Practice question

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. Initially, the conductors have charges Q₁ and Q₂ , with energy U_i = (Q₁²/2C) + (Q₂²/2C) . Upon connection, charge redistributes to equal potentials, total charge Q₁ + Q₂ splits equally ( (Q₁ + Q₂/2) each), so final energy U_f = 2 × (((Q₁ + Q₂/2))²/2C) = ((Q₁ + Q₂)²/4C) . Typically, (Q₁ +