JEE Main 12 Jan 2019 Morning Question 4
Question: In the figure shown, after the switch S is turned from position A to position B the energy dissipated in the circuit in terms of capacitance C and total charge Q is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{5}{8}\frac{Q^{2}}{C} $
B) $ \frac{1}{8}\frac{Q^{2}}{C} $
C) $ \frac{3}{8}\frac{Q^{2}}{C} $
D) $ \frac{3}{4}\frac{Q^{2}}{C} $
Show Answer
Answer:
Correct Answer: C
Solution:
Initially, the energy stored in the circuit is $ \frac{Q^{2}}{2C}. $
When the switch S is turned into position B,
the net capacitance becomes $ C+3C=4C $ and total charge Q remains the same.
So, the energy stored will be $ \frac{Q^{2}}{2(4C)}=\frac{Q^{2}}{8C}. $
So, the difference of energy is dissipated in the given situation i.e.,
$ \frac{Q^{2}}{8C}-\frac{Q^{2}}{2C}=-\frac{3}{8}\frac{Q^{2}}{C}. $
