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#conductor geometry

2 public questions tagged with this topic.

Wheatstone Bridge and Meter Bridge

Why does the resistance of a conductor increase when its cross-sectional area is reduced?

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Resistance R = rho l / A . Reducing A increases R inversely, as fewer charge carriers can pass through a smaller area, increasing opposition to current flow. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields Fewer paths for current,

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Electric Current, Drift Velocity and Mobility

What happens to the current density in a conductor if the conductor is stretched to twice its original length without ch

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Stretching doubles length ( l' = 2l ) and halves area ( A' = A/2 ) due to volume conservation. Resistance becomes R' = rho (2l) / (A/2) = 4R . Current I' = V / (4R) = I/4 . Current density j = I / A , so j' = I' / A' = (I/4) / (A/2) = I

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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