Question

Read the following.


Then $K_{eq}=?$

• A. $\frac{(K_1+K_2)(K_3+K_4)}{K_1+K_2+K_3+K_4}$
• B. $\frac{(K_1+K_4)(K_2+K_3)}{K_1+K_3+K_4+K_2}$
• C. $\frac{(K_2+K_4)(K_1+K_3)}{K_1+K_2+K_3+K_4}$
• D. $\frac{K_1{K}_2+K_3{K}_4}{K_1+K_2+K_3+K_4}$

Answer 1

The correct option is A $\frac{(K_1+K_2)(K_3+K_4)}{K_1+K_2+K_3+K_4}$

$C_1=\frac{K_1\frac{A}{2}{\epsilon }_0}{\frac{d}{2}}=\frac{K_{1}A{\epsilon }_0}{d}$

$C_2=\frac{K_{2}A{\epsilon }_0}{d}$

$C_3=\frac{K_{3}A{\epsilon }_0}{d}$

$C_4=\frac{K_{4}A{\epsilon }_0}{d}$

$C_{eq}=\frac{C_1{C}_3}{C_1+C_3}+\frac{C_2{C}_4}{C_2+C_4}$

$C_{eq}=K_{eq}=\frac{A{\epsilon }_0}{d}=[\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_4}{K_2+K_4}]+\frac{A{\epsilon }_0}{d}$

$K_{eq}=[\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_4}{K_2+K_4}]$