Resistance

To control the current flowing through wires, it is important to understand electrical resistance. In an electrical circuit, resistance is the opposition to the flow of electric current. It is measured in ohms (Ω). According to Ohm’s law, resistance is defined as the ratio of the applied voltage to the current flowing in the circuit. Therefore,

\boxed {R = \frac{V}{I}}
Where,
R = resistance
I = current
V = voltage

Resistance of a Conductor

The electrical resistance of a wire can also be determined using its length and cross-sectional area. The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area.
Mathematically, this relationship can be expressed as:

The resistance of a conductor is directly proportional to its length
R \propto L
The resistance of a conductor is inversely proportional to its cross-sectional area.
R \propto \frac{1}{A}
By removing the proportionality constant and introducing resistivity, the relation becomes
\boxed {R = \rho \frac{l}{A}}
Where,
R = resistance
ρ = resistivity of the conductor
l = length of the conductor
A = area of the cross-section of the conductor

The resistance depends on the material it is made of. Objects made of electrical insulators, like rubber, tend to have very high resistance, while objects made of electrical conductors like metals tend to have very low resistance.

Water Pipe Analogy for Electrical Resistance

When the length of the pipe is long, the resistance to the flow of water will be high.
When the cross-sectional area of the pipe is high, the resistance to the flow of water is low.

Electrical Resistance Relation with Power

Electric power (P) is the product of voltage and current. The SI unit of electric power is the watt (W). Electric power in a circuit can be calculated using the values of voltage, current, and resistance with the help of Ohm’s law.

P = VI
Where,
P = Electric Power
V = Voltage
I = Current
From Ohm's Law we know that,
V = IR
Substituting the value of V in the power equation
P = I²R
When the values of voltage and resistance are known, the power can also be written as
\boxed {P = \frac{V^2}{R}}

Effect of Temperature

The resistance of the materials changes with the change in temperature. The amount of the change differs with the types of material.

Metals: The electrical resistance of pure metals increases with an increase in temperature due to increased collisions of electrons with atoms. Therefore, metals have a positive temperature coefficient of resistance.
Example: Copper, Aluminum, Silver
Alloys: The electrical resistance of alloys also increases with temperature, but the change is very small. Hence, alloys have a low positive temperature coefficient of resistance.
Example: Nichrome
Semiconductors, Insulators & Electrolytes: The electrical resistance of semiconductors, insulators, and electrolytes decreases with an increase in temperature due to the increase in the number of charge carriers. Therefore, these materials have a negative temperature coefficient of resistance.

Factors Affecting Electrical Resistance

1. Length of the conductor: The greater the length of the wire, more will be the more resistance offered by the wire.

2. Area of a cross-section of the conductor: The resistance of the wire decreases as the area of the cross-section of the wire is increased.

3. Material of the conductor: Different materials have different resistances. For Metals, the resistance offered is very low, but for Insulators, the resistance offered is quite large.

4. Temperature of the Material: The electrical resistance of pure metals and alloys increases when the temperature increases, but for insulators, the electrical resistance decreases with an increase in temperature.

Resistivity

Electrical resistivity is a property of a material that is fundamental in nature, and it measures how strongly the material resists electric current. The SI unit of electrical resistivity is ohms meter and the symbol is row(ρ).

For Ideal Cases, where the cross-section and physical composition of the material are uniform across the sample. The resistivity can be written as:

ρ = R\frac{A}{l}

For less ideal cases, the current and electric field vary in different parts of the material. We use a General Expression,

ρ= \frac{E}{J}
Where,
ρ is the electrical resistance of metal Ω.m
E is the electric field in V · m-1
J is the current density in A · m-2

Sample Problems

Question 1: Two Wires of Length 50m and 40m, respectively, have the Same Area of cross-section and are Made Up of the Same Material. Which Wire has Higher Resistance?

Solution: We know that R α L, a wire of length 50m, will offer higher resistance. The resistance of a small wire is low, while the resistance of a long wire is high. t

Question 2: Two Wires L1 and L2, have lengths L and 2L, respectively. The area of cross-section is 2A and A, respectively. Both wires are made up of the same material. Find the ratio of resistance in Wire L₁ and L₂?

Solution: R = ρ\frac{l}{A}
So,
R_1 = ρ\frac{L}{2A} ⇢ (1)
R_2 = ρ\frac{2L}{A} ⇢ (2)
Dividing equation 1 by equation 2 above two equations,
\frac{R_1}{R_2} = \frac{ρ\frac{L}{2A}}{ρ\frac{2L}{A}}
\frac{R_1}{R_2} = \frac{1}{4}

Question 3: Calculate the resistance of a copper wire of length 5m and area of cross-section 2 × 10^-6 m². The resistivity of copper is 1.7 × 10^-8 Ωm.

Solution: Length of copper wire = 5m
Area of cross-section of copper wire = 2 \times10^-6 m²
Resistivity of copper wire = 1.7 × 10^-8Ωm
We know that,
R = ρ\frac{l}{A}
So putting the values of ρ, l, and A in the above equation,
R = \frac{(1.7 \times 10^{-8})\times (5)}{2 \times 10^{-6}}
R = 4.25 × 10^-2Ω

Question 4: Explain the relation of Power with Resistance?

Solution: Power is Voltage times Current,
P = VI ⇢ (1)
From Ohm's Law,
V = RI ⇢ (2)
Replacing the value of equation (2) in equation (1),
P = I²R
It can also be rewritten as,
P = \frac{V^2}{R}

Question 5: What current will be taken by a 1500W appliance if the supply voltage is 220V?

Solution: Power of the appliance = 1500W
Voltage supply to the appliance = 220V
From the power-current relation, we know that,
P = VI
The above equation can be rewritten as,
I = \frac{P}{V}
I = \frac{1500}{220}
I = 6.81A

Question 6: A metal wire of resistivity 6 × 10^-6 Ωm and length 20m has a resistance of 10Ω. Calculate its radius.

Solution: Resistivity of the wire = 6 × 10^-6Ωm
Length of the wire = 20m
Resistance of the wire = 10Ω
R = ρ\frac{l}{A}
The above equation can be rewritten as,
A = ρ\frac{l}{R} ⇢ (1)
Replacing the values in equation (1),
A = \frac{(6\times 10^{-6})\times (20)}{10}
A = 12 × 10^-6m²⇢ (2)
Generally, the area-cross section of a wire is a circle. So the area of a circle is,
A = πr² ⇢ (3)
r = radius of the cross-section of the wire
Putting the value of (2) in (3),
πr² = 12 × 10^-6
r = 1.95 × 10^-3m

Unsolved Problems

Question 1: A copper wire has a length of 10 m and a cross-sectional area of 1.5 × 10⁻⁶ m². If the resistivity of copper is 1.7 × 10⁻⁸ Ωm, calculate the resistance of the wire.

Question 2: Two wires made of the same material have the same length, but the cross-sectional area of the first wire is twice that of the second wire. Find the ratio of their resistances.

Question 3: A wire of resistance 8 Ω carries a current of 3 A. Calculate the voltage across the wire.

Question 4: An electric heater operates at 220 V and draws a current of 5 A. Calculate the power consumed by the heater.

Question 5: A metal wire has a resistance of 12 Ω and is connected to a 120 V supply. Find the current flowing through the wire.