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1. Electric Charges and Fields
normal
A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$ . Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$ . The charging battery is now disconnected and the capacitors are connect in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
A
zero
B
$\frac{3}{2}\,C{V^2}$
C
$\frac{25}{6}\,C{V^2}$
D
$\frac{9}{2}\,C{V^2}$
Solution
$\mathrm{V}_{\mathrm{C}}=\frac{4 \mathrm{CV}-\mathrm{CV}}{3 \mathrm{C}}=\mathrm{V}$
$\therefore \mathrm{U}_{\mathrm{f}}=\frac{1}{2} \times(\mathrm{C}+2 \mathrm{C}) \mathrm{V}^{2}=\frac{3}{2} \mathrm{\,CV}^{2}$
Standard 12
Physics
