Question

What is the equivalent capacitance of the infinite ladder network shown in figure between points $\text{A}$ and $\text{B}$.

• A. $\frac{4}{3}\mu \text{F}$
• B. $\frac{2}{3}\mu \text{F}$
• C. Infinite
• D. $(1+\sqrt{3})\mu \text{F}$

Answer 1

The correct option is A $\frac{4}{3}\mu \text{F}$

Since, the circuit is symmetrical about the line $\text{AB}$. So there is no charge will accumulate on middle capacitors $\left(1\mu \text{F}\right)$. Thus, middle capacitors become ineffective, and we can remove it.


Since, capacitors above and below the line $\text{AB}$ are in series, so their equivalent capacitance will be

$\frac{1}{C_{eq}}=\frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+.......$

Using infinite G.P.,

$\frac{1}{C_{eq}}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}(S_{\infty }=a/(1-r\left)\right)$

$\therefore C_{eq}=\frac{2}{3}$

Now, circuit can be redrawn as
Both capacitors are in parallel. So, equivalent capacitance,

$C_{AB}=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}\mu \text{F}$

Therefore, option $\left(A\right)$ is correct.