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Electricity (Basic)
> Electrical Resistance
Electrical Resistance
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According to physicist, George Simon Ohm, the voltage applied externally to the two ends of the conductor is constant when electric current flows through it at fixed physical condition. This is called as Ohm’s law.
Voltage œ current
V œ I
V = RI
Where we have,
V= Voltage (V)
I = Current (A)
R = Resistance (Ω)
Resistance is nothing but the obstruction to the flow of charge. Here, R is called the resistance of the conductor and unit of resistance is Ohm and its symbol is
Ω
.
Inverse of Ohm is called the conductance of the material and unit of inverse ohm is mho Ω
^{ 1}
Ohm’s law gives the information about the relationship between the potential differences between the conductor and the electrical current flowing through it.
Let us consider a conductor having an area A and length
l
, at a given temperature, then
R œ
l
and R œ 1/A
So,
R œ
l
/A
R = p
l
/A
Where we have p = resistivity of the material
Unit of p = Ω m
The value of p depends of the type, temperature and pressure on the conductor. Resistivity is increase with increase in temperature.
Conductivity is equal to inverse of resistivity and its symbol is
6
and unit is (
Ω
m)
^{1}
Limitation of Ohm’s law:
Value of V depends on the value of current
Value of V depends on the direction in which it is applied
It is nonlinear.
Superconductivity:
According to Kamerlingh Onnes, the resistance of certain material like mercury becomes almost zero when its temperature is lowered beyond certain fixed temperature. The material is called superconductor and this process is called superconductivity.
Examples of superconductors – Hg, Si, Se, Ge, Te
Calculation Examples
Example1:
The potential difference between two electrodes of battery is 10V and having 0.5 ohm resistance. Calculate the current flows between two electrodes?
a) 20A b) 0.5A c) 0.05A d) 0.005A
Reason:
Here we have:
R = 0.5 ohm
V = 10V
I = ?
V = R/I
10 = 0.5 / I
I = 0.5 /10
I = 0.05 A
Example2:
Twelve gauge copper wire is used for wiring. It has resistivity 1.77 ×
10
^{8}
at 20 °C. Calculate the resistance of 20 m long wires at same temperature. (12 gauge = 2.05
×
10
^{3}
m diameter)
Reason:
Here we have
l
= 20 m
p = 1.77
×
10
^{8}
d = 12 gauge = 2.05 ×
10
^{3}
m
A = 3.14r
^{2}
= 3.14
×
(d/2)
^{2}
= 3.14
× (2.05 ×
10
^{3}
/ 2)
^{ 2}
R = p
l
/A
R = ( 1.77 ×
10
^{8}
× 20 ) / ( 3.14 × (2.05 ×
10
^{3}
/ 2)
^{ 2}
)
R = ( 1.77 ×
10
^{8}
× 20 × 4 ) / ( 3.14 × (2.05 ×
10
^{3}
)
^{ 2}
)
R = 10.89 ×
10
^{2}
ohm
Example3:
The relationship between the potential differences between the conductor and the electrical current flowing through it is explained by …………….phenomenon.
a) Resistivity b) superconductivity c) Conductivity d) Ohm’s law
Answer: The relationship between the potential differences between the conductor and the electrical current flowing through it is explained by Ohm’s law.
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