A) \[25\,\mu \,F\]
B) \[20\,\mu \,F\]
C) \[40\,\mu \,F\]
D) \[5\,\mu \,F\]
Correct Answer: A
Solution :
\[{{C}_{1}}=\frac{{{\varepsilon }_{0}}\left( \frac{A}{4} \right)}{d},\,{{C}_{2}}=\frac{K{{\varepsilon }_{0}}\left( \frac{A}{2} \right)}{d},\,{{C}_{3}}=\frac{{{\varepsilon }_{0}}\left( \frac{A}{4} \right)}{d}\] \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}=\left( \frac{K+1}{2} \right)\frac{{{\varepsilon }_{0}}A}{d}\]\[=\left( \frac{4+1}{2} \right)\times 10=25\mu F\]You need to login to perform this action.
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