
Can I use two capacitors in series?
Answer
216.3k+ views
Hint: The capacitor is used to store electrical energy in an electric field. It is a two-terminal device. It is a little form of battery but they work differently. In the battery, the chemical energy converts into electric energy but in the capacitor, it stores electric energy or electrostatic energy in an electric field. It is having two electrical conductions that are separated by a distance and the space between these two electric conductors is filled by a vacuum.
Formula used: The capacitor is in series is written as,
$\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + ...$
Complete step by step solution:
The electric field in capacitors depends on the charge which is given to the conducting plates and we already know that the electric field is directly proportional to the potential difference \[(V{\text{)}}\] . It is written as;
$Q\alpha V \\
\Rightarrow Q = CV \\
\Rightarrow C = \dfrac{Q}{V}$
Here, $C$ is known as the capacitance of a capacitor.
Now, let us see what happens when we take 2 capacitors in series. The image of two capacitors in series is shown below,

Image: Two capacitors are in series.
Using the formula of 2 capacitors is written as;
$\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Therefore, we can see that, we can add two capacitors in series and when we add two capacitors in series. Its total capacitance will decrease as compared to the individual capacitance.
Hence, we can add 2 capacitance in series.
Note: Here we can take the capacitance in series as well as in parallel form. In series, the total capacitance will decrease as compared to the individual capacitance whereas in parallel capacitors, we are having total capacitance that will increase as compared to the individual capacitance.
Formula used: The capacitor is in series is written as,
$\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + ...$
Complete step by step solution:
The electric field in capacitors depends on the charge which is given to the conducting plates and we already know that the electric field is directly proportional to the potential difference \[(V{\text{)}}\] . It is written as;
$Q\alpha V \\
\Rightarrow Q = CV \\
\Rightarrow C = \dfrac{Q}{V}$
Here, $C$ is known as the capacitance of a capacitor.
Now, let us see what happens when we take 2 capacitors in series. The image of two capacitors in series is shown below,

Image: Two capacitors are in series.
Using the formula of 2 capacitors is written as;
$\dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Therefore, we can see that, we can add two capacitors in series and when we add two capacitors in series. Its total capacitance will decrease as compared to the individual capacitance.
Hence, we can add 2 capacitance in series.
Note: Here we can take the capacitance in series as well as in parallel form. In series, the total capacitance will decrease as compared to the individual capacitance whereas in parallel capacitors, we are having total capacitance that will increase as compared to the individual capacitance.
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