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} $
