Capacitors combination can be made in many ways. The combination is connected to a battery to apply a potential difference (V) and charge the plates (Q). We can define the equivalent capacitance of the combination between two points to be: C = Q/V
Two frequently used methods of combination are:
When one plate of one capacitor is connected with one plate of the other capacitor, such combination is called parallel combination.
All capacitors have the same potential difference but different charges.
We can say that : Q_{1} = C_{1}V
Q_{1} = Charge on capacitor C_{1}
C_{1} = Capacitance of capacitor C_{1 }
V = Potential across capacitor C_{1}
The charge on the capacitor is proportional to its capacitance Q µ C
Where Q = Q_{1} + Q_{2} + Q_{3} ..............
A parallel combination of three capacitors, with one plate of each capacitor connected to one side of the circuit and the other plate connected to the other side, is illustrated in Figure.
Since the capacitors are connected in parallel, they all have the same voltage V across their plates. However, each capacitor in the parallel network may store a different charge. To find the equivalent capacitance 𝐶𝑝 of the parallel network, we note that the total charge Q stored by the network is the sum of all the individual charges:
𝑄 = 𝑄_{1 }+ 𝑄_{2 }+ 𝑄_{3}
On the lefthand side of this equation, we use the relation 𝑄 = 𝐶𝑝𝑉, which holds for the entire network. On the righthand side of the equation, we use the relations
𝑄_{1 }= 𝐶_{1}𝑉 , 𝑄_{2 }= 𝐶_{2}𝑉 , and 𝑄_{3 }= 𝐶_{3}𝑉 for the three capacitors in the network.
In this way we obtain 𝐶𝑝𝑉 = 𝐶_{1}𝑉 + 𝐶_{2}𝑉 + 𝐶_{3}𝑉.
This equation, when simplified, is the expression for the equivalent capacitance of the parallel network of three capacitors: 𝐶𝑝 = 𝐶_{1 }+ 𝐶_{2 }+ 𝐶_{3}
This expression is easily generalized to any number of capacitors connected in parallel in the network.
When initially uncharged capacitors are connected as shown, then the combination is called series combination
All capacitors will have the same charge but different potential difference across them.
We can say that
V_{1} = potential across C_{1}
Q = charge on positive plate of C_{1}
C_{1} = capacitance of capacitor similarly
V_{1} : V_{2} : V_{3} = _{}
We can say that potential difference across capacitor is inversely proportional to its capacitance in series combination.
Let the capacitance of each capacitor be C_{1}, C_{2 }and C_{3 }and their equivalent capacitance is C_{eq}.
As these capacitors are connected in series, thus charge across each capacitor is same as Q. When some electrical components, let say 3, are connected in series with each other, the potential difference of the battery V gets divided across each component as
V_{1}, V_{2 }and V_{3 }as shown in the figure.
∴ V = V_{1 }+ V_{2 }+ V_{3}
Using V = Q/C
Equivalent capacitance for series combination =
In general,
Example 1: Find charge on each capacitor.
Sol. Charge on C_{1} = C_{1}V_{1} = 2 × (20  5)μC
= 30 μC
Charge on C_{2} = C_{2}V_{2} = 2 × (20  (10))μC
= 60 μC
Charge on C_{3} = C_{3}V_{3} = 4 × (20  10)μC
= 40 μC
Example 2: Find charge on each capacitor.
Sol. Charge on C_{1} = (x  10) C_{1}
Charge on C_{2} = (x  0) C_{2}
Charge on C_{3} = (x  20) C_{3}
Now from charge conservation at node x
(x  10)C_{1} (x  0)C_{2} (x  20)C_{3} = 0
⇒ 2x  20 2x 4x  80 = 0
⇒ x = 25 Therefore
so
Example 3: In the given circuit find out the charge on each capacitor. (Initially they are uncharged)
Sol. Let potential at A is 0, so at D it is 30 V, at F it is 10 V and at point G potential is 25V. Now apply Kirchhoff's I^{st} law at point E. (total charge of all the plates connected to 'E' must be same as before i.e. 0)
Therefore, (x  10) + (x  30) 2 +(x 25) 2 = 0
5x = 20
x = 4
Final charges :
Q_{2mF} = (30  4) 2 = 52 mC
Q_{1mF} = (10  4) = 6 mC
Q_{2mF} = (4  (25)) 2 = 58 mC
Example 4:
Find voltage across capacitor C_{1}.
Sol.
Now from charge conservation at node x and y
for x
(x  4)C_{1} + (x  2)C_{2} + (x  y)C_{3} = 0 ⇒
2(x  4) + 2(x  2) (x  y) 2 = 0
6x  2y  12 = 0 .....(1)
For y
(y  x)C_{3} + [y (4)]C_{4} (y  0)C_{5} = 0 ⇒ (y  x)2 (y 4) 2 y 2 = 0
= 6y  2x 8 = 0 .....(2)
eq. (1) & (2)
y =  3 Therefore
x = 7 Therefore
So potential difference = x  y =
Example 5: Three initially uncharged capacitors are connected in series as shown in circuit with a battery of emf 30V. Find out following :
(i) charge flow through the battery,
(ii) potential energy in 3 mF capacitor.
(iii) U_{total} in capacitors
(iv) heat produced in the circuit
Sol.
C_{eq} = 1 μF.
(i) Q = C_{eq} V = 30 μC
(ii) charge on 3μF capacitor = 30 μC
energy = = = 150 μJ
(iii) U_{total} = = 450 μJ
(iv) Heat produced = (30 μC) (30)  450 μJ = 450 μJ
Example 6: Two capacitors of capacitance 1 mF and 2mF are charged to potential difference 20 V and 15 V as shown in figure. If now terminal B and C are connected together terminal A with positive of battery and D with negative terminal of battery then find out final charges on both the capacitor.
Now applying kirchhoff voltage law
 40  2q  30  q =  60
3q =  10
Charge flow =  μC.
Charge on capacitor of capacitance 1μF = 20 q =
Charge on capacitor of capacitance 2μF = 30 q =
261 videos249 docs232 tests

1. How are capacitors connected in parallel? 
2. What is the formula for calculating the total capacitance in a parallel combination of capacitors? 
3. How are capacitors connected in series? 
4. What is the formula for calculating the total capacitance in a series combination of capacitors? 
5. Can capacitors be combined in both parallel and series combinations in the same circuit? 

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