Question

Two capacitors $C_1$ and $C_2$ are connected in series, assume that $C_1<C_2$. The equivalent capacitance of this arrangement is $C$, where

• A. $C<C_1/2$
• B. $C_1/2<C<C_1$
• C. $C_1<C<C_2$
• D. $C_2<C<2C_2$

Answer 1

Answer: (B)

$C_1/2<C<C_1$

We have series capacitance,
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

$⟹C=\frac{C_1{C}_2}{C_1+C_2}$
And also given that $C_2>C_1$
Hence, $2C_2>C_2>C_1>C>\frac{C_1}{2}$
Option $\left(a\right)$, $\left(c\right)$ and $\left(d\right)$ are wrong.