A) \[12\times {{10}^{-6}}J\]
B) \[9\times {{10}^{-4}}J\]
C) \[4.5\times {{10}^{-6}}J\]
D) \[2.25\times {{10}^{-6}}J\]
Correct Answer: B
Solution :
In capacitor, energy is .stored in electric field between the plates. Increase in energy \[\Delta U={{U}_{f}}-{{U}_{i}}\] \[=\frac{1}{2}C\,({{V}_{f}}^{2}-V_{i}^{2})\] \[=\frac{1}{2}C\,({{V}_{f}}^{2}-V_{i}^{2})\] Given, \[C=6\,\mu F=6\times {{10}^{-6}},{{V}_{i}}=10\] volt, \[{{v}_{f}}=20\] volt \[\therefore \] \[\Delta U=\frac{1}{2}\times 6\times {{10}^{-6}}[{{(20)}^{2}}-{{(10)}^{2}}]\] \[=3\times {{10}^{-6}}\times 300\] \[=9\times {{10}^{-4}}J\]
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