Electrical Resistance Formula
When voltage is applied across any conductor then electric charges start flowing through a conductor and the rate of flow of charge is termed as current. There the property of materials which measures how much amount of charge flow through it is known as electrical resistance and also on the basis of which the materials are classified as good or bad conductors of electricity.
Electrical Resistance
The electrical resistance is the property of using, we generate heat and have various applications in our daily life.
Electrical Resistance: The property of an electrical conductor to oppose (resist) the flow of electric current is known as electrical resistance. All materials have their own electrical resistance. It is represented by R and its SI unit is Ohm Ω (Greeks letter Omega)
Georg Simon Ohm was a German physicist who gave a law which is known as ohm’s law and also gives the relation between current, voltage and resistance. According to Ohm’s law, the current flowing through a conductor is directly proportional to the potential difference across it.
V ∝ I
or
V = IR
Therefore, from the above equations, we get the relation:
R = V/I
where,
- V is the potential difference across the conductor (in Volts),
- I is the current through the conductor (in Amperes) and
- R is the constant of proportionality called Resistance (in Ohms).
Specific Electrical Resistance (Resistivity)
Specific electrical resistance or resistivity is a fundamental property of a material that measures how strongly it resists electric current. A low resistivity indicates a material that readily allows electric current. Resistivity is defined as the resistance offered by the material per unit length for unit cross-section and is inversely proportional to resistance.
- Resistivity is commonly represented by the Greek letter ρ (rho).
- The SI unit of electrical resistivity is the Ohm-meter (Ω-m) or mho.
Mathematically, resistivity is defined as,
ρ = (R×A)/L
or
R = (ρ×L)/A
where
- R is the Resistance,
- L is the Length and
- A is the cross-sectional area of the conductor.
From the above relation, it can be seen that the resistance of a conductor is depending on the following factors:
- The cross-sectional area of the conductor
- Length of the conductor
- The material of the conductor
- The temperature of the conducting material
Materials with a low value of resistivity conduct electricity very well are conductors and insulators will have a higher value of resistivity than that of conductors. Some materials and their standard specific resistance (Resistivity) at 20°^{ }C:
Material | Resistivity (in Ω-m) |
Aluminium | 2.8 × 10^{-8} |
Copper | 1.7 × 10^{-8} |
Gold | 2.4 × 10^{-8} |
Carbon (Graphite) | 1 × 10^{-5} |
Germanium | 4.6 × 10^{-1} |
Iron | 1.0 × 10^{-7} |
Lead | 1.9 × 10^{-7} |
Nichrome | 1.1 × 10^{-6} |
Silver | 1.6 × 10^{-8} |
Hence, the two important formulae for Electrical Resistance are as follows:
- R = V/I
- R = (ρ×L)/A
where, R is the Resistance, V is the Voltage, I is the current, ρ is the resistivity, L is the Length and A is the cross-sectional area of the conductor.
Sample Problems on Electrical Resistance
Problem 1: What is the resistance of the circuit in which applied voltage is 12 V and current flowing through it is 4 A?
Solution:
According to the relation:
V = IR
or
R = V/I
Therefore,
R = 12 V/4 A
R= 3 Ω
Problem 2: What is the current flowing through the circuit in which applied voltage is 12 V and resistance of the conductor is 3 ohms?
Solution:
According to the relation:
V = IR
or
I = V/R
This implies,
I = 12 V/3 Ω
I = 4 A
Problem 3: What is the voltage applied to the circuit in which current passing through the conductor is 4 A and resistance of the conductor is 3 Ω?
Solution:
According to the relation:
V = IR
Therefore,
V = 4 A × 3 Ω
V =12 V
Problem 4: Calculate the resistance of a copper wire of length 4 m and the area of cross-section 10^{-6} m^{2}. The resistivity of copper is 1.7 × 10^{-8} Ωm.
Solution:
Using formula
R = (ρ×L)/A
R = (1.7 x 10^{-8} Ωm) × 4 m/10^{-6 }m^{2}
R = 6.8×10^{-2 }Ω
Problem 5: A copper wire of length 4 m and area of cross-section 10^{-6} m^{2} has a resistance of 6.8 × 10^{-2} ohms. Calculate the resistivity of copper.
Solution:
Using formula
ρ = (R×A)/L
ρ = (6.8 × 10^{-2}) × 10^{-6 }/ 4
ρ = 1.7 × 10^{-8 }Ωm.
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