A) \[300\,\mu J\]
B) \[450\,\mu J\]
C) \[225\,\mu J\]
D) \[245.0\,J\]
E) \[100\,\mu J\]
Correct Answer: A
Solution :
Energy \[{{E}_{1}}=\frac{1}{2}{{C}_{1}}V_{1}^{2}=\frac{1}{2}\times 1\times {{10}^{-6}}\times {{(30)}^{2}}\] \[=450\times {{10}^{-6}}J\] Common potential \[V=\frac{{{q}_{1}}+{{q}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] \[=\frac{1\times 30+0}{1+2}=10\,volt\] \[{{E}_{2}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}}\] \[=\frac{1}{2}(1+2)\times {{10}^{-6}}\times {{(10)}^{2}}\] \[=1.5\times 100\times {{10}^{-6}}\] \[=150\times {{10}^{-6}}J\] Loss of energy\[={{E}_{2}}-{{E}_{1}}=300\,\mu J\]
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