How to Calculate and Solve for Capacitance of Parallel Plates | Electrical Properties

The image of capacitance of parallel plates is shown below.

To compute for capacitance of parallel plates, three essential parameters are needed and these parameters are Permittivity of the Material (ε), Cross-sectional (A) and Distance Between the Plates (d).

The formula for calculating capacitance of parallel plates:

C = εrεoA / d

Where:

C = Capacitance of Parallel Plates
εr = Permittivity of the material
εo = vacuum permittivity = 8.8541878176E-12
A = cross-sectional area
d = distance between the plates

Let’s solve an example;
Find the capacitance of parallel plates when the permittivity of the material is 14, the cross-sectional area is 8 and the distance between the plates is 2.

This implies that;

εr = Permittivity of the material = 14
εo = vacuum permittivity = 8.854E-12
A = cross-sectional area = 8
d = distance between the plates = 2

C = εrεoA / d
C = 14 x 8.854-12 x 8 / 2
C = 9.9166e-10 / 2
C = 4.958e-10

Therefore, the capacitance of parallel plates is 4.958e-10 farads.

Continue reading How to Calculate and Solve for Capacitance of Parallel Plates | Electrical Properties

How to Calculate and Solve for Capacitance | Electrical Properties

The image of capacitance is shown below.

To compute for capacitance, two essential parameters are needed and these parameters are Charge (Q) and Voltage (V).

The formula for calculating capacitance:

C = Q / V

Where;

C = Capacitance
Q = Charge
V = Voltage

Let’s solve an example;
Find the capacitance when the charge is 12 and the voltage is 6.

This implies that;

Q = Charge = 12
V = Voltage = 6

C = Q / V
C = 12 / 6
C = 2

Therefore, the capacitance is 2 farad.

Calculating the Charge when the Capacitance and the Voltage is Given.

Q = CV

Where;

Q = Charge
C = Capacitance
V = Voltage

Let’s solve an example;
Find the charge when the capacitance is 10 and the voltage is 5.

This implies that;

C = Capacitance = 10
V = Voltage = 5

Q = CV
Q = 10(5)
Q = 50

Therefore, the charge is 50.

Continue reading How to Calculate and Solve for Capacitance | Electrical Properties

How to Calculate and Solve for Mobility of an Ionic Specie | Electrical Properties

The image of mobility of an ionic specie is shown below.

To compute for mobility of an ionic specie, five essential parameters are needed and these parameters are Valence (nI), Electrical Charge (e), Diffusion Coefficient (DI), Boltzmann’s Constant (K) and Temperature (T).

The formula for calculating the mobility of an ionic specie:

μI = nIeDI/KT

Where:

μI = Mobility of an Ionic Specie
nI = Valence
e = Electrical Charge
DI = Diffusion Coefficient
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the mobility of an ionic specie when the valence is 18, the electrical charge is 9, the diffusion coefficient is 12, the boltzmann’s constant is 6 and the temperature is 3.

This implies that;

nI = Valence = 18
e = Electrical Charge = 9
DI = Diffusion Coefficient = 12
K = Boltzmann’s Constant = 6
T = Temperature = 3

μI = nIeDI/KT
μI = (18)(9)(12)/(6)(3)
μI = 1944/18
μI = 108

Therefore, the mobility of an ionic specie is 108 m²/(V. s).

Continue reading How to Calculate and Solve for Mobility of an Ionic Specie | Electrical Properties

How to Calculate and Solve for Total Conductivity | Electrical Properties

The total conductivity is illustrated by the image below.

To compute for total conductivity, two essential parameters are needed and these parameters are Electronic Conductivity (σelectronic) and Ionic Conductivity (σionic).

The formula for calculating total conductivity:

Where:

σtotal = Total Conductivity
σelectronic = Electronic Conductivity
σionic = Ionic Conductivity

Let’s solve an example;
Find the total conductivity when the electronic conductivity is 32 and the ionic conductivity is 8.

This implies that;

σelectronic = Electronic Conductivity = 32
σionic = Ionic Conductivity = 8

σtotal = σelectronic + σionic
σtotal = 32 + 8
σtotal = 40

Therefore, the total conductivity is 40 S/m.

Calculating the Electronic Conductivity when the Total Conductivity and the Ionic Conductivity is Given.

σelectronic = σtotal – σionic

Where:

σelectronic = Electronic Conductivity
σtotal = Total Conductivity
σionic = Ionic Conductivity

Given an example;
Find the electronic conductivity when the total conductivity is 18 and the ionic conductivity is 9.

This implies that;

σtotal = Total Conductivity = 18
σionic = Ionic Conductivity = 9

σelectronic = σtotal – σionic
σelectronic = 18 – 9
σelectronic = 9

Therefore, the electronic conductivity is 9.

Continue reading How to Calculate and Solve for Total Conductivity | Electrical Properties

How to Calculate and Solve for Electron Mobility of Metal | Electrical Properties

The image of electron mobility of metal is shown below.

To compute for electron mobility of metal, two essential parameters are needed and these parameters are Hall Coefficient (RH) and Electrical Conductivity (σ).

The formula for calculating electron mobiltiy of metal:

μe = |RH

Where:

μe = Electron Mobility of Metals
RH = Hall Coefficient
σ = Electrical Conductivity

Let’s solve an example;
Find the electron mobility of metal when the hall coefficient is 28 and the electrical conductivity is 7.

This implies that;

RH = Hall Coefficient = 28
σ = Electrical Conductivity = 7

μe = |RH
μe = |28|(7)
μe = (28)(7)
μe = 196

Therefore, the electron mobility of metals is 196 m²/(V.s).

Calculating the Hall Coefficient when the Electron Mobility of Metals and the Electrical Conductivity is Given.

RH = μe / σ

Where:

RH = Hall Coefficient
μe = Electron Mobility of Metals
σ = Electrical Conductivity

Let’s solve an example;
Find the hall coefficient when the electron mobility of metals is 24 and the electrical conductivity is 12.

This implies that;

μe = Electron Mobility of Metals = 24
σ = Electrical Conductivity = 12

RH = μe / σ
RH = 24 / 12
RH = 2

Therefore, the hall coefficient is 2.

Continue reading How to Calculate and Solve for Electron Mobility of Metal | Electrical Properties

How to Calculate and Solve for Hall Voltage | Electrical Properties

The image of hall voltage is shown below.

To compute for hall voltage, four essential parameters are needed and these parameters are Hall Coefficient (RH), Current (Ix), Magnetic Field (Bz) and Thickness (d).

The formula for calculating hall voltage:

VH = RHIxBz/d

Where:

VH = Hall Voltage
RH = Hall Coefficient
Ix = Current
Bz = Magnetic Field
d = Thickness

Let’s solve an example;
Find the hall voltage when the hall coefficient is 21, the current is 7, the magnetic field is 4 and the thickness is 2.

This implies that;

RH = Hall Coefficient = 21
Ix = Current = 7
Bz = Magnetic Field = 4
d = Thickness = 2

VH = RHIxBz/d
VH = (21)(7)(4)/2
VH = (588)/2
VH = 294

Therefore, the hall voltage is 294 V.

Calculating the Hall Coefficient when the Hall Voltage, the Current, the Magnetic Field and the Thickness is Given.

RH = VH x d / Ix x Bz

Where:

RH = Hall Coefficient
VH = Hall Voltage
Ix = Current
Bz = Magnetic Field
d = Thickness

Let’s solve an example;
Find the hall coefficient when the hall voltage is 10, the current is 8, the magnetic field is 4 and the thickness is 2.

This implies that;

VH = Hall Voltage = 10
Ix = Current = 8
Bz = Magnetic Field = 4
d = Thickness = 2

RH = VH x d / Ix x Bz
RH = 10 x 2 / 8 x 4
RH = 20 / 32
RH = 0.625

Therefore, the hall coefficient is 0.625.

Continue reading How to Calculate and Solve for Hall Voltage | Electrical Properties

How to Calculate and Solve for Electrical Conductivity for Extrinsic p-type Semiconductor | Electrical Properties

The electrical conductivity for extrinsic p-type semiconductor is illustrated by the image below.

To compute for electrical conductivity for extrinsic p-type semiconductor, three essential parameters are needed and these parameters are Number of Holes (p), Hole Mobility (μh) and Electrical Charge (e).

The formula for calculating electrical conductivity for extrinsic p-type semiconductor:

σ ≅ p|e|μh

Where:

σ = Electrical Conductivity for Extrinsic p-type Semiconductor
p = Number of Holes
μh = Hole Mobility
e = Electrical Charge

Let’s solve an example;
Find the electrical conductivity for extrinsic p-type semiconductor when the number of holes is 24, the hole mobility is 12 and the electrical charge is 4.

This implies that;

p = Number of Holes = 24
μh = Hole Mobility = 12
e = Electrical Charge = 4

σ ≅ p|e|μh
σ ≅ (24)|4|(12)
σ ≅ (24)(4)(12)
σ ≅ 1152

Therefore, the electrical conductivity for extrinsic p-type semiconductor is 1152 S/m.

Calculating the Number of Holes when the Electrical Conductivity for extrinsic p-type semiconductor, the Hole Mobility and the Electrical Charge is Given.

p = σ / μh(e)

Where:

p = Number of Holes
σ = Electrical Conductivity for Extrinsic p-type Semiconductor
μh = Hole Mobility
e = Electrical Charge

Let’s solve an example;
Find the number of holes when the electrical conductivity for extrinsic p-type semiconductor is 42, the hole mobility is 3 and the electrical charge is 1.

This implies that;

σ = Electrical Conductivity for Extrinsic p-type Semiconductor = 42
μh = Hole Mobility = 3
e = Electrical Charge = 1

p = σ / μh(e)
p = 42 / 3(1)
p = 42 / 3
p = 14

Therefore, the number of holes is 14.

Continue reading How to Calculate and Solve for Electrical Conductivity for Extrinsic p-type Semiconductor | Electrical Properties

How to Calculate and Solve for Electrical Conductivity for Extrinsic n-type Semiconductor | Electrical Properties

The image of electrical conductivity for extrinsic n-type semiconductor is shown below.

To compute for electrical conductivity for extrinsic n-type semiconductor, three essential parameters are needed and these parameters are Number of Free Conducting Electrons (n), Electron Mobility (μe) and Electrical Charge (e).

The formula for calculating the electrical conductivity for extrinsic n-type semiconductor:

σ ≅ n|e|μe

Where:

σ = Electrical Conductivity for Extrinsic n-type Semiconductor
n = Number of Free Conducting Electrons
μe = Electron Mobility
e = Electrical Charge

Let’s solve an example;
Find the electrical conductivity for extrinsic n-type semiconductor when the number of free conducting electrons is 12, the electron mobility is 6 and the electrical charge is 4.

This implies that;

n = Number of Free Conducting Electrons = 12
μe = Electron Mobility = 6
e = Electrical Charge = 4

σ ≅ n|e|μe
σ ≅ (12)|4|(6)
σ ≅ (12)(4)(6)
σ ≅ 288

Therefore, the electrical conductivity for extrinsic n-type semiconductor is 288 S/m.

Calculating the Number of Free Conducting Electrons when the Electrical Conductivity for Extrinsic n-type Semiconductor, the Electron Mobility and the Electrical Charge is Given.

n = σ / μe(e)

Where:

n = Number of Free Conducting Electrons
σ = Electrical Conductivity for Extrinsic n-type Semiconductor
μe = Electron Mobility
e = Electrical Charge

Let’s solve an example;
Given that, the electrical conductivity for extrinsic n-type semiconductor is 40, the electron mobility is 4 and the electrical charge is 2. Find the number of free conducting electrons?

This implies that;

σ = Electrical Conductivity for Extrinsic n-type Semiconductor = 40
μe = Electron Mobility = 4
e = Electrical Charge = 2

n = σ / μe(e)
n = 40 / 4(2)
n = 40 / 8
n = 5

Therefore, the number of free conducting electrons is 5.

Continue reading How to Calculate and Solve for Electrical Conductivity for Extrinsic n-type Semiconductor | Electrical Properties

How to Calculate and Solve for Electrical Conductivity for Intrinsic Semiconductor | Electrical Properties

The image of electrical conductivity for intrinsic semiconductor is shown below.

To compute for electrical conductivity for intrinsic semiconductor, five essential parameters are needed and these parameters are Number of Holes (p), Number of Free Conducting Electrons (n), Hole Mobility (μh), Electron Mobility (μe) and Electrical Charge (e).

The formula for calculating electrical conductivity for intrinsic semiconductor:

σ = n|e|μe + p|e|μh

Where:

σ = Electrical Conductivity for Intrinsic Semiconductor
p = Number of Holes
n = Number of Free Conducting Electrons
μh = Hole Mobility
μe = Electron Mobility
e = Electrical Charge

Let’s solve an example;
Find the electrical conductivity for intrinsic semiconductor when the number of holes is 14, the number of free conducting electrons is 12, the hole mobility is 8, the electron mobility is 4 and the electrical charge is 2.

This implies that;

p = Number of Holes = 14
n = Number of Free Conducting Electrons = 12
μh = Hole Mobility = 8
μe = Electron Mobility = 4
e = Electrical Charge = 2

σ = n|e|μe + p|e|μh
σ = (12)|2|(4) + (14)|2|(8)
σ = (12)(2)(4) + (14)(2)(8)
σ = 96 + 224
σ = 320

Therefore, the electrical conductivity for intrinsic semiconductor is 320 S/m.

Continue reading How to Calculate and Solve for Electrical Conductivity for Intrinsic Semiconductor | Electrical Properties

How to Calculate and Solve for Electron Mobility | Electrical Properties

The electron mobility is illustrated by the image below.

To compute for electron mobility, three essential parameters are needed and these parameters are Electrical Conductivity (σ), Number of Free Conductivity Electrons (n) and Electrical Charge (e).

The formula for calculating electron mobility:

μe = σ/n|e|

Where:

σ = Electron Mobility
σ = Electrical Conductivity
n = Number of Free Conductivity Electrons
e = Electrical Charge

Let’s solve an example;
Find the electron mobility when the electrical conductivity is 28, the number of free conductivity electrons is 7 and the electrical charge is 2.

This implies that;

σ = Electrical Conductivity = 28
n = Number of Free Conductivity Electrons = 7
e = Electrical Charge = 2

μe = σ/n|e|
μe = 28/7|2|
μe = 28/7(2)
μe = 28/14
μe = 2

Therefore, the electron mobility is 2 m²/(V.s).

Continue reading How to Calculate and Solve for Electron Mobility | Electrical Properties