Two capacitors of equal capacitance $(C_1 = C_2)$ are shown in the figure. Initially, while the switch $S$ is open, one of the capacitors is uncharged and the other carries charge $Q_0$. The energy stored in the charged capacitor is $U_0$. Sometimes after the switch is closed, the capacitors $C_1$ and $C_2$ carry charges $Q_1$ and $Q_2$, respectively; the voltages across the capacitors are $ V_1$ and $V_2$; and the energies stored in the capacitors are $U_1$ and $U_2$. Which of the following statements is INCORRECT ?
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In the above diagram, $C_{1}$ and $C_{2}$ are connected in parallel.
Hence, both get the same voltage. $\therefore V_{1}=V_{2}$ is correct.
We know that $Q_{1}=C_{1} V_{1}$ and $Q_{2}=C_{2} V_{2}$ $\because C_{1}=C_{2}, \quad Q_{1}=Q_{2}$
We know that energy $U_{1}=\frac{1}{2} C_{1} V_{1}^{2}$ and $U_{2}=\frac{1}{2} C_{2} V_{2}^{2}$
By closing the switch, the $C_{1}$ or $V_{1}$ doesn't get affected.
Hence $U_{0}=U_{1}$ and not $U_{1}+U_{2}$
Similarly, By closing the switch, the $Q_{1}$ or $V_{1}$ doesn't get affected.
Hence $Q_{0}=Q_{1}$
$\because Q_{0}=Q_{1}=Q_{2}, \quad Q_{0}=\frac{1}{2}\left(Q_{1}+Q_{2}\right)$
Thus $D$ is the only wrong statement.
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