V ₃ / R ₃ = 32 V / 8 Ω = 4 A

-i _x = i ₁ + i ₂ + i ₃ + i _o = 16 + 8 + 4 + 2 A = 30 A

Finally, the dissipated power P is:

P = (i _x) ². R _eq = 30 A x 1.07 Ω = 32.1 W

References

1. Alexander, C. 2006. Fundamentals of electrical circuits. 3rd. Edition. McGraw Hill.
2. Conversion megaampere / millivolt to absiemens Calculator. Recovered from: pinkbird.org.
3. García, L. 2014. Electromagnetism. 2nd. Edition. Industrial University of Santander. Colombia.
4. Knight, R. 2017. Physics for Scientists and Engineering: a Strategy Approach. Pearson.
5. Roller, D. 1990. Physics. Electricity, Magnetism and Optics. Volume II. Editorial Reverté.
6. Wikipedia. Electrical conductance. Recovered from: es.wikipedia.org.
7. Wikipedia. Siemens. Recovered from: es.wikipedia.org.