Why does a conductor’s current density increase when its length is halved while keeping the potential difference constan
**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Resistance R = rho l / A . Halving length ( l' = l/2 ) halves R ( R' = R/2 ). Current I = V / R , so I' = V / (R/2) = 2I . Current density j = I / A , so j' = 2I / A = 2j . Applying I = n
Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination