Conductivity is the reciprocal of the resistivity

Materials Science · FE Reference Handbook section

Materials Science

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10 exam-style examples

~45 min

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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
$A = 130 ft^{2}$
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
$q = 2,145 Btu/hr$

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
$L = 1.5 m, A = 1.8e-5 m^{2}$
$\varepsilon_r = 2.0, A_p = 0.0180 m^{2}, d = 0.00090 m$
$V = 146 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.5)/1.8e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.0)(0.0180)/0.00090 = 3.542e-10 F$
Formula
$Q = C V$
Substituting
$Q = 3.542e-10(146) = 5.171e-8 C$

Answer:

R = 0.000 \Omega, 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
$A = 177 ft^{2}$
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
$q = 7,169 Btu/hr$

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
$L = 2.5 m, A = 1.4e-5 m^{2}$
$\varepsilon_r = 4.0, A_p = 0.0170 m^{2}, d = 0.00030 m$
$V = 30 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(2.5)/1.4e-5 = 0.1786 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.0)(0.0170)/0.00030 = 2.007e-9 F$
Formula
$Q = C V$
Substituting
$Q = 2.007e-9(30) = 6.021e-8 C$

Answer:

R = 0.179 \Omega, 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
$A = 365 ft^{2}$
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
$q = 51,100 Btu/hr$

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
$L = 2.0 m, A = 1.0e-5 m^{2}$
$\varepsilon_r = 7.0, A_p = 0.0090 m^{2}, d = 0.00040 m$
$V = 85 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(2.0)/1.0e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.0)(0.0090)/0.00040 = 1.395e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.395e-9(85) = 1.185e-7 C$

Answer:

R = 0.000 \Omega, 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
$A = 260 ft^{2}$
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
$q = 12,920 Btu/hr$

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
$L = 1.0 m, A = 4.0e-6 m^{2}$
$\varepsilon_r = 7.0, A_p = 0.0030 m^{2}, d = 0.00020 m$
$V = 162 V$

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
$R = \rho L / A$
Substituting
$R = 0.00e+0(1.0)/4.0e-6 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.0)(0.0030)/0.00020 = 9.297e-10 F$
Formula
$Q = C V$
Substituting
$Q = 9.297e-10(162) = 1.506e-7 C$

Answer:

R = 0.000 \Omega, 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
$A = 244 ft^{2}$
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
$q = 7,027 Btu/hr$

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
$L = 0.5 m, A = 2.0e-6 m^{2}$
$\varepsilon_r = 2.5, A_p = 0.0090 m^{2}, d = 0.00070 m$
$V = 43 V$

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
$R = \rho L / A$
Substituting
$R = 1.00e-6(0.5)/2.0e-6 = 0.2500 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.5)(0.0090)/0.00070 = 2.846e-10 F$
Formula
$Q = C V$
Substituting
$Q = 2.846e-10(43) = 1.224e-8 C$

Answer:

R = 0.250 \Omega, 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