A \( 12 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \) and \( 8 \, \Omeg
**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Total resistance: R = 4 + 8 = 12 Ω . Current: I = (V/R) = (12/12) = 1 A . Power: P = I² R = 1² × 4 = 4 W . 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
Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases