Current Electricity — Exam Practice

Answer Between an applied field's influence, electrons undergo frequent random collisions with the lattice ions. Their thermal motion is fast but random and averages to zero displacement; only the small additional velocity gained between collisions, in the field's direction, survives as a net drift.

Answer Between collisions, an electron accelerates under the field: a = eE/m. Averaged over the relaxation time τ (mean time between collisions), the drift velocity is v_d = aτ = eEτ/m.

Consider a conductor of cross-section A and free electron density n. In time dt, electrons within a distance v_ddt of any cross-section pass through it. Charge crossing: dq = n(Av_ddt)e. So current I = dq/dt = nAev_d.

Answer (a) Junction rule: the sum of currents entering a junction equals the sum leaving (charge conservation). Loop rule: the sum of potential changes around any closed loop is zero (energy conservation).

Applying both rules to the four-arm bridge circuit, and setting the galvanometer current to zero (balance condition), the loop equations reduce to P/Q = R/S.

(b) S = QR/P = (20 × 15)/10 = 30 Ω.

Solution Resistivity ρ is an intrinsic material property (unlike resistance R = ρL/A, which does depend on geometry). It also varies with temperature, since collision frequency between electrons and the lattice changes with temperature.

Solution "Balance" is defined precisely as the condition where no current flows through the galvanometer branch — the bridge's two midpoints are at the same potential.

Solution An ideal cell delivers its full EMF as terminal voltage regardless of current drawn — only possible if internal resistance is exactly zero, so no energy is lost inside the cell itself.

Solution In series, both EMFs and internal resistances simply add: total EMF = nε, total internal resistance = nr.

Solution Volume constant: A₁L₁ = A₂L₂. With L₂ = 2L₁, this gives A₂ = A₁/2.

R = ρL/A, so R₂/R₁ = (L₂/L₁) × (A₁/A₂) = 2 × 2 = 4.

R₂ = 4 × 10 = 40 Ω. (A useful shortcut to remember: stretching to n times the length, at constant volume, always multiplies resistance by n².)

Solution Charge can't accumulate or vanish at a junction — whatever current flows in must flow out. (The loop rule, by contrast, is a consequence of energy conservation.)

Solution ε = I(R + r) → 2 = 1 × (R + 0.5) → R = 1.5 Ω.