Resistivity of a material within a resistor

Example 1

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor

A conductor of resistivity 1.00e-6 Ω·m is 2.0 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 = 3.0), plate area 0.0190 m² and spacing 0.00080 m is charged to 167 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.0 m, A = 1.4e-5 m^{2}$
$\varepsilon_r = 3.0, A_p = 0.0190 m^{2}, d = 0.00080 m$
$V = 167 V$

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
$R = 1.00e-6(2.0)/1.4e-5 = 0.1429 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.0)(0.0190)/0.00080 = 6.308e-10 F$
Formula
$Q = C V$
Substituting
$Q = 6.308e-10(167) = 1.054e-7 C$

Answer:

R = 0.143 \Omega, C = 6.31e-10 F, Q = 1.05e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.10e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 2

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (2)

A conductor of resistivity 0.00e+0 Ω·m is 2.0 m long with a cross-section of 1.5e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.0), plate area 0.0010 m² and spacing 0.00040 m is charged to 155 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 2.0 m, A = 1.5e-5 m^{2}$
$\varepsilon_r = 2.0, A_p = 0.0010 m^{2}, d = 0.00040 m$
$V = 155 V$

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
$R = 0.00e+0(2.0)/1.5e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.0)(0.0010)/0.00040 = 4.427e-11 F$
Formula
$Q = C V$
Substituting
$Q = 4.427e-11(155) = 6.862e-9 C$

Answer:

R = 0.000 \Omega, C = 4.43e-11 F, Q = 6.86e-9 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.21e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 3

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (3)

A conductor of resistivity 1.00e-6 Ω·m is 1.5 m long with a cross-section of 1.6e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.0), plate area 0.0120 m² and spacing 0.00080 m is charged to 130 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 1.5 m, A = 1.6e-5 m^{2}$
$\varepsilon_r = 4.0, A_p = 0.0120 m^{2}, d = 0.00080 m$
$V = 130 V$

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
$R = 1.00e-6(1.5)/1.6e-5 = 0.0938 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.0)(0.0120)/0.00080 = 5.312e-10 F$
Formula
$Q = C V$
Substituting
$Q = 5.312e-10(130) = 6.906e-8 C$

Answer:

R = 0.094 \Omega, C = 5.31e-10 F, Q = 6.91e-8 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 → Resistivity of a material within a resistor

Example 4

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (4)

A conductor of resistivity 0.00e+0 Ω·m is 2.5 m long with a cross-section of 1.7e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 6.5), plate area 0.0100 m² and spacing 0.00030 m is charged to 51 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 2.5 m, A = 1.7e-5 m^{2}$
$\varepsilon_r = 6.5, A_p = 0.0100 m^{2}, d = 0.00030 m$
$V = 51 V$

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
$R = 0.00e+0(2.5)/1.7e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.5)(0.0100)/0.00030 = 1.918e-9 F$
Formula
$Q = C V$
Substituting
$Q = 1.918e-9(51) = 9.784e-8 C$

Answer:

R = 0.000 \Omega, C = 1.92e-9 F, Q = 9.78e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 2.95e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 5

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (5)

A conductor of resistivity 1.00e-6 Ω·m is 4.5 m long with a cross-section of 1.3e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.0), plate area 0.0130 m² and spacing 0.00010 m is charged to 65 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 4.5 m, A = 1.3e-5 m^{2}$
$\varepsilon_r = 3.0, A_p = 0.0130 m^{2}, d = 0.00010 m$
$V = 65 V$

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
$R = 1.00e-6(4.5)/1.3e-5 = 0.3462 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.0)(0.0130)/0.00010 = 3.453e-9 F$
Formula
$Q = C V$
Substituting
$Q = 3.453e-9(65) = 2.244e-7 C$

Answer:

R = 0.346 \Omega, C = 3.45e-9 F, Q = 2.24e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.15e-9 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 6

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (6)

A conductor of resistivity 1.00e-6 Ω·m is 4.0 m long with a cross-section of 1.3e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 2.0), plate area 0.0170 m² and spacing 0.00040 m is charged to 130 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 4.0 m, A = 1.3e-5 m^{2}$
$\varepsilon_r = 2.0, A_p = 0.0170 m^{2}, d = 0.00040 m$
$V = 130 V$

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
$R = 1.00e-6(4.0)/1.3e-5 = 0.3077 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(2.0)(0.0170)/0.00040 = 7.526e-10 F$
Formula
$Q = C V$
Substituting
$Q = 7.526e-10(130) = 9.784e-8 C$

Answer:

R = 0.308 \Omega, C = 7.53e-10 F, Q = 9.78e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 3.76e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 7

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (7)

A conductor of resistivity 0.00e+0 Ω·m is 2.0 m long with a cross-section of 1.1e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 4.5), plate area 0.0060 m² and spacing 0.00050 m is charged to 170 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 2.0 m, A = 1.1e-5 m^{2}$
$\varepsilon_r = 4.5, A_p = 0.0060 m^{2}, d = 0.00050 m$
$V = 170 V$

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
$R = 0.00e+0(2.0)/1.1e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(4.5)(0.0060)/0.00050 = 4.781e-10 F$
Formula
$Q = C V$
Substituting
$Q = 4.781e-10(170) = 8.128e-8 C$

Answer:

R = 0.000 \Omega, C = 4.78e-10 F, Q = 8.13e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.06e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 8

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (8)

A conductor of resistivity 1.00e-6 Ω·m is 2.0 m long with a cross-section of 8.0e-6 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 6.0), plate area 0.0070 m² and spacing 0.00070 m is charged to 59 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.0 m, A = 8.0e-6 m^{2}$
$\varepsilon_r = 6.0, A_p = 0.0070 m^{2}, d = 0.00070 m$
$V = 59 V$

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
$R = 1.00e-6(2.0)/8.0e-6 = 0.2500 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(6.0)(0.0070)/0.00070 = 5.312e-10 F$
Formula
$Q = C V$
Substituting
$Q = 5.312e-10(59) = 3.134e-8 C$

Answer:

R = 0.250 \Omega, C = 5.31e-10 F, Q = 3.13e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 8.85e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 9

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (9)

A conductor of resistivity 1.00e-6 Ω·m is 2.0 m long with a cross-section of 1.1e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 7.0), plate area 0.0060 m² and spacing 0.00050 m is charged to 144 V — find its capacitance and stored charge.

Given

ρ = 1.00e-6 Ω·m
$L = 2.0 m, A = 1.1e-5 m^{2}$
$\varepsilon_r = 7.0, A_p = 0.0060 m^{2}, d = 0.00050 m$
$V = 144 V$

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
$R = 1.00e-6(2.0)/1.1e-5 = 0.1818 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(7.0)(0.0060)/0.00050 = 7.437e-10 F$
Formula
$Q = C V$
Substituting
$Q = 7.437e-10(144) = 1.071e-7 C$

Answer:

R = 0.182 \Omega, C = 7.44e-10 F, Q = 1.07e-7 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 1.06e-10 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Example 10

Resistivity of a conductor and capacitance of a parallel plate — Resistivity of a material within a resistor (10)

A conductor of resistivity 0.00e+0 Ω·m is 4.5 m long with a cross-section of 1.2e-5 m². Find its resistance. A parallel plate capacitor with the same material as dielectric (ε_r = 3.5), plate area 0.0040 m² and spacing 0.00050 m is charged to 171 V — find its capacitance and stored charge.

Given

ρ = 0.00e+0 Ω·m
$L = 4.5 m, A = 1.2e-5 m^{2}$
$\varepsilon_r = 3.5, A_p = 0.0040 m^{2}, d = 0.00050 m$
$V = 171 V$

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
$R = 0.00e+0(4.5)/1.2e-5 = 0.0000 \Omega$
Formula
$C = \varepsilon_{0} \varepsilon_r A / d$
Substituting
$C = (8.854\times10^{-12})(3.5)(0.0040)/0.00050 = 2.479e-10 F$
Formula
$Q = C V$
Substituting
$Q = 2.479e-10(171) = 4.239e-8 C$

Answer:

R = 0.000 \Omega, C = 2.48e-10 F, Q = 4.24e-8 C

Why the other options are there

R = 0.000000 Ω (L and A swapped)
C = 7.08e-11 F (relative permittivity omitted)

Reference: FE Reference Handbook — Materials Science → Resistivity of a material within a resistor

Resistivity of a material within a resistor

Core formulas for this FE topic

Worked exam-style examples

Resistivity of a material within a resistor

Core formulas for this FE topic

t=Lt= Lt=L

ρ=resistivityofthematerial\rho = resistivity of the materialρ=resistivityofthematerial

R=resistanceoftheresistorR = resistance of the resistorR=resistanceoftheresistor

A=cross−sectionalareaoftheresistorA = cross-sectional area of the resistorA=cross−sectionalareaoftheresistor

L=lengthoftheresistorL = length of the resistorL=lengthoftheresistor

Worked exam-style examples