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Ratio of drift velocities, series and parallel

Short answer

In series, same I, v_d = I/(nAe) → v₁/v₂ = A₂/A₁ = 2:1. In parallel, same length so same electric field, v_d = μE equal, so v₁:v₂ = 1:1.

Two copper wires of same length l and cross sectional areas A and 2A are connected to a battery. What will be the ratio of drift velocities when the wires are in (a) series and (b) parallel?

Wire 1 has area A and wire 2 has area 2A. Both have the same length l and are copper, so the same n and resistivity ρ. Drift velocity is v_d = I/(nAe) = J/(ne).

(a) In series

The same current I flows in both wires.

v₁ = I/(n A e) ;  v₂ = I/(n 2A e)
v₁ / v₂ = 2A / A = 2

Answer: v₁ : v₂ = 2 : 1 (the thinner wire has twice the drift speed).

(b) In parallel

The same potential difference V is across both wires, and they have the same length, so the same electric field E = V/l.

v_d = μ E   (same material, same μ, same E)
v₁ = v₂

Check with currents: I₁ = V/R₁ with R₁ = ρl/A, so I₁ = VA/(ρl) and I₂ = 2VA/(ρl). Then v₁ = I₁/(nAe) = V/(ρ l n e) and v₂ = I₂/(n 2A e) = V/(ρ l n e). Equal.

Answer: v₁ : v₂ = 1 : 1 (equal drift speeds).

✓ Answer checked by Claude Opus 5.5 (high effort)

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