Electrostatics Question 269

Question: An infinite number of identical capacitors each of capacitance $ 1\mu F $ are connected as in . Then the equivalent capacitance between $ A $ and $ B $ is [EAMCET 1990]

Options:

A) Fraction of stored energy after 1 second is 16/25

B) Potential difference between the plates after 2 seconds will be 32 V

C) Potential difference between the plates after 2 seconds will be 20 V

D) Fraction of stored energy after 1 second is 4/5

Show Answer

Answer:

Correct Answer: A , B

Solution:

By using $ V=V _{0}{{e}^{-t/CR}}\Rightarrow 40=50{{e}^{-1/CR}}\Rightarrow {{e}^{-1/CR}}=4/5 $ Potential difference after 2 sec

$ {V}’=V _{0}{{e}^{-2/CR}}=50{{({{e}^{-1/CR}})}^{2}}=50{{( \frac{4}{5} )}^{2}}=32V $

Fraction of energy after 1 sec $ = $ $ \frac{\frac{1}{2}C{{(V _{f})}^{2}}}{\frac{1}{2}C{{(V _{i})}^{2}}}={{( \frac{40}{50} )}^{2}}=\frac{16}{25} $

Fraction of stored energy after 1 second is 16/25 (a, b)

By using $ V=V _{0}{{e}^{-t/CR}}\Rightarrow 40=50{{e}^{-1/CR}}\Rightarrow {{e}^{-1/CR}}=4/5 $

Potential difference after 2 sec $ {V}’=V _{0}{{e}^{-2/CR}}=50{{({{e}^{-1/CR}})}^{2}}=50{{( \frac{4}{5} )}^{2}}=32V $ Fraction of energy after 1 sec $ = $

$ \frac{\frac{1}{2}C{{(V _{f})}^{2}}}{\frac{1}{2}C{{(V _{i})}^{2}}}={{( \frac{40}{50} )}^{2}}=\frac{16}{25} $



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