The capacitor with capacitance C₁ initially carries a charge Q (see figure below). While, thecapacitor with capacitance C₂ is initially uncharged. The switch S is closed at time t = 0, and thecharges begin to flow in the circuit. Find the charge flown through the wires as a function of time,q(t).QC₁A) (0) (1 - e-(C₁+C₂)t/RC₁C₂)C₁+C₂B) (02₂) (1 - e-(C₁+C₂)t/RC₁C₂)C₁+C₂QC₁ C₂C) (043) (1 — e¯(C₁+C₂)t/RC₂C₂)D) Q(1 + e-(₁+₂)t/RC1C₂)E) None of the above.+QC₁RC₂

The capacitors are finally connected in parallel, and the total charge (= q 0) is divided among them…

Answered: The capacitor with capacitance C₁…

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