Capacitors Capacities 2 PF, 3 PF, And 6 PF Are Connected In Series. The Combination Is Connected To A 100 V Battery. What Is The Energy Stored In The System?

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In this article, we will explore the concept of capacitors connected in series and delve into the calculation of the total energy stored within such a system. Specifically, we will address the scenario involving three capacitors with capacitances of 2 pF (picofarads), 3 pF, and 6 pF, respectively, connected in series to a 100 V (volt) battery. We will meticulously walk through the steps to determine the equivalent capacitance of the series combination, and subsequently, compute the total energy stored in the entire system. Understanding how capacitors behave in series circuits is crucial for various applications in electronics, ranging from energy storage to signal filtering. So, let's dive into the physics behind this circuit configuration and unlock the secrets of energy storage in capacitors.

When capacitors are connected in series, they form a chain where the same amount of charge flows through each capacitor. Unlike parallel connections, the voltage across each capacitor in a series connection is not the same, but the sum of the voltages across each capacitor equals the total voltage applied across the series combination. This fundamental characteristic stems from the conservation of charge. The key concept to grasp here is the determination of the equivalent capacitance for capacitors in series. The reciprocal of the equivalent capacitance (C{C}eq) is equal to the sum of the reciprocals of the individual capacitances. Mathematically, this is expressed as:

1Ceq=1C1+1C2+1C3+...{ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... }

Where C1{C_1}, C2{C_2}, C3{C_3} and so on, represent the individual capacitances in the series circuit. This formula is the cornerstone for simplifying series capacitor networks and allows us to treat the entire combination as a single equivalent capacitor for calculations. Once the equivalent capacitance is known, we can then proceed to determine the total charge stored in the system and, subsequently, the total energy stored. Before we jump into the calculations for our specific problem, it's important to remember that the unit for capacitance is the farad (F), but in practical circuits, we often encounter smaller units like microfarads (μF), nanofarads (nF), and picofarads (pF). Therefore, careful attention must be paid to unit conversions when performing calculations. The relationship between these units is as follows: 1 μF = 10^-6 F, 1 nF = 10^-9 F, and 1 pF = 10^-12 F. With this understanding of series capacitors and the crucial formula for equivalent capacitance, we are well-equipped to tackle the problem at hand and calculate the total energy stored in the system.

The first crucial step in determining the energy stored in our series capacitor network is to calculate the equivalent capacitance. As we discussed earlier, the formula for the equivalent capacitance (C{C}eq) of capacitors in series is given by the reciprocal of the sum of the reciprocals of individual capacitances. In our specific scenario, we have three capacitors with capacitances C1{C_1} = 2 pF, C2{C_2} = 3 pF, and C3{C_3} = 6 pF. To apply the formula effectively, we first need to ensure that all capacitance values are in the same unit, which in this case is picofarads (pF). Now, let's plug these values into the formula:

1Ceq=12 pF+13 pF+16 pF{ \frac{1}{C_{eq}} = \frac{1}{2 \text{ pF}} + \frac{1}{3 \text{ pF}} + \frac{1}{6 \text{ pF}} }

To add these fractions, we need to find a common denominator, which in this case is 6. So, we rewrite the equation as:

1Ceq=36 pF+26 pF+16 pF{ \frac{1}{C_{eq}} = \frac{3}{6 \text{ pF}} + \frac{2}{6 \text{ pF}} + \frac{1}{6 \text{ pF}} }

Adding the fractions on the right side, we get:

1Ceq=3+2+16 pF=66 pF{ \frac{1}{C_{eq}} = \frac{3 + 2 + 1}{6 \text{ pF}} = \frac{6}{6 \text{ pF}} }

This simplifies to:

1Ceq=11 pF{ \frac{1}{C_{eq}} = \frac{1}{1 \text{ pF}} }

Taking the reciprocal of both sides, we find the equivalent capacitance:

Ceq=1 pF{ C_{eq} = 1 \text{ pF} }

Therefore, the equivalent capacitance of the series combination of the three capacitors is 1 pF. This value represents the capacitance of a single capacitor that would store the same amount of charge as the entire series network when subjected to the same voltage. With the equivalent capacitance now determined, we can proceed to calculate the total charge stored in the system, which is a necessary step before we can finally calculate the total energy stored. Remember, the equivalent capacitance effectively simplifies the entire series network into a single capacitor for the purpose of calculations, making the subsequent steps more straightforward. In the next section, we will use this equivalent capacitance to determine the total charge stored when the capacitor network is connected to the 100 V battery.

Now that we have calculated the equivalent capacitance (Ceq{C_{eq}}) of the series combination to be 1 pF, the next crucial step is to determine the total charge (Q{Q}) stored in the system when connected to the 100 V battery. The fundamental relationship between charge, capacitance, and voltage is given by the equation:

Q=C×V{ Q = C \times V }

Where:

  • Q{Q} represents the charge stored in coulombs (C).
  • C{C} represents the capacitance in farads (F).
  • V{V} represents the voltage in volts (V).

In our case, we have the equivalent capacitance Ceq{C_{eq}} = 1 pF and the voltage V{V} = 100 V. However, before we can plug these values into the equation, we need to ensure that the capacitance is expressed in farads, the standard unit for capacitance. We know that 1 pF = 10−12{10^{-12}} F, so we convert the equivalent capacitance:

Ceq=1 pF=1×10−12 F{ C_{eq} = 1 \text{ pF} = 1 \times 10^{-12} \text{ F} }

Now, we can substitute the values of Ceq{C_{eq}} and V{V} into the equation for charge:

Q=(1×10−12 F)×(100 V){ Q = (1 \times 10^{-12} \text{ F}) \times (100 \text{ V}) }

Q=100×10−12 C{ Q = 100 \times 10^{-12} \text{ C} }

Q=1×10−10 C{ Q = 1 \times 10^{-10} \text{ C} }

Therefore, the total charge stored in the system is 1×10−10{1 \times 10^{-10}} coulombs. This charge is distributed across the three capacitors in the series combination, with each capacitor holding the same amount of charge due to the nature of series connections. Knowing the total charge stored and the equivalent capacitance, we are now in a position to calculate the total energy stored in the capacitor network. The energy stored in a capacitor is directly related to the capacitance and the square of the voltage, or alternatively, can be expressed in terms of charge and capacitance. In the next section, we will utilize the values we have calculated for equivalent capacitance and total charge to determine the total energy stored in the system, completing our analysis of the series capacitor network.

With the equivalent capacitance (Ceq{C_{eq}}) and the total charge (Q{Q}) stored in the system determined, we can now proceed to calculate the total energy (U{U}) stored in the capacitor network. There are two primary formulas we can use to calculate the energy stored in a capacitor. The first formula expresses energy in terms of capacitance and voltage:

U=12CV2{ U = \frac{1}{2} C V^2 }

And the second formula expresses energy in terms of charge and capacitance:

U=12Q2C{ U = \frac{1}{2} \frac{Q^2}{C} }

Since we have already calculated both the equivalent capacitance and the total charge stored, we can use either formula to find the total energy stored. For simplicity, let's use the second formula, which involves the charge and capacitance:

U=12Q2Ceq{ U = \frac{1}{2} \frac{Q^2}{C_{eq}} }

We have Q=1×10−10{Q = 1 \times 10^{-10}} C and Ceq=1×10−12{C_{eq} = 1 \times 10^{-12}} F. Substituting these values into the formula, we get:

U=12(1×10−10 C)21×10−12 F{ U = \frac{1}{2} \frac{(1 \times 10^{-10} \text{ C})^2}{1 \times 10^{-12} \text{ F}} }

First, let's square the charge:

(1×10−10 C)2=1×10−20 C2{ (1 \times 10^{-10} \text{ C})^2 = 1 \times 10^{-20} \text{ C}^2 }

Now, substitute this back into the energy equation:

U=121×10−20 C21×10−12 F{ U = \frac{1}{2} \frac{1 \times 10^{-20} \text{ C}^2}{1 \times 10^{-12} \text{ F}} }

Divide the numerator by the denominator:

U=12×10−20+12C2F{ U = \frac{1}{2} \times 10^{-20 + 12} \frac{\text{C}^2}{\text{F}} }

U=12×10−8 J{ U = \frac{1}{2} \times 10^{-8} \text{ J} }

U=0.5×10−8 J{ U = 0.5 \times 10^{-8} \text{ J} }

U=5×10−9 J{ U = 5 \times 10^{-9} \text{ J} }

Therefore, the total energy stored in the system is 5×10−9{5 \times 10^{-9}} joules, which can also be expressed as 5 nanojoules (nJ). This energy is stored in the electric field created between the plates of the capacitors. The energy storage capacity of a capacitor network is a crucial parameter in many electronic applications, and understanding how to calculate this energy is essential for circuit design and analysis. We have successfully determined the total energy stored in the series capacitor network by first calculating the equivalent capacitance, then finding the total charge stored, and finally applying the energy formula. This step-by-step approach demonstrates the interconnectedness of these electrical concepts and highlights the importance of a solid understanding of the underlying principles.

In summary, we have successfully calculated the total energy stored in a series network of three capacitors with capacitances 2 pF, 3 pF, and 6 pF connected to a 100 V battery. The process involved several key steps, each building upon the previous one. First, we calculated the equivalent capacitance of the series combination using the formula for capacitors in series. This equivalent capacitance allowed us to treat the entire network as a single capacitor for subsequent calculations. Next, we determined the total charge stored in the system using the relationship between charge, capacitance, and voltage. It was crucial to ensure that all units were consistent, particularly converting picofarads to farads before applying the formula. Finally, with the equivalent capacitance and total charge known, we calculated the total energy stored in the system using the formula that relates energy to charge and capacitance. The result showed that the total energy stored in the series capacitor network is 5×10−9{5 \times 10^{-9}} joules, or 5 nanojoules. This methodical approach highlights the importance of understanding the fundamental principles of capacitor behavior in series circuits and the relationships between charge, capacitance, voltage, and energy. The ability to calculate the energy stored in capacitor networks is essential in various fields, including electronics, electrical engineering, and physics. From designing energy storage systems to analyzing circuit performance, these calculations play a crucial role. The principles and methods discussed in this article provide a solid foundation for understanding more complex capacitor networks and their applications. By mastering these concepts, one can effectively analyze and design circuits that utilize capacitors for a wide range of purposes. The journey from understanding the basics of series capacitors to calculating the energy stored demonstrates the power of applying fundamental principles to solve practical problems in electrical circuits.