A) 1.5 and 1.33
B) 1.33 and 1.5
C) 3.0 and 2.67
D) 2.67 and 3.0
Correct Answer: A
Solution :
The total energy before connection \[=\frac{1}{2}\times 4\times {{10}^{-6}}\times {{(50)}^{2}}+\frac{1}{2}\times 2\times {{10}^{-6}}\times {{(100)}^{2}}\] \[=1.5\times {{10}^{-2}}J\] When connected in parallel \[4\times 50+2\times 100=6\times V\] Þ \[V=\frac{200}{3}\] Total energy after connection \[=\frac{1}{2}\times 6\times {{10}^{-6}}\times {{\left( \frac{200}{3} \right)}^{2}}=1.33\times {{10}^{-2}}J\]
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