Capacitors are devices that can store electric charge and energy in an electric field. They consist of two conductive plates separated by an insulator, such as air, paper, or ceramic.
The amount of charge that a capacitor can store depends on the voltage applied across its plates, the area of the plates, the distance between them, and the type of insulator.
The capacitance of a capacitor is a measure of how much charge it can store per unit of voltage, and it is expressed in farads (F).
Capacitors can be connected in different ways to form circuits, such as in series or in parallel. In a series connection, the capacitors are connected end to end, so that the same current flows through all of them.
In a Parallel Connection
the capacitors are connected side by side, so that the same voltage is applied across all of them. The total capacitance of a circuit depends on how the capacitors are connected.
We will focus on a series connection of capacitors, and we will try to answer the question: what is the capacitance C2 of the second capacitor? To do this, we will need to use some basic formulas and concepts from electric circuits and capacitor theory.
Let us Consider the Following Circuit Diagram
where three capacitors C1, C2, and C3 are connected in series across a battery with voltage V.
The first thing we need to know is that the charge stored on each capacitor is the same, since the current is the same for all of them. Let us call this charge Q.
The second thing we need to know is that the voltage across each capacitor is proportional to its capacitance, according to the formula:
V = Q/C
where V is the voltage, Q is the charge, and C is the capacitance. Therefore, the voltage across the first capacitor is V1 = Q/C1, the voltage across the second capacitor is V2 = Q/C2, and the voltage across the third capacitor is V3 = Q/C3.
The third thing we need to know is that the total voltage across the series connection is equal to the sum of the voltages across each capacitor, according to Kirchhoff’s voltage law. This means that:
V = V1 + V2 + V3
Substituting the expressions for V1, V2, and V3, we get:
V = Q/C1 + Q/C2 + Q/C3
Simplifying, we get:
Q/V = 1/C1 + 1/C2 + 1/C3
This is the formula for the total capacitance of a series connection of capacitors, which we can call CT. Therefore, we can write:
CT = Q/V = 1/C1 + 1/C2 + 1/C3
Now, we have everything we need to find the capacitance C2 of the second capacitor.
We can rearrange the formula for CT to isolate C2, as follows:
1/C2 = 1/CT – 1/C1 – 1/C3
C2 = 1 / (1/CT – 1/C1 – 1/C3)
To calculate C2, we need to know the values of CT, C1, and C3. These values may be given in the problem statement, or they may be measured using a multimeter or other device.
For example, let us assume that CT = 0.05 F, C1 = 0.1 F, and C3 = 0.2 F. Plugging these values into the formula for C2, we get:
C2 = 1 / (1/0.05 – 1/0.1 – 1/0.2)
C2 = 1 / (20 – 10 – 5)
C2 = 1 / 5
C2 = 0.2 F
Therefore, the capacitance C2 of the second capacitor is 0.2 F. This means that the second capacitor can store 0.2 coulombs of charge per volt of voltage.
We can check our answer by using the formula for CT again, and substituting the values of C1, C2, and C3. We should get the same value of CT that we started with. Let us see:
CT = 1/C1 + 1/C2 + 1/C3
CT = 1/0.1 + 1/0.2 + 1/0.2
CT = 10 + 5 + 5
CT = 20
CT = 1/20
CT = 0.05 F
This matches the value of CT that we assumed, so our answer is correct.
In conclusion, we have learned how to find the capacitance of a capacitor in a series connection of capacitors, using some basic formulas and concepts from electric circuits and capacitor theory.
We have also learned how to check our answer using the formula for the total capacitance of a series connection.