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Combination of capacitors is considered one of the most asked concept.
61 Questions around this concept.
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 $Q_2 $ on the $2 \mu \mathrm{F}$ capacitor changes as C is varied from $1 \mu \mathrm{F}$ to $3 \mu \mathrm{F}$. $Q_2$ as a function of ' C ' is given properly by (figures are drawn schematically and are not to scale)
Three capacitances, each of 3 F, are provided. These cannot be combined to provide the resultant capacitance of :
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A capacitor is made of a flat plate of area $A$ and a second plate having a stair-like structure as shown in the figure. If the area of each stair is $\frac{A}{3}$ and the height is $\mathrm{d}$, the capacitance of the arrangement is:
Two capacitors of capacity C1 and C2 are connected in parallel then the equivalent capacitance is:
The capacitance between x and y is:
Four capacitors of each of capacity 3μF are connected as shown in the adjoining figure. The ratio of equivalent capacitance between A and B and between A and C will be
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Four condensers are joined as shown in the adjoining figure. The capacity of each is 4μF. The equivalent capacity between the points A and B will be
Two identical parallel plate capacitors are connected in series to a battery of 50V. A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively
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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