| A parallel plate capacitor has a plate area \[A=200\,c{{m}^{2}}\]and a plate separation of \[2\,mm.\] |
 |
| A dielectric slab of dielectric constant \[k=7\]and width 1 mm fills the capacitor. Percentage of energy stored in dielectric is |
A) 12.5%
B) 25%
C) 50%
D) 100%
Correct Answer: A
Solution :
| \[\frac{{{U}_{D}}}{{{U}_{Total}}}=\frac{{{U}_{D}}}{{{U}_{gap}}+{{\mathsf{U}}_{\mathsf{D}}}}\] |
| \[=\frac{{{u}_{D}}\times {{\left( vol \right)}_{D}}}{\{{{u}_{gap}}\times {{\left( vol \right)}_{gap}}+\left( {{u}_{D}} \right){{\left( vol \right)}_{D}}\}}\]\[=\frac{\frac{1}{2}{{\in }_{o}}\frac{E_{gap}^{2}}{K}A.l}{\frac{1}{2}{{\in }_{o}}\,E_{gap}^{2}A\left( d-l \right)+\frac{1}{2}{{\in }_{o}}\frac{E_{gap}^{2}}{K}A.l}\]\[=\frac{l/K}{\left( d-l \right)+l/K}=\frac{l}{\left( d-l \right)K+l}\] |
| \[=\frac{1}{1\times 7+1}=\frac{1}{8}\] |
| \[\therefore \] Per centage of energy stored |
| \[=\frac{1}{8}\times 100=125%\] |
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