A capacitor of capacitance 900,μ F is charged by a 100,V battery. capacitor is disconnected from ba

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2. Electric Potential and Capacitance

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A capacitor of capacitance $900\,\mu F$ is charged by a $100\,V$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-2}\,J$. The value of $x$ is $..............$

A

$224$

B

$223$

C

$222$

D

$225$

(JEE MAIN-2023)

Solution

$C =900\,\mu F$

$Q = CV =900 \times 10^{-6} \times 100=9 \times 10^{-2}=90\,MC$

Now

Common potential will be developed across both capacitors by $kVL$

Total charge on left plates of capacitors should be conserved.

$90\,mc +0=2\,cv _0$

$cv _0=45\,mc$

Heat dissipated $= U _{ i }- U _{ f }$ [Change in energy stored in the capacitors]

$=\frac{1}{2} \frac{(90\,mc )^2}{900\,\mu F }-2 \times \frac{1}{2} \frac{(45\,mc )^2}{900\,\mu F }\left[ U =\frac{ Q ^2}{2 c }\right]$

$=\frac{1}{2 \times 900 \times 10^{-6}}(8100-4050) \times 10^{-6}$

$=2.25\,Joule$

OR

Heat $=\frac{1}{2} \frac{ C _1 C _2}{ C _1+ C _2}\left( V _1- V _2\right)^2$

$=\frac{1}{2} \frac{ C ^2}{2 C }(100-0)^2$

$=\frac{1}{2} \frac{900 \times 10^{-6}}{2} \times 10^4=\frac{9}{4} \text { Joule }=2.25 \text { Joule }$

Standard 12

Physics

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