NEET Solved Paper 2016 Question 5
Question: A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is:
Options:
A) 0%
B) 20%
C) 75%
D) 80%
Show Answer
Answer:
Correct Answer: D
Solution:
Initial energy stored in capacitor 2 μF $ U _{i}=\frac{1}{2}2{{(V)}^{2}}=V^{2} $
Final voltage after switch 2 is ON $ V _{f}=\frac{C _1V _1}{C _1+C _2}=\frac{2V}{10}=0.2,V $
Final energy in both the capacitors $ U _{f}=\frac{1}{2}(C _1+C _2)V_f^{2}=\frac{1}{2}10{{( \frac{2V}{10} )}^{2}}=0.2V^{2} $
So energy dissipated $ =\frac{V{{}^{2}}-0.2V^{2}}{V^{2}}\times 100=80% $
BETA
