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. $C < \dfrac{{{C_1}}}{2}$
B. $\dfrac{{{C_1}}}{2} < C < {C_1}$
C. ${C_1} < C < {C_2}$
D. ${C_2} < C < 2{C_2}$

Answer 1

Hint One way to solve this problem is to allocate two values to the two capacitances following the condition ${C_1} < {C_2}$. Then substituting these values in the series formula for capacitance, we can formulate the appropriate relation.
Formula used:
$\begin{gathered}
  \dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \\
   \Rightarrow C = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} \\
\end{gathered} $
Where $C$ is the resultant capacitance when two capacitors ${C_1}$ and ${C_2}$ are connected in series.

Complete step by step answer
A capacitor is a system of conductors and dielectric that can store electric charge. It consists of two conductors containing equal and opposite charges and has a potential difference $V$ between them.
The potential difference between the conductors is proportional to the charge on the capacitor and is given by the relation $Q = CV$where $Q$ is the charge on the positive conductor and $C$ is called the capacitance.
When two capacitors of capacitance ${C_1}$ and ${C_2}$ are connected in parallel, the resultant capacitance is given by,
$C = {C_1} + {C_2}$
And when two capacitors are connected in series then the resultant capacitance is given by,
$\begin{gathered}
  \dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \\
   \Rightarrow C = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} \\
\end{gathered} $
Now, in the given question we are being told that two capacitors ${C_1}$ and ${C_2}$are connected in series and ${C_1} < {C_2}$
Now, in order to solve this question we can assume that ${C_1} = 1\mu F$ and ${C_2} = 2\mu F$
Substituting these values in the series formula we get,
$C = \dfrac{{1 \times 2}}{{1 + 2}} = \dfrac{2}{3}$
$ \Rightarrow C \approx 0.67$
Now, from this information we can establish the relation
$\dfrac{{{C_1}}}{2} < C < {C_1} < {C_2} < 2{C_2}$

Therefore, the correct option is B.

Note In an electric circuit, the capacitor behaves as an open loop in the steady state. Which means that the voltage applied in the circuit appears across the capacitor and zero current flows through it.