A) \[\text{12 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{J}\]
B) \[\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{J}\]
C) \[\text{4}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{J}\]
D) \[\text{2}\text{.25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{J}\]
Correct Answer: B
Solution :
In capacitor, energy is stored in electric field between the plates. Increase in energy \[\Delta U={{U}_{f}}-{{U}_{i}}\] \[=\frac{1}{2}C{{V}_{f}}^{2}-\frac{1}{2}C{{V}_{i}}^{2}\] \[=\frac{1}{2}C({{V}_{f}}^{2}-{{V}_{i}}^{2})\] Given, \[C=6\mu F=6\times {{10}^{-6}},{{V}_{i}}=10\,volt,\] \[{{V}_{f}}=20\,\text{volt}\] \[\therefore \] \[\Delta U=\frac{1}{2}\times 6\times {{10}^{-6}}[{{(20)}^{2}}-{{(10)}^{2}}]\] \[=3\times {{10}^{-6}}\times 300\] \[=9\times {{10}^{-4}}J\]
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