A copper wire of radius r and length l has resistance R. A second copper wire has radius 2r and the same length l. The two wires are connected in parallel. What is the resultant resistance of this parallel combination in terms of R?

Introduction / Context:
This problem tests your understanding of how resistance depends on length, area and resistivity of a conductor, and how resistances combine when connected in parallel. It uses two copper wires of different radii but the same length and material. You need to find the equivalent resistance of their parallel combination, expressed in terms of the known resistance R of the thinner wire.

Given Data / Assumptions:

• First copper wire: radius r, length l, resistance R.
• Second copper wire: radius 2r, same length l, same material (same resistivity).
• The two wires are connected in parallel across the same potential difference.
• Copper is assumed to be homogeneous with constant resistivity.

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 cylindrical wires, A = pi * r^2. If radius changes, area and therefore resistance change. For parallel combination of two resistances R1 and R2, the equivalent resistance R_eq is given by 1 / R_eq = 1 / R1 + 1 / R2, or R_eq = (R1 * R2) / (R1 + R2). We first find the resistance of the thicker wire in terms of R, then apply the parallel formula.

Step-by-Step Solution:
Step 1: Write the resistance of the first wire: R1 = R = rho * l / (pi * r^2). Step 2: For the second wire, the radius is 2r, so its area is A2 = pi * (2r)^2 = 4 * pi * r^2. Step 3: Its resistance R2 is R2 = rho * l / A2 = rho * l / (4 * pi * r^2) = (1/4) * rho * l / (pi * r^2) = R/4. Step 4: Now use the parallel resistance formula: R_eq = (R1 * R2) / (R1 + R2). Step 5: Substitute R1 = R and R2 = R/4 to get R_eq = (R * R/4) / (R + R/4). Step 6: Simplify: numerator = R^2 / 4; denominator = (5R/4). So R_eq = (R^2 / 4) * (4 / 5R) = R / 5.

Verification / Alternative check:
As a consistency check, the equivalent resistance of two resistors in parallel is always less than the smaller of the two. Here, the smaller resistance is R2 = R/4. Our result R_eq = R/5 is indeed smaller than R/4, so it passes this important sanity check. Additionally, if both wires had the same resistance R, the parallel result would be R/2, so getting a value smaller than R/4 is reasonable because one branch is R and the other is R/4, providing more overall conductance.

Why Other Options Are Wrong:
5R: This is much larger than either individual resistance and cannot be the result of a parallel combination, which must be smaller than the smallest branch resistance.

5R/4 or 4R/5: Both are greater than R/4, again impossible for a parallel combination of R and R/4. Such values might arise from incorrect algebra or using a series formula instead of the parallel formula.

Common Pitfalls:
A common mistake is to confuse series and parallel formulas, accidentally adding resistances instead of using the reciprocal relationship. Another pitfall is failing to square the radius when computing the area, leading to an incorrect factor for the resistance of the thicker wire. Always remember that area is proportional to r^2, so doubling the radius makes the area four times larger and the resistance four times smaller, not just twice smaller.

Final Answer:
The resultant resistance of the parallel combination of the two copper wires is R/5.