A) \[\frac{1}{k}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{2{{k}_{3}}}\]
B) \[\frac{1}{k}=\frac{1}{{{k}_{1}}+{{k}_{2}}}+\frac{1}{2{{k}_{3}}}\]
C) \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}+2{{k}_{3}}\]
D) \[k={{k}_{1}}+{{k}_{2}}+2{{k}_{3}}\]
Correct Answer: B
Solution :
\[{{C}_{1}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}\frac{A}{2}}{\left( \frac{d}{2} \right)}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A}{d}\] \[{{C}_{2}}=\frac{{{K}_{2}}{{\varepsilon }_{0}}\frac{A}{2}}{\left( \frac{d}{2} \right)}=\frac{{{K}_{2}}{{\varepsilon }_{0}}A}{d}\] and \[{{C}_{3}}=\frac{{{K}_{3}}{{\varepsilon }_{0}}A}{\left( \frac{d}{2} \right)}=\frac{2{{K}_{3}}{{\varepsilon }_{0}}A}{d}\] \[\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}+{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]\[=\frac{1}{\frac{{{\varepsilon }_{0}}A}{d}({{K}_{1}}+{{K}_{2}})}+\frac{1}{\frac{{{\varepsilon }_{0}}}{d}\times 2{{K}_{3}}}\] \[\frac{1}{{{C}_{eq}}}=\frac{d}{{{\varepsilon }_{0}}A}\left[ \frac{1}{{{K}_{1}}+{{K}_{2}}}+\frac{1}{2{{K}_{3}}} \right]\] \[{{C}_{eq}}={{\left[ \frac{1}{{{K}_{1}}+{{K}_{2}}}+\frac{1}{2{{K}_{3}}} \right]}^{-1}}.\frac{{{\varepsilon }_{0}}A}{d}\] So \[{{K}_{eq}}={{\left[ \frac{1}{{{K}_{1}}+{{K}_{2}}}+\frac{1}{2{{K}_{3}}} \right]}^{-1}}\]You need to login to perform this action.
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