| A capacitor of \[2\mu F\] is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is: [NEET - 2016] |
 |
A) 0%
B) 20%
C) 75%
D) 80%
Correct Answer: D
Solution :
| [d] Initial energy stored in capacitor \[2\mu F\] |
| \[{{U}_{i}}=\frac{1}{2}2{{(V)}^{2}}={{V}^{2}}\] |
| Final voltage after switch 2 is ON |
| \[{{V}_{f}}=\frac{{{C}_{1}}{{V}_{1}}}{{{C}_{1}}+{{C}_{2}}}=\frac{2V}{10}=0.2\,V\] |
| Final energy in both the capacitors |
| \[{{U}_{f}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}})V_{f}^{2}\,\,\,\,=\frac{1}{2}10{{\left( \frac{2V}{10} \right)}^{2}}=0.2{{V}^{2}}\] |
| So energy dissipated |
| \[=\frac{V{{}^{2}}-0.2{{V}^{2}}}{{{V}^{2}}}\times 100=80%\] |
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