[JEE Main Online Paper Held On 12-Jan-2019 Morning]
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) \[\frac{3}{4}\frac{{{Q}^{2}}}{C}\]
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.,\[\frac{{{Q}^{2}}}{8C}-\frac{{{Q}^{2}}}{2C}=-\frac{3}{8}\frac{{{Q}^{2}}}{C}.\]
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