Two metallic copper wires have the same length, but the radius of wire 1 is half of the radius of wire 2. The resistance of wire 1 is R. If both wires are connected in series, what is the total resistance of the combination?

Introduction / Context:
This problem tests your understanding of how resistance depends on the dimensions of a conductor and how resistances combine in series. For metallic wires of the same material and same length, resistance changes with the cross sectional area, which in turn depends on the radius of the wire. Once the individual resistances are known, the total resistance for a series connection is simply the sum. Such questions are common in basic electricity and help build intuition for designing circuits and choosing conductor sizes.

Given Data / Assumptions:
• Both wires are made of copper (same resistivity). • Both wires have the same length L. • Radius of wire 1 is r1 and radius of wire 2 is r2, with r1 = r2 / 2. • Resistance of wire 1 is given as R. • The two wires are connected in series, so the same current flows through both.

Concept / Approach:
The resistance of a uniform wire is given by R = rho * L / A, where rho is resistivity, L is length, and A is cross sectional area. For a circular wire, A = pi * r^2, so resistance is inversely proportional to r^2 when material and length are fixed. If one wire has half the radius of another but the same length and material, its area is one quarter, so its resistance is four times larger. After finding the relationship between the resistance of wire 1 and wire 2, we simply add the two resistances in series to find the total resistance.

Step-by-Step Solution:
Step 1: Write the resistance formula for a wire: R = rho * L / A. Step 2: For circular cross section, area A = pi * r^2. Step 3: Let the radius of wire 2 be r2. Then the radius of wire 1 is r1 = r2 / 2. Step 4: Area of wire 1: A1 = pi * (r2 / 2)^2 = pi * r2^2 / 4. Step 5: Area of wire 2: A2 = pi * r2^2. Step 6: Since resistivity rho and length L are the same for both, R1 / R2 = A2 / A1 = (pi * r2^2) / (pi * r2^2 / 4) = 4. Step 7: Therefore, R1 = 4 * R2. Given R1 = R, we get 4 * R2 = R so R2 = R / 4. Step 8: In series, total resistance R_total = R1 + R2 = R + R / 4 = 5R / 4.

Verification / Alternative check:
We can quickly check with simple numbers. Assume rho * L / (pi * r2^2) = 1 ohm so that wire 2 has resistance 1 ohm. Then wire 1, with one quarter the area, would have R1 = 4 ohm. The question states R1 = R, so R = 4 ohm and R2 = 1 ohm = R / 4. The total series resistance is 4 + 1 = 5 ohm = 5R / 4. This numerical example matches the algebraic result and confirms the reasoning.

Why Other Options Are Wrong:
Option a (2R): This would correspond to R2 = R, which ignores the dependence on the radius squared. Option b (R/2): This suggests the total resistance is less than that of wire 1 alone, which is impossible for series connection with a positive resistance wire. Option d (3R/4): This is even smaller than R and contradicts the idea that adding a resistance in series increases the total resistance. Option e (R/4): This equals the resistance of the thinner wire divided by 4 and does not follow from the geometry.

Common Pitfalls:
Students often forget that resistance is inversely proportional to the square of the radius, not directly proportional to radius. Another common mistake is to assume that having the same length automatically means equal resistance, ignoring the change in cross sectional area. In series combinations, some learners also incorrectly apply formulas for parallel resistors. To avoid errors, always relate area to radius carefully and then use the correct series rule, R_total = R1 + R2.

Final Answer:
The total resistance of the series combination of the two copper wires is 5R/4.