Conductivity is the reciprocal of the resistivity

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Conductivity is the reciprocal of the resistivity within Materials Science. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what conductivity is the reciprocal of the resistivity describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: percent versus fraction in composition and strain.

Lecture

Why this section exists. Conductivity is the reciprocal of the resistivity is the part of Materials Science that lets you connect a steel, concrete or polymer specimen under test to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a definition, a phase-diagram read, or a one-line property calculation. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. percent versus fraction in composition and strain. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: conductivity is the reciprocal of the resistivity.
Wikimedia Commons, public domain

Materials Science — Conductivity is the reciprocal of the resistivity: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a steel, concrete or polymer specimen under test. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Photoelectric effect-electrons are emitted from matter (metals and nonmetallic solids, liquids or gases) as a consequence of
their absorption of energy from electromagnetic radiation of very short wavelength and high frequency.
Piezoelectric effect-the electromechanical and the electrical state in crystalline materials.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Steady heat flow through a wall — Conductivity is the reciprocal of the resistivity

A 0.8 ft thick concrete wall of area 130 ft² has k = 0.6 Btu/(hr·ft·°F) and a 22°F temperature difference. What is the heat flow rate?

Given

k = 0.6 Btu/hr·ft·°F
A = 130 ft²
ΔT = 22°F
t = 0.8 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.6 (130) (22) /0.8$
Evaluate
$q = 2, 145 B t u / h r$

Answer: q ≈ 2,145 Btu/hr

Why the other options are there

1,373 Btu/hr (thickness multiplied)
13.2 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 2

Resistivity of a conductor and capacitance of a parallel plate — Conductivity is the reciprocal of the resistivity

A conductor of resistivity 0.00e+0 Ω·m is 1.5 m long with a cross-section of 1.8e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.0), plate area 0.0180 m² and spacing 0.00090 m is charged to 146 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 1.5 m, A = 1.8e-5 m²
ε_r = 2.0, A_p = 0.0180 m², d = 0.00090 m
V = 146 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (1.5) /1.8 e - 5 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (2.0) (0.0180) /0.00090 = 3.542 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 3.542 e - 10 (146) = 5.171 e - 8 C$

Answer: R = 0.000 Ω, C = 3.54e-10 F, Q = 5.17e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.77e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 3

Steady heat flow through a wall — Conductivity is the reciprocal of the resistivity (2)

A 0.8 ft thick concrete wall of area 177 ft² has k = 0.6 Btu/(hr·ft·°F) and a 54°F temperature difference. What is the heat flow rate?

Given

k = 0.6 Btu/hr·ft·°F
A = 177 ft²
ΔT = 54°F
t = 0.8 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.6 (177) (54) /0.8$
Evaluate
$q = 7, 169 B t u / h r$

Answer: q ≈ 7,169 Btu/hr

Why the other options are there

4,588 Btu/hr (thickness multiplied)
32.4 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 4

Resistivity of a conductor and capacitance of a parallel plate — Conductivity is the reciprocal of the resistivity (2)

A conductor of resistivity 1.00e-6 Ω·m is 2.5 m long with a cross-section of 1.4e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.0), plate area 0.0170 m² and spacing 0.00030 m is charged to 30 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 2.5 m, A = 1.4e-5 m²
ε_r = 4.0, A_p = 0.0170 m², d = 0.00030 m
V = 30 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (2.5) /1.4 e - 5 = 0.1786Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (4.0) (0.0170) /0.00030 = 2.007 e - 9 F$
Formula
$Q = C V$
Substituting
$Q = 2.007 e - 9 (30) = 6.021 e - 8 C$

Answer: R = 0.179 Ω, C = 2.01e-9 F, Q = 6.02e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 5.02e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 5

Steady heat flow through a wall — Conductivity is the reciprocal of the resistivity (3)

A 1.0 ft thick concrete wall of area 365 ft² has k = 2.0 Btu/(hr·ft·°F) and a 70°F temperature difference. What is the heat flow rate?

Given

k = 2.0 Btu/hr·ft·°F
A = 365 ft²
ΔT = 70°F
t = 1.0 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 2.0 (365) (70) /1.0$
Evaluate
$q = 51, 100 B t u / h r$

Answer: q ≈ 51,100 Btu/hr

Why the other options are there

51,100 Btu/hr (thickness multiplied)
140.0 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 6

Resistivity of a conductor and capacitance of a parallel plate — Conductivity is the reciprocal of the resistivity (3)

A conductor of resistivity 0.00e+0 Ω·m is 2.0 m long with a cross-section of 1.0e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 7.0), plate area 0.0090 m² and spacing 0.00040 m is charged to 85 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 2.0 m, A = 1.0e-5 m²
ε_r = 7.0, A_p = 0.0090 m², d = 0.00040 m
V = 85 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (2.0) /1.0 e - 5 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (7.0) (0.0090) /0.00040 = 1.395 e - 9 F$
Formula
$Q = C V$
Substituting
$Q = 1.395 e - 9 (85) = 1.185 e - 7 C$

Answer: R = 0.000 Ω, C = 1.39e-9 F, Q = 1.19e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.99e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 7

Steady heat flow through a wall — Conductivity is the reciprocal of the resistivity (4)

A 1.3 ft thick concrete wall of area 260 ft² has k = 1.9 Btu/(hr·ft·°F) and a 34°F temperature difference. What is the heat flow rate?

Given

k = 1.9 Btu/hr·ft·°F
A = 260 ft²
ΔT = 34°F
t = 1.3 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.9 (260) (34) /1.3$
Evaluate
$q = 12, 920 B t u / h r$

Answer: q ≈ 12,920 Btu/hr

Why the other options are there

21,835 Btu/hr (thickness multiplied)
64.6 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 8

Resistivity of a conductor and capacitance of a parallel plate — Conductivity is the reciprocal of the resistivity (4)

A conductor of resistivity 0.00e+0 Ω·m is 1.0 m long with a cross-section of 4.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 7.0), plate area 0.0030 m² and spacing 0.00020 m is charged to 162 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
L = 1.0 m, A = 4.0e-6 m²
ε_r = 7.0, A_p = 0.0030 m², d = 0.00020 m
V = 162 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 0.00 e + 0 (1.0) /4.0 e - 6 = 0.0000Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (7.0) (0.0030) /0.00020 = 9.297 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 9.297 e - 10 (162) = 1.506 e - 7 C$

Answer: R = 0.000 Ω, C = 9.30e-10 F, Q = 1.51e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.33e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 9

Steady heat flow through a wall — Conductivity is the reciprocal of the resistivity (5)

A 1.5 ft thick concrete wall of area 244 ft² has k = 0.9 Btu/(hr·ft·°F) and a 48°F temperature difference. What is the heat flow rate?

Given

k = 0.9 Btu/hr·ft·°F
A = 244 ft²
ΔT = 48°F
t = 1.5 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.9 (244) (48) /1.5$
Evaluate
$q = 7, 027 B t u / h r$

Answer: q ≈ 7,027 Btu/hr

Why the other options are there

15,811 Btu/hr (thickness multiplied)
43.2 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Example 10

Resistivity of a conductor and capacitance of a parallel plate — Conductivity is the reciprocal of the resistivity (5)

A conductor of resistivity 1.00e-6 Ω·m is 0.5 m long with a cross-section of 2.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.5), plate area 0.0090 m² and spacing 0.00070 m is charged to 43 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
L = 0.5 m, A = 2.0e-6 m²
ε_r = 2.5, A_p = 0.0090 m², d = 0.00070 m
V = 43 V

Find

Resistance R, capacitance C and charge Q

Start with the thinking

Resistivity is a material property; resistance also depends on geometry.
Charge held by a capacitor is simply Q = C V once the capacitance is known.

Step-by-step solution

Formula
$R = ρ L / A$
Substituting
$R = 1.00 e - 6 (0.5) /2.0 e - 6 = 0.2500Ω$
Formula
$C = ε_{0} ε_{r} A / d$
Substituting
$C = (8.854 \times 1 0^{- 12}) (2.5) (0.0090) /0.00070 = 2.846 e - 10 F$
Formula
$Q = C V$
Substituting
$Q = 2.846 e - 10 (43) = 1.224 e - 8 C$

Answer: R = 0.250 Ω, C = 2.85e-10 F, Q = 1.22e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.14e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Conductivity is the reciprocal of the resistivity

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a steel, concrete or polymer specimen under test, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Conductivity is the reciprocal of the resistivity contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a definition, a phase-diagram read, or a one-line property calculation.
Unit rule: percent versus fraction in composition and strain.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

percent versus fraction in composition and strain
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.