Question

Effective capacitance between A and B of the
network shown in the figure is

Answer 1

Given that,

All capacitor is equal.

Effective capacitance between A and B of the  network

Firstly we will simplify of the figure

According to simplify the figure,

C₂ and C₃ in parallel

We need to calculate the equivalent capacitor

Using formula of parallel

$C=C_{2}+C_{3}$

Put the value into the formula

$C'=C+C$

$C'=2C$

C₁, C' and C₄ is connected in series

We need to calculate the equivalent capacitor

Using formula of series

$\dfrac{1}{C''}=\dfrac{1}{C_{1}}+\dfrac{1}{C'}+\dfrac{1}{C_{4}}$

Put the value into the formula

$\dfrac{1}{C''}=\dfrac{1}{C}+\dfrac{1}{2C}+\dfrac{1}{C}$

$C''=\dfrac{2C}{5}$

C₆, C'' and C₅ is connected in series

We need to calculate the equivalent capacitor

Using formula of series

$\dfrac{1}{C'''}=\dfrac{1}{C_{6}}+\dfrac{1}{C''}+\dfrac{1}{C_{5}}$

Put the value into the formula

$\dfrac{1}{C'''}=\dfrac{1}{C}+\dfrac{5}{2C}+\dfrac{1}{C}$

$\dfrac{1}{C'''}=\dfrac{9}{2C}$

$C'''=\dfrac{2C}{9}$

C₇ and C₈ is connected in series

We need to calculate the equivalent capacitor

Using formula of series

$\dfrac{1}{C''''}=\dfrac{1}{C_{7}}+\dfrac{1}{C_{8}}$

Put the value into the formula

$\dfrac{1}{C''''}=\dfrac{1}{C}+\dfrac{1}{C}$

$C''''=\dfrac{C}{2}$

Now, C'''' and C''' are connected in parallel

We need to calculate the effective capacitance between A and B

Using formula of parallel

$C_{AB}=C'''+C''''$

Put the value into the formula

$C_{AB}=\dfrac{C}{2}+\dfrac{2C}{9}$

$C_{AB}= \dfrac{13C}{18}$

Hence, The effective capacitance between A and B is $\dfrac{13C}{18}$

Option is not match.