Redistribution of charges and concept of common potential

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Question

A $5 μ F$ capacitor is charged fully by a 220V supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5 μ F$ capacitor. If the energy change during the charge redistribution is $\frac{X}{100} J$ then value of X to the nearest integer is_______.

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Solution

Energy lost during charge redistribution
$\begin{array}{l} = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2} \\ = \frac{1}{2} \frac{5 \times 2.5}{5 + 2.5} \times {(220 - 0)}^{2} \times 10^{- 6} \\ (∵ F o r u n c h a r g e d c a p a c i \tan c e V = 0) \\ = \frac{5}{6} \times {(2.2)}^{2} \times 10^{- 2} = \frac{X}{100} \\ \Rightarrow x = \frac{5 \times {(2.2)}^{2}}{6} = 4.03 \end{array}$

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Redistribution of charges and concept of common potential

Remember concepts with our Masterclasses.

A 5μF capacitor is charged fully by a 220V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5μF capacitor. If the energy change during the charge redistribution is X100J then value of X to the nearest integer is_______.

Energy lost during charge redistribution =12C1C2C1+C2V1−V22=125×2.55+2.5×220−02×10−6∵For uncharged capacitance V=0​=56×2.22×10−2=X100​⇒x=5×2.226=4.03

Ready to Test Your Skills?

free study material

A $5 μ F$ capacitor is charged fully by a 220V supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5 μ F$ capacitor. If the energy change during the charge redistribution is $\frac{X}{100} J$ then value of X to the nearest integer is_______.