Question

Find the capacitance of the infinite ladder between point $A$ and $B$

• A. $4.3\mu F$
• B. $10\mu F$
• C. $8.6\mu F$
• D. $3\mu F$

Answer 1

Answer: (A)

Let $C$ be the capacitance of infinite ladder
$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{5}=\frac{5+C}{5C}$
$C^{\prime }=\frac{5C}{5+C}$
$C^{\prime \prime }=2+C^{\prime }$
$C^{\prime \prime }=2+\frac{5C}{5+C}$
$C^{\prime \prime }=C$ (Addition of one or more $5\mu F$ and $2\mu F$ do not change the total capacitance. So, overall capacitance should be $C$ )
$C=\frac{10+2C+5C}{5+C}$
$5C+C^2=10+7C$
$C^2=10+2C$
$C^2-2C-10=0$
$D=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\therefore C=\frac{2\pm \sqrt{44}}{2}$
$C=\frac{2+6.6}{2},C=\frac{2-6.6}{2}$
$C=\frac{8.6}{2},C=\frac{-4.6}{2}$
$C=-2.3,C=4.3$
-ve value neglected
So, total capacitance is $4.3\mu F$