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Combination Of Capacitors - Parallel And Series - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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A parallel plate capacitor is made by stacking  n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential of each one can be made zero. Then :

 

 

In the given circuit, charge Q2 on the 2\muF capacitor changes as C is varied from 1\muF to 3\muF. Q2 as a function of ‘C’ is given properly by (figures are drawn schematically and are not to scale)

Three capacitances, each of 3 \muF, are provided. These cannot be combined to provide the resultant capacitance of :

 

Concepts Covered - 1

Combination of capacitors

Combination of capacitors:-

Capacitors can be combined in two ways.

1. Series

2. Parallel

Series combination:

If capacitors are connected in such a way that we can proceed from one point to another by only one path passing through all capacitors then all these capacitors are said to be in series.

  

Here three capacitors are connected in series and are connected across a battery of potential difference  ‘V’.

Charge:  'q' given by battery deposits at first plate of first capacitor. Due to induction it attract  '–q' on the opposite plate.The pairing +ve q charges are repelled to first plate of Second capacitor which in turn induce -q on the opposite plate. Same action is repeated to all the capacitors and in this way all capacitors get q charge. As a result ; the charge given by battery q, every capacitor gets charge q.

Potential difference:   V is the sum of potentials across all capacitors. Therefore

$
\begin{gathered}
V=v_1+v_2+v_3 \\
v_1=\frac{q_1}{c_1}, v_2=\frac{q_2}{c_2}, v_3=\frac{q_3}{c_3}
\end{gathered}
$


Equivalence equation: The equivalent capacitance for the combination of capacitance in series can be calculated as

$
C_e=\frac{q}{V}
$


Or,

$
\begin{aligned}
& 1 / \mathrm{C}_{\mathrm{e}}=\mathrm{V} / \mathrm{q} \\
&=\left(\mathrm{v}_1+\mathrm{v}_2+\mathrm{v}_3\right) / \mathrm{q} \\
&=\mathrm{v}_1 / \mathrm{q}+\mathrm{v}_2 / \mathrm{q}+\mathrm{v}_3 / \mathrm{q} \\
& 1 / \mathrm{C}_{\mathrm{e}}=1 / \mathrm{C}_1+1 / \mathrm{C}_2+1 / \mathrm{C}_3 \\
& \qquad C=\frac{c_1 c_2}{c_1+c_2}, \text { and } v_1=\frac{c_2}{c_1+c_2} \cdot V
\end{aligned}
$
 

 

Parallel  combination: If capacitors are connected in such a way that there are many paths to go from one point to other. All these paths are parallel and capacitance of each path is said to be connected in parallel.

 

Here three capacitors are connected in parallel and are connected across a battery of potential difference  ‘V’.

The potential difference across each capacitor is equal and it is same as P.D. across Battery. The charge given by source is divided and each capacitor gets some charge. The total charge $q=q_1+q_2+q_3$.

Therefore, each capacitor has charge

$
\mathrm{q}_1=\mathrm{C}_1 \mathrm{~V}_1, \mathrm{q}_2=\mathrm{C}_2 \mathrm{~V}_2, \mathrm{q}_3=\mathrm{C}_3 \mathrm{~V}_3
$


Equivalent Capacitance : We know that, $q=q_1+q_2+q_3$ when divided by v both sides,

$
\frac{q}{v}=\frac{\mathrm{q}_1}{v}+\frac{\mathrm{q}_2}{v}+\frac{\mathrm{q}_3}{v}
$


Therefore the equivalence capacitance will be:

$
\mathbf{C}=c_1+c_2+c_3
$


The equivalent capacitance in parallel increases, and it is more than largest in parallel. In parallel combination V

$
v=\frac{q_1}{c_1}=\frac{q_2}{c_2}=\frac{q_3}{c_3}
$

is same therefore . In parallel combination $q \propto c$. Larger capacitance larger is charge.

Charge distribution: $q_1=c_1 v, q_2=c_2 v, q_3=c_3 v$.

In 2 capacitor system charge on one capacitor

$
q=\frac{c_1}{c_1+c_2+\ldots} \cdot q
$


Capacitors in parallel give $\mathrm{C}=\mathrm{nc}$.

 

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Combination of capacitors

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