Two wires are made having same length and area of cross-section A. Wire 1 is made of copper and wire 2 is made of aluminium. It is given that the electrical conductivity of copper is more than that of aluminium. In this context, which one of the following statements is correct?

Question

Prepp Team · Accepted Answer

Understanding Electrical Resistance and Conductivity

This question asks us to compare the electrical resistance of two wires made of different materials but having the same physical dimensions. Wire 1 is made of copper, and Wire 2 is made of aluminium. We are given that copper has higher electrical conductivity than aluminium.

Let's first understand the key properties involved: electrical conductivity and electrical resistance.

Electrical Conductivity (σ): This property measures how easily electric current flows through a material. A material with high conductivity allows current to flow easily.
Electrical Resistance (R): This property measures how much a material opposes the flow of electric current. A material with high resistance opposes current flow strongly.

Conductivity and resistivity are inversely related. Resistivity (ρ) is the inverse of conductivity (\(\rho = \frac{1}{\sigma}\)). This means that a material with higher conductivity has lower resistivity, and vice versa.

Calculating and Comparing Resistance

The resistance (R) of a wire depends on its material's resistivity (ρ), its length (L), and its cross-sectional area (A). The formula for resistance is:

\(R = \rho \frac{L}{A}\)

In this problem, both wires have the same length (L) and the same area of cross-section (A).

For Wire 1 (Copper): Resistance \(R_1 = \rho_{Cu} \frac{L}{A}\)
For Wire 2 (Aluminium): Resistance \(R_2 = \rho_{Al} \frac{L}{A}\)

We are given that the electrical conductivity of copper (\(\sigma_{Cu}\)) is more than that of aluminium (\(\sigma_{Al}\)):

\(\sigma_{Cu} > \sigma_{Al}\)

Since resistivity is the inverse of conductivity (\(\rho = \frac{1}{\sigma}\)), if copper has higher conductivity, it must have lower resistivity compared to aluminium:

\(\rho_{Cu} = \frac{1}{\sigma_{Cu}}\) and \(\rho_{Al} = \frac{1}{\sigma_{Al}}\)

As \(\sigma_{Cu} > \sigma_{Al}\), it follows that \(\frac{1}{\sigma_{Cu}} < \frac{1}{\sigma_{Al}}\).

Therefore, the resistivity of copper is less than the resistivity of aluminium:

\(\rho_{Cu} < \rho_{Al}\)

Now, let's compare the resistances using the formula \(R = \rho \frac{L}{A}\). Since L and A are the same for both wires, the resistance is directly proportional to the resistivity. As \(\rho_{Cu} < \rho_{Al}\), we can conclude:

\(\rho_{Cu} \frac{L}{A} < \rho_{Al} \frac{L}{A}\)

So, \(R_1 < R_2\).

This means the resistance of Wire 1 (copper) is lower than the resistance of Wire 2 (aluminium).

Property	Copper (Wire 1)	Aluminium (Wire 2)	Comparison
Length (L)	L	L	Same
Area (A)	A	A	Same
Conductivity (σ)	Higher	Lower	σ_Cu > σ_Al
Resistivity (ρ)	Lower	Higher	ρ_Cu < ρ_Al
Resistance (R)	Lower	Higher	R₁ < R₂

Analyzing the Given Options

Let's examine each statement based on our finding that the resistance of Wire 1 (copper) is lower than the resistance of Wire 2 (aluminium), i.e., \(R_1 < R_2\) or \(R_2 > R_1\).

Option 1: The resistance of wire 1 will be higher than that of wire 2.
This states \(R_1 > R_2\). This contradicts our finding \(R_1 < R_2\). So, this statement is incorrect.
Option 2: The resistance of wire 2 will be higher than that of wire 1.
This states \(R_2 > R_1\), which is equivalent to \(R_1 < R_2\). This matches our finding. So, this statement is correct.
Option 3: The resistance of both the wires will be the same.
This states \(R_1 = R_2\). This contradicts our finding \(R_1 < R_2\). So, this statement is incorrect.
Option 4: If same current is flown through both the wires, the power dissipated in both the wires will be the same.
The power dissipated (P) in a wire when a current (I) flows through it is given by \(P = I^2 R\). If the same current (I) flows through both wires, the power dissipated will be:
For Wire 1 (Copper): \(P_1 = I^2 R_1\)
For Wire 2 (Aluminium): \(P_2 = I^2 R_2\)
Since we found that \(R_1 < R_2\), it follows that \(I^2 R_1 < I^2 R_2\) (assuming I is non-zero). Therefore, \(P_1 < P_2\).
The power dissipated in Wire 1 will be less than in Wire 2. So, the power dissipated will NOT be the same. This statement is incorrect.

Based on our analysis, only the statement that the resistance of wire 2 (aluminium) will be higher than that of wire 1 (copper) is correct.

Revision Table: Electrical Properties

Property	Description	Relationship with Resistance (R)
Resistivity (ρ)	Intrinsic property of material opposing current flow.	\(R \propto \rho\) (directly proportional for constant L, A)
Conductivity (σ)	Intrinsic property of material allowing current flow.	\(\rho = 1/\sigma\), so \(R \propto 1/\sigma\) (inversely proportional for constant L, A)
Length (L)	Length of the conductor.	\(R \propto L\) (directly proportional for constant ρ, A)
Area (A)	Cross-sectional area of the conductor.	\(R \propto 1/A\) (inversely proportional for constant ρ, L)

Additional Information: Why Copper is Preferred

Copper is widely used for electrical wiring in homes and electronics despite aluminium being lighter and cheaper. This is primarily because copper has significantly higher electrical conductivity (and lower resistivity) than aluminium. For wires of the same length and thickness, a copper wire will have much lower resistance than an aluminium wire. Lower resistance leads to less power loss (heat generation) when current flows, which is more efficient and safer, especially for transmitting electricity over distances or handling high currents. While aluminium is used in some applications like power transmission lines (where weight is a major factor), copper remains the standard for most wiring applications due to its superior electrical properties, ductility, and corrosion resistance.