Unraveling the Mystery of Capacitors in Parallel and Series
Introduction
The behavior of capacitors in parallel and series configurations can sometimes present intriguing challenges. This article aims to delve into the details of two capacitors with unusual and counterintuitive characteristics: their equivalent capacitances when connected in parallel and series are 9 picofarads (pF) and 1.5 pF, respectively. We will explore the underlying principles and solve for the capacitance of each capacitor, using fundamental formulas that govern the behavior of capacitors in both series and parallel configurations.
The Principles of Parallel and Series Capacitors
Before diving into the specific problem, it's essential to understand the basic principles of how capacitors behave in parallel and series configurations.
Parallel Connection
In a parallel connection, the equivalent capacitance (Ctotal) is the sum of the individual capacitances. This is represented by the formula:
Ctotal C1 C2
This means that the total capacitance in a parallel combination increases with the number of capacitors and their values.
In contrast, when capacitors are connected in series, the equivalent capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. This is expressed as:
1/Ctotal 1/C1 1/C2
This formula shows that the equivalent capacitance in a series combination is less than the capacitance of any single capacitor in the series, making it a less effective storage device in such configurations.
Solving for the Capacitance of Each Capacitor
Given the specific characteristics of the capacitors in question, we know that:
The equivalent capacitance in parallel: Ctotal 9 pF The equivalent capacitance in series: 1/Ctotal 1.5 pFLet's denote the capacitance of the two capacitors as C1 and C2. Using the formulas for parallel and series connections, we can set up the following equations:
Parallel Connection Equation
From the parallel connection equation:
C1 C2 9 pF
From the series connection equation:
1/(1/C1 1/C2) 1.5 pF
Rewriting this:
1/C1 1/C2 1/(1.5 pF) 0.6667 pF-1
Step-by-Step Solution
To solve for C1 and C2, we follow these steps:
Step 1: Equations Setup
We now have two equations:
C1 C2 9 1/C1 1/C2 0.6667Let's denote C1 as x and C2 as y. So our equations become:
x y 9
1/x 1/y 0.6667
Step 2: Solve for y in terms of x
From the first equation, we can express y in terms of x:
y 9 - x
Substituting this into the second equation:
1/x 1/(9 - x) 0.6667
Step 3: Simplify and Solve the Equation
Multiplying through by x(9 - x) to clear the denominators:
(9 - x) x 0.6667x(9 - x)
9 0.6667x(9 - x)
9 6x - 0.6667x^2
0.6667x^2 - 6x 9 0
Multiplying through by 3 to clear the decimal:
2x^2 - 18x 27 0
This can be simplified using the quadratic formula x (-b ± √(b^2 - 4ac)) / (2a), where a 2, b -18, and c 27:
x (18 ± √(18^2 - 4 * 2 * 27)) / (2 * 2)
x (18 ± √(324 - 216)) / 4
x (18 ± √108) / 4
x (18 ± 10.39) / 4
So, x (28.39 / 4) or (7.61 / 4)
x 7.1 or 1.9
Thus, C1 7.1 pF and C2 1.9 pF (or vice versa).
Conclusion
The unique characteristics of these capacitors, with a large parallel capacitance and a small series capacitance, present an intriguing problem to explore. By applying the principles of parallel and series connections and solving a system of equations, we can determine the individual capacitances of the capacitors. The solution reveals how these capacitors behave in different configurations, offering valuable insights into their properties and behavior.