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Two metallic wires A and B are made using copper. The radius of wire A is r while its length is l. A dc voltage V is applied across the wire A, causing power dissipation, P. The radius of wire B is 2r and its length is 2l and the same dc voltage V is applied across it causing power disspation P 1. Which one of the following is the correct relationship between P and P1?
Explanation
The correct answer is Option 2: P = P1/2. This relationship is derived from the formula for electrical power dissipation, P = V²/R, and the formula for resistance, R = ρ(l/A), where A = πr².
For wire A, the resistance is R_A = ρl / πr². Therefore, the power dissipation is P = V² / (ρl / πr²) = (V²πr²) / ρl.
For wire B, the radius is 2r and the length is 2l. Its resistance is R_B = ρ(2l) / π(2r)² = ρ(2l) / 4πr² = ρl / 2πr². Comparing the two, we find R_B = R_A / 2.
Substituting R_B into the power formula for wire B: P1 = V² / R_B = V² / (R_A / 2) = 2(V² / R_A) = 2P. Rearranging this gives P = P1/2. Since wire B has half the resistance of wire A due to its larger cross-sectional area, it dissipates twice the power under the same voltage.
SIMILAR QUESTIONS
Two metallic wires made from copper have the same length, but the radius of wire 1 is half that of wire 2. The resistance of wire 1 is R. If both the wires are joined together in series, the total resistance becomes:
- (A) 2R
- (B) 1.25R
- (C) 0.5R
- (D) 1.33R
A copper wire of radius r and length l has a resistance of R. A second copper wire with radius 2r and length l is taken and the two wires are joined in a parallel combination. The resultant resistance of the parallel combination of the two wires will be:
A) R/5
B) 5R/4
C) R/2
D) 4R/5