An infinity number of identical capacitors each of capacitance \(1 \mu \mathrm{F}\) are connected as shown in the figure. Then the equivalent capacitance between \(A\) and \(B\) is : Mark answer in \(\mu \mathrm{C}\)

Question

Aicharya · Accepted Answer

This combination form a G.P.
calculation of capacitance as arranged in series combination
\(\begin{array}{l}
\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}} \\
\quad \text { as } \mathrm{Ca}=\mathrm{C}: \mathrm{Cb}=\mathrm{C} / 2 ; \mathrm{Cc}=\mathrm{C} / 4 \\
\mathrm{~S}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \ldots \ldots \ldots
\end{array}\)
Sum of infinite G.P.
\(\begin{array}{l}
\mathrm{S}=\frac{a}{1-r} \\
\text { Here } \mathrm{a}=\text { first term and } \mathrm{r}=\text { common ratio } \\
=\frac{1}{2} \\
\qquad \frac{1}{1-\frac{1}{2}}=2 \Rightarrow C_{e q}=2 \mu F
\end{array}\)

An infinity number of identical capacitors each of capacitance \(1 \mu \mathrm{F}\) are connected as shown…

Solution