A) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{2}})({{K}_{2}}+{{K}_{4}})}{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}}}\]
B) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{2}})({{K}_{2}}+{{K}_{4}})}{2({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]
C) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{2}})({{K}_{3}}+{{K}_{4}})}{({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]
D) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{4}})({{K}_{2}}+{{K}_{3}})}{2({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]
E) None of these
Correct Answer: E
Solution :
[Bonus] \[{{C}_{eq.}}\,=\,\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\,+\,\frac{{{C}_{3}}{{C}_{4}}}{{{C}_{3}}+{{C}_{4}}}\] \[{{C}_{1}}\,=\,\frac{{{k}_{1}}\,{{\in }_{0}}\,A/2}{d/2}\,=\,\frac{{{k}_{1}}\,{{\in }_{0}}\,A}{d}\] Similarly\[{{C}_{2}}\,=\,\frac{{{k}_{2}}\,{{\in }_{0}}\,A}{d},\,\,{{C}_{3}}\,\,=\,\,\frac{{{k}_{3}}\,{{\in }_{0}}\,A\,\,}{d}\],\[{{C}_{4}}\,=\,\frac{{{k}_{4}}\,{{\in }_{0}}\,A}{d}\] \[{{C}_{eq.}}\,=\,\,\frac{{{k}_{eq.}}\,A}{d}\] \[{{C}_{eq.}}\,=\,\frac{\frac{{{k}_{1}}\,{{\in }_{0}}\,A}{d}\,\,.\,\,\frac{{{k}_{2}}\,{{\in }_{0}}\,A}{d}}{\frac{{{k}_{1}}\,{{\in }_{0}}\,A}{d}+\frac{{{k}_{2}}\,{{\in }_{0}}\,A}{d}}\,\,+\,\,\frac{\frac{{{k}_{3}}\,{{\in }_{0}}\,A}{d}.\,\frac{{{k}_{4}}\,{{\in }_{0}}\,A}{d}}{\frac{{{k}_{3}}\,{{\in }_{0}}\,A}{d}\,+\frac{{{k}_{4}}\,{{\in }_{0}}\,A}{d}}\]] \[{{C}_{eq.}}\,=\,\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\,\,\frac{{{\in }_{0}}A}{d}\,\,+\,\,\frac{{{k}_{3}}{{k}_{4}}}{{{k}_{3}}+{{k}_{4}}}\,\,\frac{{{\in }_{0}}A}{d}\] Now, \[{{k}_{eq.}}\,=\,\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\,\,+\,\,\frac{{{k}_{3}}{{k}_{4}}}{{{k}_{3}}+{{k}_{4}}}\,\,\]You need to login to perform this action.
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