Question

An infinite number of identical capacitors, each of capacitance $1\mu F$ are connected as shown in the figure. Then the equivalent capacitance between A and B is :

• A. $1\mu F$
• B. $2\mu F$
• C. $\frac{1}{2}\mu F$
• D. $\infty$

Answer 1

The correct option is B $2\mu F$

Firstly calculating the equivalent capacitance of capacitors which are in series,

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

For $C_1=C_2=.....=C_n$, we get $C_{eq}=\frac{C}{n}$

Thus for first row, we get $C_A=C$

For 2nd row, we get $C_B=\frac{C}{2}$

For third row we get $C_C=\frac{C}{4}$ and so on.

Looking at the diagram all the capacitors are in parallel, now ultimate equivalent capacitance

$C_{final}=C_A+C_B+C_C+...+C_{\infty }$

${\Rightarrow C}_{final}=C+\frac{C}{2}+\frac{C}{4}+\frac{C}{8}+...+\frac{C}{\infty }$ which represents a geometric progression

${\Rightarrow C}_{final}=\frac{C}{1-\frac{1}{2}}=2C=2\mu F$