A) \[1.26\times {{10}^{-4}}J\]
B) \[2.57\times {{10}^{-4}}J\]
C) \[1.26\times {{10}^{-6}}J\]
D) \[2.57\times {{10}^{-6}}J\]
Correct Answer: A
Solution :
Store energy in capacitor of \[3\,\mu F\] \[{{U}_{1}}=\frac{1}{2}\times {{C}_{1}}{{V}^{2}}=\frac{1}{2}\times 3\times {{(6)}^{2}}\times {{10}^{-6}}\] \[=54\times {{10}^{-6}}J\] Store energy in capacitor of \[4\,\,\mu F\] \[{{U}_{2}}=\frac{1}{2}{{C}_{2}}{{V}^{2}}\] \[=\frac{1}{2}4\times {{(6)}^{2}}\times {{10}^{-6}}\] \[=72\times {{10}^{-6}}J\] When both capacitors are connected in series \[{{C}_{eq}}=\frac{{{C}_{1}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{3\times 4}{3+4}=\frac{12}{7}\mu F\] Energy lost \[=\frac{1}{2}{{C}_{eq}}\,{{({{V}_{1}}-{{V}_{2}})}^{2}}=\frac{1}{2}\times \frac{12}{7}\times {{(0)}^{2}}\times {{(10)}^{-6}}\] Total energy \[={{U}_{1}}+{{U}_{2}}\] \[=1.26\times {{10}^{-4}}J\]
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