A wire of copper having length l and area of cross-section A is taken and a current I is flown through it. The power dissipated in the wire is P. If we take an aluminum wire having same dimensions and pass the same current through it, the power dissipated will be
P
< P
> P
2P
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CDS-1 – 2018
For the copper wire, P = I² * R_copper = I² * (ρ_copper * l / A). We are given this power is P.
For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I² * R_aluminum = I² * (ρ_aluminum * l / A).
To compare P_aluminum and P, we need to compare the resistivities of aluminum (ρ_aluminum) and copper (ρ_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, ρ_aluminum > ρ_copper.
Since P_aluminum is proportional to ρ_aluminum (for fixed I, l, A) and P is proportional to ρ_copper, and ρ_aluminum > ρ_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire.
– Resistance depends on the material’s resistivity and the wire’s dimensions (R = ρl/A).
– Aluminum has higher resistivity than copper.