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 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 :
[c] Initially the energy stored in the circuit is |
When the switch S is turned into position B the net capacitance becomes and total charge Q remains the same. |
So the energy stored will be |
So the difference of energy is dissipated in the given situation i.e. |
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