Capacitors in Series and Parallel
Consider an example; we are designing a circuit and want to use a capacitor but the capacitor with a required calculated capacitance is not available, then what to do to solve this issue? There’s an immediate solution, we can connect Capacitors in Series and Parallel to achieve the required value of capacitance.
Here in this article we will learn what happen when we connect Capacitors in Series or Capacitors in Parallel.
Here in this case all the current or charge passing through the first capacitor has no other available path to pass the charge, so the charge from one capacitor passes to the linked second capacitor and same like charge passes to third, fourth and so on.
As shown in the figure, the positive terminal of the DC battery is attached to the right side plate of the capacitor C1 and negative terminal of the DC battery is attached to the left side plate of the capacitor C3.
When a voltage is applied to the circuit, the negative charges in the right side plate of the capacitor C1 are attracted to the positive terminal of the battery. This results a shortage of negative charges in the right side plate of C1. As a result, the right side plate of the capacitor C1 is positively charged.
Similarly, the positive charges in the left side plate of the capacitor C3 are attracted to the negative terminal of the battery. This results a shortage of positive charges in the left side plate of C3. As a result, the left side plate of the capacitor C3 is negatively charged.
The negative charges in the left side plate of the capacitor C3 repel the negative charges in the right side plate of capacitor C3. This result the negative charges to flow from the right side plate of the capacitor C3 to left side plate of the capacitor C2. As a result, the right side plate of the capacitor C3 is positively charged and the left side plate of the capacitor C2 is negatively charged.
The negative charges in the left side plate of the capacitor C2 repel the negative charges in the right side plate of capacitor C2. This result the negative charges to flow from the right side plate of the capacitor C2 to left side plate of the capacitor C1. As an outcome, the right side plate of the capacitor C2 is positively charged and the left side plate of the capacitor C1 is negatively charged.
Therefore, all the three capacitors become charged.
We all know that the flow of charge is called as current. Since the same current is flowing through all the three capacitors, so each capacitor will hold or carry the same charge. That means if one capacitor carries a charge of 4C then the remaining capacitors also carries the same 4C charge.
So if we observe the charge on one of the capacitor, then we should find the charge on all the remaining capacitors connected with it.
To find the charge on each capacitor, first we have to find the total capacitance or equivalent capacitance.
The total capacitance can be given or calculated by;
1/C = (1/C1) + (1/C2) + (1/C3) = (1/1) + (1/2) + (1/3) = 1.83F
C = 1/1.83 = 0.546 F
By using the formula C = Q / V; we can simply find the charge stored on the equivalent capacitor;
Q = C V = 0.546 x 10 = 5.46
The charge on each of the specific capacitors in series is same as the charge on the equivalent capacitor.
So if the charge on the equivalent capacitor was 5.46 Coulombs, then the charge on each of the individual capacitors in series is going to be 5.46 Coulombs.
Therefore,
Charge on C1 = 5.46 C
Charge on C2 = 5.46 C
Charge on C3 = 5.46 C
Though, in series capacitor circuit, the voltage across each individual capacitor is dissimilar.
We can simply find the voltage across each individual capacitor by using a formula; C = Q / V
The capacitance and charge on the each individual capacitor are now we know. So we need to calculate the unknown voltage.
V = Q / C
The voltage across capacitor (C1) is V1 = Q / C1 = 5.46 / 1 = 5.46 V
The voltage across capacitor (C2) is V2 = Q / C2 = 5.46 / 2 = 2.73 V
The voltage across capacitor (C3) is V3 = Q / C3 = 5.46 / 3 = 1.82 V
The total voltage in a series capacitor circuit is equal to the sum of all the individual voltages summed together.
I.e. V = V1 + V2 + V3 = 5.46 + 2.73 + 1.82 = 10 V
Each capacitor will get the different charge when we apply a voltage to the parallel circuit,. The capacitor with high capacitance will get more charge while the capacitor with less capacitance will get less of the charge. For example, the four farad capacitor (4F) will get more charge than the two farad capacitor (2F) does get.
Ct = C1 + C2 + C3 = 1 + 2 + 3 = 6F
In the circuit diagram, the upper plates of the three capacitors are directly attached to the positive terminal of the battery and the lower plates of the three capacitors are directly attached to the negative terminal of the battery. Therefore, the voltage across all the three capacitors is same which is equal to the DC battery voltage (10 V).
But, in parallel capacitor circuit, the charge stored on each of the capacitor will be different.
By using the capacitance formula, we can simply find the charge stored on each of the capacitor;
C = Q / V
Q = C × V
The charge stored at capacitor (C1) is Q1 = C1 × V = 1 × 10 = 10 Coulomb
The charge stored at capacitor (C2) is Q2 = C2 × V = 2 × 10 = 20 Coulomb
The charge stored at capacitor (C3) is Q3 = C3 × V = 3 × 10 = 30 Coulomb
The total charge stored in the parallel capacitor circuit is equal to the sum of all the individual capacitor charges summed together.
Qt = Q1 + Q2 + Q3 = 10 + 20 + 30 = 60 Coulomb
Capacitors in Series
Capacitors in Series are a circuit in which all the capacitors are linked one after another in the same track or path so that the same charge or current flows through each capacitor.Here in this case all the current or charge passing through the first capacitor has no other available path to pass the charge, so the charge from one capacitor passes to the linked second capacitor and same like charge passes to third, fourth and so on.
Capacitors in Series Formula
Adding capacitors in series i.e. the total capacitance of capacitors in series circuit is achieved by adding up the reciprocals (1/C) of the capacitance values of each individual capacitor.Example:
Below figure shows a series capacitor circuit. The circuit involves three capacitors that are connected in series and a DC voltage source V1.Capacitors in Series |
The capacitance of three capacitors are C1 = 1F, C2 = 2F, C3 = 3F and DC voltage source V1 = 10V.
As shown in the figure, the positive terminal of the DC battery is attached to the right side plate of the capacitor C1 and negative terminal of the DC battery is attached to the left side plate of the capacitor C3.
When a voltage is applied to the circuit, the negative charges in the right side plate of the capacitor C1 are attracted to the positive terminal of the battery. This results a shortage of negative charges in the right side plate of C1. As a result, the right side plate of the capacitor C1 is positively charged.
Similarly, the positive charges in the left side plate of the capacitor C3 are attracted to the negative terminal of the battery. This results a shortage of positive charges in the left side plate of C3. As a result, the left side plate of the capacitor C3 is negatively charged.
The negative charges in the left side plate of the capacitor C3 repel the negative charges in the right side plate of capacitor C3. This result the negative charges to flow from the right side plate of the capacitor C3 to left side plate of the capacitor C2. As a result, the right side plate of the capacitor C3 is positively charged and the left side plate of the capacitor C2 is negatively charged.
The negative charges in the left side plate of the capacitor C2 repel the negative charges in the right side plate of capacitor C2. This result the negative charges to flow from the right side plate of the capacitor C2 to left side plate of the capacitor C1. As an outcome, the right side plate of the capacitor C2 is positively charged and the left side plate of the capacitor C1 is negatively charged.
Therefore, all the three capacitors become charged.
We all know that the flow of charge is called as current. Since the same current is flowing through all the three capacitors, so each capacitor will hold or carry the same charge. That means if one capacitor carries a charge of 4C then the remaining capacitors also carries the same 4C charge.
So if we observe the charge on one of the capacitor, then we should find the charge on all the remaining capacitors connected with it.
To find the charge on each capacitor, first we have to find the total capacitance or equivalent capacitance.
The total capacitance can be given or calculated by;
1/C = (1/C1) + (1/C2) + (1/C3) = (1/1) + (1/2) + (1/3) = 1.83F
C = 1/1.83 = 0.546 F
By using the formula C = Q / V; we can simply find the charge stored on the equivalent capacitor;
Q = C V = 0.546 x 10 = 5.46
The charge on each of the specific capacitors in series is same as the charge on the equivalent capacitor.
So if the charge on the equivalent capacitor was 5.46 Coulombs, then the charge on each of the individual capacitors in series is going to be 5.46 Coulombs.
Therefore,
Charge on C1 = 5.46 C
Charge on C2 = 5.46 C
Charge on C3 = 5.46 C
Though, in series capacitor circuit, the voltage across each individual capacitor is dissimilar.
We can simply find the voltage across each individual capacitor by using a formula; C = Q / V
The capacitance and charge on the each individual capacitor are now we know. So we need to calculate the unknown voltage.
V = Q / C
The voltage across capacitor (C1) is V1 = Q / C1 = 5.46 / 1 = 5.46 V
The voltage across capacitor (C2) is V2 = Q / C2 = 5.46 / 2 = 2.73 V
The voltage across capacitor (C3) is V3 = Q / C3 = 5.46 / 3 = 1.82 V
The total voltage in a series capacitor circuit is equal to the sum of all the individual voltages summed together.
I.e. V = V1 + V2 + V3 = 5.46 + 2.73 + 1.82 = 10 V
Capacitors in Parallel
The circuit with Capacitors in Parallel is an electronic circuit where all the capacitors are attached side by side in different paths so that the same charge or current will not flow through each capacitor.Each capacitor will get the different charge when we apply a voltage to the parallel circuit,. The capacitor with high capacitance will get more charge while the capacitor with less capacitance will get less of the charge. For example, the four farad capacitor (4F) will get more charge than the two farad capacitor (2F) does get.
Capacitors in Parallel Formula
Connecting capacitors in parallel will increases the size of the capacitor's plates without increasing the distance between the capacitors. Therefore the total capacitance of the parallel capacitor circuit is found by simply adding up the capacitance values of the each capacitor.Example:
Below figure shows a parallel capacitor. The circuit contains three capacitors that are connected in parallel and a DC voltage source V1.Capacitors in Parallel |
If the capacitance values of the three capacitors are; C1 = 1F, C2 = 2F, C3 = 3F and DC voltage source V1 = 10 V then the total capacitance is;
Ct = C1 + C2 + C3 = 1 + 2 + 3 = 6F
In the circuit diagram, the upper plates of the three capacitors are directly attached to the positive terminal of the battery and the lower plates of the three capacitors are directly attached to the negative terminal of the battery. Therefore, the voltage across all the three capacitors is same which is equal to the DC battery voltage (10 V).
But, in parallel capacitor circuit, the charge stored on each of the capacitor will be different.
By using the capacitance formula, we can simply find the charge stored on each of the capacitor;
C = Q / V
Q = C × V
The charge stored at capacitor (C1) is Q1 = C1 × V = 1 × 10 = 10 Coulomb
The charge stored at capacitor (C2) is Q2 = C2 × V = 2 × 10 = 20 Coulomb
The charge stored at capacitor (C3) is Q3 = C3 × V = 3 × 10 = 30 Coulomb
The total charge stored in the parallel capacitor circuit is equal to the sum of all the individual capacitor charges summed together.
Qt = Q1 + Q2 + Q3 = 10 + 20 + 30 = 60 Coulomb
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