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STD 12 - 2. Electric potential and capacitance — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electric potential and capacitance4 Marks
MCQ
A capacitor $C$ is fully charged with voltage $V _{0}$ After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{ C }{2} .$ The energy loss in the process after the charge is distributed between the two capacitors is$.........$$CV _{0}^{2}$
  • ✓
    $0.166$
  • B
    $0.5$
  • C
    $0.33$
  • D
    $0.25$
✓

Answer

Correct option: A.
$0.166$
a
$\frac{ CV _{0}- q }{ C }=\frac{ q }{ C / 2}=\frac{2 q }{ C }$

$V _{0}=\frac{3 q }{ C } \Rightarrow q =\frac{ CV _{0}}{3}$

$U _{ i }=\frac{1}{2} CV _{0}^{2}$

$U _{ f }=\frac{\left(\frac{2 CV _{0}}{3}\right)^{2}}{2 C }+\frac{\left(\frac{ CV _{0}}{3}\right)^{2}}{2\left(\frac{ C }{2}\right)}$

$=\frac{1}{2} CV _{0}^{2}\left[\frac{4}{9}+\frac{2}{9}\right]=\frac{1}{2} CV _{0}^{2}\left(\frac{2}{3}\right)$

Heat loss $=\frac{1}{2} CV _{0}^{2}-\left(\frac{2}{3}\right)\left(\frac{1}{2} CV _{0}^{2}\right)$

$=\frac{1}{6} CV _{0}^{2}$

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