MCQ
Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2 V$ . The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :
  • A
    $\frac{1}{4} CV ^2$
  • B
    $2 CV ^2$
  • C
    $\frac{1}{2} CV ^2$
  • D
    $\frac{3}{4} CV ^2$

Answer

$V_{C}=\frac{q_{\text {net }}}{C_{\text {net }}}$
$=\frac{CV+2 CV}{2 C}$
$V_{C}=\frac{3 V}{2}$
Loss of energy
$=\frac{1}{2} CV^2+\frac{1}{2} C(2 V)^2-\frac{1}{2} 2 C\left(\frac{3 V}{2}\right)^2$
$=\left(\frac{CV^2}{4}\right)$

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