The document discusses several factors that affect resistance in conductors: - Resistance increases with length and decreases with cross-sectional area. It is directly proportional to length and inversely proportional to area. - Resistance also depends on the material used and increases with temperature for most metals. Different materials have different resistivities. - Examples are given of how resistance changes with these factors in various applications like variable resistors. Formulas are provided relating resistance, length, area, resistivity, and temperature.

1. Factors Affecting
Resistance
Reported By: Ralph Lery Guerrero
Kevin Roxas
Marco Lauro Delos Santos

2. Resistance
• Is defined as an obstacle to
the flow of electric current.
• Is the opposition offered by
any object to the passage of
an electric current through
it.

3. Length
• Resistance of a conductor is directly
proportional to the length of the wire i.e.,
longer the wire greater will be the
resistance and shorter the wire smaller will
be the resistance. If L represents the length
of the uniform wire, then
RL

4. Length
R1/R2=l1/l2
Where
R1= resistance of the first conductor(in ohms)
R2= resistance of the second conductor(in
ohms)
l1=length of the first conductor (in cm)
l2=length of the second conductor (in cm)

5. Length
• Resistivity- is the resistance per unit
length of a specific substance to
electric force.

6. Length
• The volume control of your radio is a
variable resistor. As you turn the
volume control knob, the effective the
length of the resistance changes.
• Resistance and the amount of current
passing through the circuit change,
making the sound from the speaker
soft and loud.

7. Diameter (Cross-sectional area)
• Resistance of a conductor is inversely
proportional to the area of the cross-
section of the uniform wire. That means,
thinner the wire, greater the resistance
and thicker the wire, lower the
resistance. If A is the area of cross-
section of the uniform wire, then,

8. Diameter (Cross-sectional area)
If the cross-sectional area is
doubled, twice as many electrons
became available to flow so that
the current is doubled.

9. Diameter (Cross-sectional area)
R1/R2=d22/d12
Some connecting wires are made of stranded thin
wires.
Stranding wires has the same effect as increasing
cross-sectional area.
The wire’s gauge number specifies its size. The
smaller gauge number, the thicker the wire.
Hence, the lesser the resistance of the
conductor.

10. Diameter (Cross-sectional area)
Problem
An iron wire conductor with a
diameter of 0.8 mm has a
resistance of 0.4 ohm Ω. Calculate
the resistance of a n iron wire 0.4
mm in diameter.
Find:
R2

11. Diameter (Cross-sectional area)
Given:
R1=0.4Ω
d1=0.8mm
d2=0.4mm
Solution:
R1/R2=d22/d12
0.4Ω/R2=(0.4mm)2/(0.8mm)2
R2=(0.4Ω)(0.64mm)2/0.16mm2
R2=1.60Ω

12. Temperature
• The resistance of a metallic
conductor increases as the
temperature increases e.g.
copper
• The resistance of a
semiconductor/insulator
decreases as the temperature
increases

13. Temperature
Resistance increases with the temperature.

14. Kind of Material (Resistivity)
• The material also affects the
resistance of a conductor by a fixed
amount for different materials. This
is known as resistivity ().
R = L  = constant of proportionality
A

15. Kind of Material (Resistivity)
Resistors- conductors whose
resistance is constant when
the temperature is constant.

16. Kind of Material (Resistivity)
Material Resistivity (W m)
Conductors Silver 1.60 x 10-8
Copper 1.62 x 10-8
Aluminium 2.63 x 10-8
Tungsten 5.20 x 10-8
Nickel 6.84 x 10-8
Iron 10.0 x 10-8
Chromium 12.9 x 10-8
Mercury 94.0 x 10-8
Manganese 1.84 x 10-6
Constantan (alloy of Cu
Alloys 49 x 10-6
and Ni)
Manganin (alloy of Cu, Mn
44 x 10-6
and Ni)
Nichrome (alloy of Ni, Cr,
100 x 10-6
Mn and Fe)
Insulators Glass 1010 - 1014
Hard rubber 1013 - 1016
Ebonite 1015 - 1017
Diamond 1012 - 1013
Paper (dry) 1012

17. Kind of Material (Resistivity)
Problem
Given: Find:
L=15 m R
Diameter=0.085 cm
= 1.6x10-8 Ωm
T=20°C
Solutions:
Conversion: d= 0.085cm/100 cm x
1m
d= (8.5x10-4m)

18. Kind of Material (Resistivity)
a.) r=1/2d
r=8.5x10-4 m/2
r= 4.25x10-4 m
b.) A=πr2
A=(3.14)(4.25x10-4 m)2
A=5.67x10-7

19. Kind of Material (Resistivity)
c.) R = L
A
R=(1.6x10-8Ωm)(15m)
5.67x10-7m2
R=0.4Ω

20. Signing Off … By: RLUG
THANK YOU! =))