Two metallic wires made from copper have same length but the radius of wire 1 is half of that of wire 2. The resistance of wire 1 is R. If both the wires are joined together in series, the total resistance becomes
2R
R/2
5R/4
3R/4
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CDS-1 – 2018
For wire 1, with radius r1 and length l, the resistance R1 = ρ * l / (πr1²). We are given R1 = R.
For wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = ρ * l / (πr2²) = ρ * l / (π(2r1)²) = ρ * l / (π * 4r1²) = (1/4) * [ρ * l / (πr1²)].
Since R = ρ * l / (πr1²), we have R2 = R/4.
When the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2.
R_total = R + R/4 = 5R/4.
– When conductors are in series, their resistances add up (R_total = R1 + R2 + …).