Understanding Power Dissipation in Wires

The question asks us to compare the power dissipated in a copper wire and an aluminum wire when they have the same dimensions (length and area of cross-section) and the same current flows through them. We are given that the power dissipated in the copper wire is P.

Power dissipation in a wire is primarily due to the resistance of the wire and the current flowing through it. This is known as Joule heating or \(I^2R\) loss.

The formula for power dissipated (P) is given by:

\begin{equation*} P = I^2 R \end{equation*}

where:

• \(I\) is the current flowing through the wire.
• \(R\) is the resistance of the wire.

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

\begin{equation*} R = \rho \frac{L}{A} \end{equation*}

where:

• \(\rho\) (rho) is the resistivity of the material.
• \(L\) is the length of the wire.
• \(A\) is the area of cross-section of the wire.

In this problem, we are told that the aluminum wire has the "same dimensions" as the copper wire. This means they have the same length (\(L\)) and the same area of cross-section (\(A\)). We are also told that the "same current" (\(I\)) is passed through both wires.

So, for both the copper and aluminum wires, \(L\), \(A\), and \(I\) are constant.

The only factor that changes between the two wires is the material, which means their resistivity (\(\rho\)) will be different. To compare the power dissipation, we need to know the relative resistivities of copper and aluminum.

Copper is known to be a better electrical conductor than aluminum. Electrical conductivity is the reciprocal of resistivity. Therefore, aluminum has a higher resistivity than copper.

Let's denote the resistivity of copper as \(\rho_{Cu}\) and the resistivity of aluminum as \(\rho_{Al}\). We know that \(\rho_{Al} > \rho_{Cu}\).

Now, let's look at the resistance formula again:

\begin{equation*} R = \rho \frac{L}{A} \end{equation*}

Since \(L\) and \(A\) are the same for both wires, the resistance is directly proportional to the resistivity (\(R \propto \rho\)).

Because \(\rho_{Al} > \rho_{Cu}\), it follows that the resistance of the aluminum wire (\(R_{Al}\)) will be greater than the resistance of the copper wire (\(R_{Cu}\)):

\begin{equation*} R_{Al} > R_{Cu} \end{equation*}

Now let's consider the power dissipation formula:

\begin{equation*} P = I^2 R \end{equation*}

Since the current (\(I\)) is the same for both wires, the power dissipated is directly proportional to the resistance (\(P \propto R\)).

The power dissipated in the copper wire is given as \(P = I^2 R_{Cu}\). The power dissipated in the aluminum wire, let's call it \(P_{Al}\), is \(P_{Al} = I^2 R_{Al}\).

Since \(R_{Al} > R_{Cu}\) and \(I\) is the same, we can conclude that \(I^2 R_{Al} > I^2 R_{Cu}\).

Therefore, the power dissipated in the aluminum wire (\(P_{Al}\)) will be greater than the power dissipated in the copper wire (\(P\)):

\begin{equation*} P_{Al} > P \end{equation*}

This means the power dissipated in the aluminum wire will be greater than P.

Revision Table: Electrical Properties

Additional Information: Copper vs. Aluminum Wires

While copper is generally preferred for electrical wiring due to its lower resistivity and better conductivity, aluminum is sometimes used, especially in larger power transmission applications, because it is lighter and less expensive than copper for the same conductivity (though requiring a larger cross-sectional area). However, aluminum has some disadvantages, such as being less ductile, more prone to oxidation, and having different thermal expansion properties, which can lead to issues at connections if not properly managed. The fact that aluminum has higher resistivity than copper means that for the same dimensions and current, it will heat up more due to higher power dissipation.