Question

A number of capacitors, each of equal capacitance $C$, are arranged as shown in the figure. The equivalent capacitance between A and B is

• A. $n^{2}C$
• B. $(2n+1)C$
• C. $\frac{(n-1)n}{2}C$
• D. $\frac{(n+1)n}{2}C$

Answer 1

The correct option is D $\frac{(n+1)n}{2}C$

As the capacitors are connected in parallel in all the respectives branches . The equivalent capacitance in respective brances can be given by $C,2C,3C,4C........nC$
Further all the branches are also connected in parallel thus, the equivalent capacitance of the whole circuit can be given by
$C_{eq}=C+2C+3C+4C...........+nC$
This is nothing but the sum of first n natural numbers
$\therefore C_{eq}=\frac{n(n+1)}{2}C$