Why does a battery’s emf remain constant even when it delivers a large current?
**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. Emf is the potential difference across the battery terminals when no current flows, determined by the chemical reactions inside. It’s an intrinsic property, independent of current, though terminal voltage drops due to internal resistance. 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
Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination
