Animated Solution for Physics - Current Electricity: A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then, for the two wires to have the same resistance, the ratio $l_B/l_A$ of their respective lengths must be

Imagine you are holding two different wires, let's call them Wire A and Wire B. The problem presents a fascinating scenario where these two wires, despite having different physical and material properties, offer the exact same resistance to the flow of electric current. Our mission is to find out how their lengths compare.

Let's break down the given conditions. We are told that Wire B is made of a material that is twice as resistive as the material of Wire A. Mathematically, we can write this as ${\rho }_B=2{\rho }_A$.

Furthermore, Wire B is thicker. Its diameter is twice the diameter of Wire A, which means $d_B=2d_A$. The most crucial piece of information, however, is that their electrical resistances are perfectly balanced: $R_A=R_B$.

To solve this, we need to recall the fundamental formula that governs the resistance of a cylindrical conductor. The resistance $R$ depends on the resistivity $\rho$, the length $l$, and the cross-sectional area $A$.

Since the cross-section of a wire is a circle, its area can be expressed in terms of its diameter $d$ as $A=\frac{\pi d^2}{4}$. Substituting this into our resistance formula gives us a more useful form for this specific problem:

Now, we bring our conditions together. Since the resistances are equal, we can equate the expressions for both wires:

This equation might look a bit intimidating, but let's take a breath. Notice how the constants $4$ and $\pi$ are present on both sides? They gracefully cancel out, leaving us with a much cleaner relationship:

It's time to substitute the specific relationships we identified earlier. We replace ${\rho }_B$ with $2{\rho }_A$ and $d_B$ with $2d_A$. Watch out for the trap here—the entire diameter $d_B$ is squared, so the factor of $2$ becomes a $4$:

Now, the magic happens. The resistivity ${\rho }_A$ and the squared diameter $d_A^2$ cancel out completely from both sides. We are left with a beautifully simple equation:

We are looking for the ratio of the length of Wire B to the length of Wire A. By rearranging our final equation, we get:

And there we have it! To compensate for being made of a more resistive material and having a larger cross-section, Wire B must be exactly twice as long as Wire A to maintain the same electrical resistance.