Two capacitors of equal capacitance (C₁ = C₂) are shown in the figure. Initially, while the switch S is open, one of the capacitors is uncharged and the other carries charge Q₀. The energy stored in the charged capacitor is U₀. Sometime after the switch is closed, the capacitors C₁ and C₂ carry charges Q₁ and Q₂, respectively; the voltages across the capacitors are V₁ and V₂; and the energies stored in the capacitors are U₁ and U₂. Which of the following statements is INCORRECT ?

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Q₁ = Q₂

U₀ = U₁ + U₂

$Q_{0} = \frac{1}{2} (Q_{1} + Q_{2})$

V₁= V₂

answer is D.

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Detailed Solution

S is open :- $U_{0} = \frac{Q_{0}}{2 C_{1}} = \frac{1}{2} C_{1} E^{2}; Q_{0} = C_{1} E$

S is closed :- $Q_{1} = C_{1} E; Q_{2} = C_{2} E$ since C₁ = C₂ we know Q₁ = Q₂

$∴ Q_{flow} = Q_{1} + Q_{2} = (C_{1} + C_{2}) E = 2 C_{1} E = 2 Q_{0}$

When the switch is closed, C₁ and C₂ will be parallel and P.D across them is equal to E.

$\begin{array}{l} ∴ V_{1} - V_{2} = E; U_{1} = \frac{1}{2} C_{1} E^{2}; U_{2} = \frac{1}{2} C_{2} E^{2} \\ ∴ U_{1} + U_{2} = \frac{1}{2} (C_{1} + C_{2}) E^{2} \Rightarrow U_{o} \neq U_{1} + U_{2} \end{array}$

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