Analyzing Power Dissipation in Metallic Wires

This problem involves understanding how the physical dimensions of a conductor affect its electrical resistance and, consequently, the power dissipated when a voltage is applied across it. We are given two metallic wires, A and B, made of the same material (copper), with different lengths and radii, and the same DC voltage applied across them.

Key Concepts for Solving the Problem

• Resistance of a wire: The resistance \(R\) of a wire is directly proportional to its length \(l\) and inversely proportional to its cross-sectional area \(A\). The formula is \(R = \rho \frac{l}{A}\), where \(\rho\) is the resistivity of the material. Since both wires are copper, \(\rho\) is the same for A and B.
• Area of cross-section: For a cylindrical wire with radius \(r\), the cross-sectional area is \(A = \pi r^2\).
• Power Dissipation: When a voltage \(V\) is applied across a resistor with resistance \(R\), the power dissipated is given by \(P = \frac{V^2}{R}\).

Step-by-Step Calculation of Resistance and Power

Let's analyze wire A and wire B separately.

Wire A:

• Radius: \(r_A = r\)
• Length: \(l_A = l\)
• Cross-sectional Area: \(A_A = \pi r_A^2 = \pi r^2\)
• Resistance: \(R_A = \rho \frac{l_A}{A_A} = \rho \frac{l}{\pi r^2}\)
• Voltage Applied: \(V_A = V\)
• Power Dissipation: \(P = \frac{V_A^2}{R_A} = \frac{V^2}{\rho \frac{l}{\pi r^2}} = \frac{V^2 \pi r^2}{\rho l}\)

Wire B:

• Radius: \(r_B = 2r\)
• Length: \(l_B = 2l\)
• Cross-sectional Area: \(A_B = \pi r_B^2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2\)
• Resistance: \(R_B = \rho \frac{l_B}{A_B} = \rho \frac{2l}{4\pi r^2} = \rho \frac{l}{2\pi r^2}\)
• Voltage Applied: \(V_B = V\)
• Power Dissipation: \(P_1 = \frac{V_B^2}{R_B} = \frac{V^2}{\rho \frac{l}{2\pi r^2}} = \frac{V^2 2\pi r^2}{\rho l}\)

Comparing Power Dissipation P and P₁

Now we have expressions for P and P₁:

• \(P = \frac{V^2 \pi r^2}{\rho l}\)
• \(P_1 = \frac{V^2 2\pi r^2}{\rho l}\)

Let's compare these two expressions. We can see that the expression for P₁ is exactly two times the expression for P:

\(P_1 = 2 \times \left( \frac{V^2 \pi r^2}{\rho l} \right)\)

Substituting the expression for P into this equation:

\(P_1 = 2P\)

We are looking for the relationship between P and P₁, starting with P. Dividing both sides by 2 gives:

\(P = \frac{P_1}{2}\)

This means the power dissipated in wire A (P) is half the power dissipated in wire B (P₁).

Let's summarize the resistances and power dissipations:

From the table, we see that \(R_B = \frac{1}{2} R_A\). Since \(P = V^2/R\), if resistance decreases, power increases for the same voltage. Specifically, \(P_1 = \frac{V^2}{R_B} = \frac{V^2}{(1/2)R_A} = 2 \frac{V^2}{R_A} = 2P\). Thus, \(P = P_1/2\).

Conclusion on Power Dissipation Relationship

The relationship between the power dissipated in wire A (P) and wire B (P₁) is \(P = P_1/2\). This is because wire B, despite being longer, has a significantly larger cross-sectional area due to its increased radius, resulting in lower resistance compared to what its length alone would suggest. The lower resistance of wire B leads to higher power dissipation when the same voltage is applied.

Revision Table: Electrical Properties and Power

Additional Information: Resistivity and Conductor Materials

Resistivity (\(\rho\)) is an intrinsic property of a material that quantifies how strongly it resists electric current flow. Materials with low resistivity are good conductors (like copper, silver, gold), while materials with high resistivity are poor conductors or insulators (like rubber, glass). The resistivity of a material is affected by temperature. For most conductors, resistivity increases with increasing temperature. In this problem, since both wires are made of copper and implicitly at the same temperature, their resistivity \(\rho\) is considered constant.

The relationship \(R = \rho \frac{l}{A}\) highlights why thick, short wires have lower resistance than thin, long wires made of the same material. A larger cross-sectional area provides more paths for electron flow, reducing resistance, while a longer length means electrons have to travel further, increasing resistance.

In this specific problem, wire B has double the length (\(l_B = 2l\)) and double the radius (\(r_B = 2r\)) compared to wire A. The length factor tends to increase resistance by a factor of 2, while the radius factor affects the area (\(A \propto r^2\)). Since the radius is doubled, the area increases by a factor of \((2)^2 = 4\). The combined effect on resistance is \(R_B \propto \frac{2l}{4A_A} = \frac{1}{2} \frac{l}{A_A}\). So, \(R_B\) is half of \(R_A\). With resistance halved and voltage kept constant, the power dissipated (\(V^2/R\)) doubles.