A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea
**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. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 4 = (20/6) = (10/3) ≈ 3.33 Ω . Total current: I = (V/Rₑq) = (24/(10/3)) = 24 × (3/10) = 7.2 A . 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
Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination