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Capcitors Combination can be explained as the series combinations and parallel combinations of the capacitors when one terminal is connected to the other by using different ways.
- To simplify this, whenever we combine a number of capacitors having different capacitances say C1, C2, and so on to obtain a resultant Capacitance having the required value of C is known as the combination of Capacitors.
- A capacitor consists of two conductors of any shape separated by a dielectric medium such that it can store charge.
- The combination may vary depending on the required value of the Resultant capacitance.
Key Terms: Capacitors, Series combination of capacitors, Principle of capacitors, insulators, Conductors, Parallel combination of capacitors
Read More: Equipotential Surfaces
How Capacitors are connected?
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Capacitors’ combinations can be made by different methods. The combination is attached to a battery to apply a potential difference (V) and to charge the plates (Q). The equivalent capacitance of the combination between two points can be explained as:
C = \(\frac{Q}{V}\)
Two frequently used methods of the combination include:
- Parallel combination
- Series combination
Capacitors in Series
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Capacitors are said to be combined in series only when the 2nd plate of the 1st capacitor has been connected to the 1st plate of the 2nd capacitor.
The capacitors initially carry 0 charges on them, but as the potential difference is applied across the capacitors, the capacitors start getting charged.
- Assume that the left-hand side plate of the 1st capacitor is connected to the +ve side of the battery, there will be a charge induced in capacitor 1 say, +Q.
- Now, as the charge will try to attract the electrons from the 2nd plate, the 2nd plate will be charged negatively with charge -Q. As the plate will carry positive as well as negative charge.
- But the negative will be attracted towards the 1st plate and the +ve charge (+Q) will be repelled to the 1st plate of the 2nd capacitor.
- This will be followed until all the plates of all the capacitors connected in the series have received a charge on them.
- Though the charge on each plate of the capacitor will be the same, the potential drop across each will be different because of the difference in their capacitance.
Series Combination of Capacitors
The potential drop across each capacitor can be found out by:
V= Q/Ceq……(1)
Where,
- V = Potential difference applied
- Q = charge on each plate of the capacitor
- Ceq = Equivalent capacitance
The equivalent capacitance can be found by using the following equation:
As, V = V1 + V2 + V3
1/Ceq =(V1 + V2 + V3)/Q ;
1/Ceq =V1/Q + V2/Q + V3/Q = 1/C1 + 1/C2 + 1/C3
1/Ceq = 1/C1 + 1/C2 + 1/C3
In brief, the reciprocal of the equivalent capacitors connected in series is the sum of the reciprocals of each capacitor in series.
NOTE: The equivalent capacitance will always be less than the minimum valued capacitance connected in series.
Also Read:
| Other Concepts Related to Combination of Capacitators | ||
|---|---|---|
| Screw Gauge | Ampere | Zener Diode |
| Beer Lambert Law | Dielectric Constant | Difference Between Distance and Displacement |
| Types of Capacitor | Energy Stored in a Capacitor | Cylindrical Capacitor Formula |
Capacitors in Parallel
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Consider 3 capacitors having different capacitances that have been connected in parallel, and a potential difference is applied across the combination.
From the image below it can be seen that each capacitor is connected to the source of power supply directly therefore, the potential difference between each capacitor is equal to the potential difference applied i. e. V.
- Now, initially, we have seen that the capacitor plates have no charge on them but, when we apply a potential difference they start getting charged.
- The charge on each capacitor will be different due to their difference in capacitance.
Parallel Combination of Capacitors
For example, say a total of +Q charge is released from the battery, the charge stored in each capacitor will depend on their capacitance which can be found out using the following equation.
From equation (1), Q = CV;
Therefore, the charge on each capacitor can be found by using: Q1 = C1V, Q2 = C2V, and so on.
The total charge Q1 + Q2 + Q3 = Q, as discussed.
Therefore, C1V + C2V + C3V = CV;
The equivalent capacitance can be found by the following equation:
(C1 + C2 +C3) V = CV
C = C1 + C2 + C3
NOTE: The equivalent capacitance will always be greater than the largest valued capacitor in the combination.
What are Conductors and Insulators?
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Conductors are those substances from which electric charge can travel with ease. On the other hand, electric charges cannot pass easily in insulators. This particular point is one of the most important differences between the two.
Capacitors. The potential V of a conductor is dependent upon the charge Q given to it. As per the observations, the potential of a conductor is proportional to the charge on it.
Q ∝V or Q = CV
The proportionality constant ‘C’ is referred to as the capacitance of the conductor. Thus, the numerical formula for the same will be C = Q/V
The capacity of a conductor is defined as the ratio between the charge of the conductor and the overall potential. If V equals 1, then C would be equal to Q. The capacity of a conductor is the charge needed to increase it via a unit potential.
Units used
- S.I Unit: Farad (coulomb/volt). The capacity of a conductor will be 1 farad in case a charge of 1 coulomb is needed to increase its potential by 1 volt.
- C.G.S – stat farad (stat-coulomb/stat-volt). The capacity of a conductor will be 1 stat farad if a charge of 1 statcoulomb is needed, to increase its potential through 1 statvolt.
The capacitance of an Isolated Spherical Conductor
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Consider a spherical conductor of radius ‘r’ that is isolated from other charged bodies and is located in the air. Let's assume that we charge ‘q’ to it. For doing the calculations, the charge ‘q’ is supposed to be concentrated at the center of the sphere.
The capacity of a conductor is defined as follows:
- Charge on the sphere = q
- Potential of the surface of the sphere = (1/4πε0r) (q/r)
- Capacitance, C = charge/potential = [q/(q/4πε0r)] = 4πε0r
But 1/4πε0r = 9 x 109; .
So, 4πε0r = 1/9 x 109
Therefore, C = r/9 x 109
In this equation, ‘C’ is in farad, and ‘r’ is taken in the meter. The capacitor or a condenser has an arrangement to provide more capacity in a smaller space.
Read More: Unit of Capacitance
Explain the principles of Capacitors
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The principle of a conductor states that an earthed conductor when placed in the neighborhood of a charged conductor, the capacity of the system will increase manifolds.
- Capacitors will get connected in series if the second plate of one gets connected with the first plate of the next and so on.
- This result makes the first plate of the first capacitor and the second plate of the last capacitors free.
- Let’s assume that these plates have zero charges when zero potential difference is applied across the two capacitors, in the presence of a non-zero potential difference the charge positive Q will be on the positive plate of capacitor 2.
Since the negative plate of capacitor 1 carries a negative Q charge, the positive plate needs to carry a positive charge Q. Similarly, the positive plate of capacitor 2 carries a charge +Q, and the negative plate must carry a charge -Q. The net result is that both capacitors will have the same stored charge Q. The potential drops, V1 and V2, across the two capacitors will be different. However, the total of these drops will be the same to that that of the potential drop V applied across the input and output wires.
V = V1+V2. The equivalent capacitance of the pair of capacitors will be Ceq equals Q/V.
Therefore,
1/Ceq = V/Q = (V1+V2)/Q = V1/Q + V2/Q = 1/C1 + 1/C2
The reciprocal of the resultant capacity of many capacitors, attached in series, will remain the same as the sum of the reciprocals of their individual capacities.
Energy Stored in Series Combination of Capacitors
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The equivalent capacitance in the series combination of capacitors is given by
1/C = 1/C1 + 1/C2 + 1/C3 +……...+ 1/Cn
Energy stored in a capacitor is given by
U = Q2/2C
Hence, energy stored in a series combination of capacitors is given by
U = Q2/2 x [1/C1 + 1/C2 + 1/C3 +……...+ 1/Cn]
⇒ U = Q2/2C1 + Q2/2C2 + Q2/2C3 +…….+ Q2/2Cn
⇒ U = U1 + U2 + U3 +……...+ Un
Energy Stored in Parallel Combination of Capacitors
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The equivalent capacitance in the series combination of capacitors is given by
C = C1 + C2 + C3 +……...+ Cn
Energy stored in a capacitor is given by
U = V2C/2
Hence, energy stored in a series combination of capacitors is given by
U = V2/2 x [C1 + C2 + C3 +……...+ Cn]
⇒ U = V2C1/2 + V2C2/2 + V2C3/2 +…….+ V2Cn/2
⇒ U = U1 + U2 + U3 +……...+ Un
Also Read:
Previous Year Questions
Things to Remember
- Whenever we combine a number of capacitors having different capacitance say C1, C2, and so on to obtain a resultant Capacitance having the required value of C is known as the combination of Capacitors.
- Conductors are those substances from which electric charge can travel with ease. On the other hand, electric charges cannot pass easily in insulators.
- The equivalent capacitance will always be less than the minimum valued capacitance connected in series.
- Capacitors’ combinations can be made by different methods.
- The capacitors initially carry 0 charges on them, but as the potential difference is applied across the capacitors, the capacitors start getting charged.
- S.I Unit: Farad (coulomb/volt). The capacity of a conductor will be 1 farad in case a charge of 1 coulomb is needed to increase its potential by 1 volt.
- C.G.S – stat farad (stat-coulomb/stat-volt). The capacity of a conductor will be 1 stat farad if a charge of 1 statcoulomb is needed, to increase its potential through 1 statvolt.
Sample Questions
Ques. Two capacitors of the capacitance of 6 muFand 12 muF are connected in series with a battery. The voltage across the 6 muF capacitor is 2 V. Compute the total battery voltage. (3 Marks)
Ans. C1=6μF ,V1 = 2V ; charge on 6μF capacitor is
q1 = C1V1 = (6μF×2V) =12μC
As we have seen that the charge on each capacitor is the same in the series combination, therefore, the charge on the 12μF capacitor is also 12μC
q2=C2V2
12μC=(12μF)V2
⇒V2=1V
Therefore Net potential difference V=2V1+V2 = 4 + 1= 5V V = 5 V
Result V = 5 V
Ques. Four capacitors each having a capacitance of 12μF are connected to a 500 V supply as shown in the figure. Determine-
(a) equivalent capacitance of the network and
(b) charge on each capacitor. (3 Marks)
Ans. (a.) Here C1, C2, and C3 are in series, therefore their equivalent capacitance
1C′=1C1+1C2+1C3
C′=C3=123=4μF
Now C’ and C4 are in parallel combination.
Therefore Cnet = C’ + C4
= 4muF + 12 muF = 16 μF
Cnet = 16 μF
(b.) C’ and C4 are in parallel, and a 500 V potential difference is applied across them.
Therefore Charge on C’ (equivalent capacitance)
q1=C′V=(4μF)×500=2000μC
Therefore, C1, C2 and C3 capacitors each will have 2000 μC, and charge on C4=C4×V=12×500 = 6000 μC
Ques. A system of capacitors, as shown, has a total energy of 160 mJ stored in it. Find out the equivalent capacitance of this system and the value of Z. (3 Marks)
Ans. Assumption, Ceq = C
According to question
½ CV2=160×10−3J
⇒1/2 × C × (200)2=160×10−3
⇒ Equivalent capacitance C =8μF
Since Equivalent capacitor of 7μF and 3μF connected in parallel = 7 + 3 = 10μF
Therefore, 10μF, 10μF (combination of 7μF and 3μF), and 15μF are in series combination. Therefore, If C′ is their equivalent capacitance
then 1/C’= 1/10 + 1/10 + 1/15
=(3+3+2)/30 = 8/30
or C′=15 / 4 μF
Therefore equivalent capacitance =15/ 4 + Z = C = 8μF
Z= 8 – 15/4
= (32−15) / 4
Z=17/4μF
Hence, Z = 4.25 μF
Ques. Two parallel plate capacitors X and Y have the same area of the plates and the same separation between them. X contains air between the plates whereas the capacitor Y contains a dielectric medium of εr=4 between its plates.
(a)if the equivalent of the combination is 4μF, calculate the capacitance of each capacitor, and
(b)the potential difference between the plates of X and Y. (5 Marks)
Ans. Let C be the capacitance of x because Cx =C * Cy= KC = 4 C
(a) According to the problem,
Ceq =4μF
Ceq=C1×C2 / (C1+C2)
4μF = C×4C / (C + 4C) = 4C / 5
Cχ = C = 5μF
Cy = 4C = 4×5μF = 20μF
(b)Charge across the combination
q = CeqV = (4μF) × 12 = 48μC
The potential difference across Cx
V1= qCx = 48μC / 5μF = 9.6V
The potential difference across Cy
V2 = qCy = 48μC/20μF = 2.4V
Ques. The equivalent capacitance of 3 identical capacitors connected in series is 1μF. What will be their equivalent capacitance if connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the same source. (3 Marks)
Ans. C = 1μF
In series combination,
CS=C / n
In parallel combination
⇒ Cp = nC
According to the problem,
C = nCS = 3×1μF = 3μF
For each capacitor in parallel combination
CP = nC = 3 × 3 = 9μF
Cp = 9muF
u1= ½ CsV2
u2 = ½ CpV2
Therefore u1:u2 = 1:9
Ques. The equivalent capacitance of the system of capacitors shown below in the figure is
(i) C/2
(ii) 2C
(iii) C
(iv) None of these (1 Mark)
Ans. (ii) 2C
Ques. The effective capacitance of a capacitor will be reduced when the capacitors are connected in
(i) series
(ii) parallel
(iii) series-parallel combination
(iv) none of the above (1 Mark)
Ans. 1. series
Ques. Two identical capacitors are connected in parallel combination, and charged to a potential V, and are then separated to be connected in series I .e. the positive plate of one is connected to the negative of the other, then
(i) The charges on the free plates connected together are destroyed
(ii) The charges on the free plates are enhanced
(iii) The energy stored in the system increases
(iv) The potential difference in the free plates becomes (1 Mark)
Ans. (iv) The potential difference in the free plates becomes
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