Two cells of emf \( 3 \, \text{V} \) and \( 7 \, \text{V} \) with internal resistances \( 0.5 \, \Omega \) and \( 1.5 \,
**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Equivalent emf: εₑq = 3 + 7 = 10 V . Total resistance: Rtₒtₐl = 0.5 + 1.5 + 10 = 12 Ω . Current: I = (εₑq/Rtₒtₐl) = (10/12) ≈ 0.83 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 P = I²R, evaluation yields 0.83 A,
Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors