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• It states that the current flowing in a circuit
is directly proportional to the applied
voltage and inversely proportional to the
resistance, provided the temperature is
remains constant
V = I R
Where:
V = the applied voltage (V)
I = current flowing in a circuit (A)
R = resistance (Ω)
Formula:
In series circuit:
V = V1 + V2 + V3 + . . . + Vn
I = I1 = I2 = I3 = . . . = In
R = R1 + R2 + R3 + . . . + Rn
In parallel circuit:
V = V1 = V2 = V3 = . . . = Vn
I = I1 + I2 + I3 + . . . + In
1/R = 1/R1 + 1/R2 + 1/R3 + . . . + 1/Rn
Example
• Determine the voltage which must be
applied to a 2kΩ resistor in order to that a
current of 10mA may flow.
Given:
R = 2kΩ = 2000Ω
I = 10mA = 10 x 10-3
A
V = ?
Solution:
V = IR
V = (10 x 10-3
A)(2000Ω)
V = 20 V
R = ρL / A
where:
ρ = resistivity (Ω*m)
R = resistance (Ω)
A = cross-sectional area (m2
)
L = length of the material
Resistance and Resistivity
Resistivity Table
Example
A copper wire of unknown length has a
diameter of 0.635 cm and a resistance of
0.28 Ω. By several successive passes through a
drawing dies the diameter of the wire is reduced
to 0.127 cm. Assuming that the resistivity of the
copper remains unchanged in the drawing
process. Calculate the resistance of the reduced
size wire.
Solution:
First, find the unknown length:
ρ = 1.72x10-8
Ωm
L= R A / ρ
= (0.28Ω)((π(6.35x10-3
m)2
)/4)
1.72x10-8
Ωm
= 515.55 m
R = ρL / A
(1.72x10-8
Ωm)(515.55m)
= π(1.27x10-3
m)2
4
= 7 Ω
Electrical Conductivity
• Electrical conductivity is the inverse of the
electrical resistivity.
• It is represented by the Greek letter σ
where :
σ = 1 / ρ
Conductance
• Reciprocal of resistance (G)
• Measured in siemens (S)
G = 1 / R
Ex: Find the conductance of a conductor of
resistance 10 Ω.
G = 1 / R = 1 / 10 Ω
= 0.1 S
In addition to Ohm’s Law
J = σ E
where:
J = current density (I/A)
σ = conductivity
E = voltage difference bet. two points
divided by the L separating them
E = V / L
• If the speed of the charges is small
enough (and normally in matter this is the
case), the magnetic force will be tiny
compared to the electric force
• This is Ohm’s law, which is true in the
special case of materials where the
conducting charge responds linearly to the
force and the charges are not too fast.
Electronic and Ionic Conduction
Electronic conduction
- within most solid materials a current
arises from the flow of electrons.
Ionic Conduction
- for ionic materials a net motion of
charged ions is possible that produces a
current.
• For example, you are familiar with the
table salt (NaCl). Salt consists of positive
sodium ions (Na+
) and negative chloride
ions (Cl-
). On the other hand the element
sodium is a silvery gray metal composed
of neutral atoms which react vigorously
with water or air. Chlorine as an element
is a neutral greenish-yellow, poisonous,
diatomic gas (Cl2)
Elestrical Conduction

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Elestrical Conduction

  • 1.
  • 2. • It states that the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance, provided the temperature is remains constant
  • 3. V = I R Where: V = the applied voltage (V) I = current flowing in a circuit (A) R = resistance (Ω) Formula:
  • 4. In series circuit: V = V1 + V2 + V3 + . . . + Vn I = I1 = I2 = I3 = . . . = In R = R1 + R2 + R3 + . . . + Rn
  • 5.
  • 6. In parallel circuit: V = V1 = V2 = V3 = . . . = Vn I = I1 + I2 + I3 + . . . + In 1/R = 1/R1 + 1/R2 + 1/R3 + . . . + 1/Rn
  • 7.
  • 8. Example • Determine the voltage which must be applied to a 2kΩ resistor in order to that a current of 10mA may flow. Given: R = 2kΩ = 2000Ω I = 10mA = 10 x 10-3 A V = ?
  • 9. Solution: V = IR V = (10 x 10-3 A)(2000Ω) V = 20 V
  • 10. R = ρL / A where: ρ = resistivity (Ω*m) R = resistance (Ω) A = cross-sectional area (m2 ) L = length of the material Resistance and Resistivity
  • 12. Example A copper wire of unknown length has a diameter of 0.635 cm and a resistance of 0.28 Ω. By several successive passes through a drawing dies the diameter of the wire is reduced to 0.127 cm. Assuming that the resistivity of the copper remains unchanged in the drawing process. Calculate the resistance of the reduced size wire.
  • 13. Solution: First, find the unknown length: ρ = 1.72x10-8 Ωm L= R A / ρ = (0.28Ω)((π(6.35x10-3 m)2 )/4) 1.72x10-8 Ωm = 515.55 m
  • 14. R = ρL / A (1.72x10-8 Ωm)(515.55m) = π(1.27x10-3 m)2 4 = 7 Ω
  • 15. Electrical Conductivity • Electrical conductivity is the inverse of the electrical resistivity. • It is represented by the Greek letter σ where : σ = 1 / ρ
  • 16. Conductance • Reciprocal of resistance (G) • Measured in siemens (S) G = 1 / R Ex: Find the conductance of a conductor of resistance 10 Ω. G = 1 / R = 1 / 10 Ω = 0.1 S
  • 17. In addition to Ohm’s Law J = σ E where: J = current density (I/A) σ = conductivity E = voltage difference bet. two points divided by the L separating them E = V / L
  • 18. • If the speed of the charges is small enough (and normally in matter this is the case), the magnetic force will be tiny compared to the electric force • This is Ohm’s law, which is true in the special case of materials where the conducting charge responds linearly to the force and the charges are not too fast.
  • 19. Electronic and Ionic Conduction Electronic conduction - within most solid materials a current arises from the flow of electrons. Ionic Conduction - for ionic materials a net motion of charged ions is possible that produces a current.
  • 20. • For example, you are familiar with the table salt (NaCl). Salt consists of positive sodium ions (Na+ ) and negative chloride ions (Cl- ). On the other hand the element sodium is a silvery gray metal composed of neutral atoms which react vigorously with water or air. Chlorine as an element is a neutral greenish-yellow, poisonous, diatomic gas (Cl2)