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Two identical capacitors are connected in series and their equivalent capacitance is \(1.0 \mu \mathrm{F}\). What is each one's capacitance value? Repeat the calculation if, instead, they were connected in parallel.

Short Answer

Expert verified
Each capacitor has a capacitance of 2.0 μF when in series, and 4.0 μF in parallel.

Step by step solution

01

Understanding Capacitance in Series

When capacitors are connected in series, the reciprocal of the total capacitance is equal to the sum of the reciprocals of each capacitor's capacitance. For two identical capacitors with capacitance \(C\) each, the equation is: \[ \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} \] Given \(C_{eq} = 1.0 \mu F\), we substitute to find \(C\).
02

Solving for Individual Capacitance in Series

Rearrange the equation \(\frac{1}{C_{eq}} = \frac{2}{C}\) to solve for \(C\). Multiply both sides by \(C\) and divide by \(2\) to obtain: \[ C = \frac{2}{C_{eq}} \] Substituting \(C_{eq} = 1.0 \mu F\), we find: \[ C = \frac{2}{1.0 \mu F} = 2.0 \mu F \] Thus, each capacitor has a capacitance of \(2.0 \mu F\).
03

Understanding Capacitance in Parallel

For capacitors connected in parallel, the total capacitance is the sum of the capacitances of the individual capacitors. For two identical capacitors each with capacitance \(C\), the equation is \[ C_{eq} = C + C = 2C \]. We aim to find \(C\) given the total \(C_{eq}\).
04

Solving for Individual Capacitance in Parallel

In parallel configuration, we already know that the total capacitance is just the sum \(2C\). The problem does not specify a new total capacitance for this part, so we assume it still seeks individual capacitance from a given total. Using the equation \(C_{eq} = 2C\): \[ \text{If } C = 2.0 \mu F, \, C_{eq} = 2 \times 2.0 \mu F = 4.0 \mu F \] Thus, if each capacitor is \(2.0 \mu F\) in parallel, equivalent capacitance is \(4.0 \mu F\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance in Series
When you connect capacitors in series, the total effect is somewhat counterintuitive compared to resistors. You might think adding more components will increase the capacity, but in series, it actually lowers it. This happens because when capacitors are connected one after another, the charge has to be shared among them, effectively reducing the combined capacitance.

The key equation for capacitors in series is:
  • The reciprocal of the total capacitance ( \(C_{eq}\)) is the sum of the reciprocals of each individual capacitance ( \(C\)).
  • This is written as: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots\)
For two identical capacitors, this simplifies to: \[ \frac{1}{C_{eq}} = \frac{2}{C} \]Solving for one capacitor, you find: \[ C = \frac{2}{C_{eq}} \]This means, if the equivalent capacitance \(C_{eq}\) is known (for instance, 1.0 \(\mu F\)), each capacitor has a higher capacitance (in this case, 2.0 \(\mu F\)). By understanding this, you can see how sharing the load affects overall capacitance in a series circuit.
Capacitance in Parallel
In contrast to capacitors in series, capacitors in parallel combine their capacitance directly. When placed side by side, each capacitor can store its full charge independently while still being part of the same circuit.

Here's how it works:
  • The total capacitance ( \(C_{eq}\)) is simply the sum of all individual capacitances.
  • The equation is: \(C_{eq} = C_1 + C_2 + \cdots\)
For two identical capacitors: \[ C_{eq} = C + C = 2C \]If each capacitor in this setup has the capacitance you found in series (2.0 \(\mu F\)), then the equivalent capacitance becomes: \[ C_{eq} = 4.0 \mu F \]This shows that, when combined in parallel, capacitors act like buckets placed side by side, each holding its total amount separately. It's a much more direct addition, accentuating how parallel connections maximize the system's energy storage.
Equivalent Capacitance
Equivalent capacitance is an essential concept when dealing with complex electric circuits. It allows you to simplify and reassess the circuit's total capacitance by reducing multiple interconnected capacitors into a single, simple value.

With capacitors:
  • In series, equivalent capacitance decreases, as capacitors must share charge along the line.
  • In parallel, equivalent capacitance increases since capacitors support and reinforce each other's charge capacity.
Understanding the formula for equivalent capacitance in both configurations helps you adapt to different setups:
  • In series: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) reminds you that sharing weakens the overall capacity.
  • In parallel: \(C_{eq} = C_1 + C_2\) reinforces how capacitors together boost total capacitance.
Equipped with this knowledge, solving circuit problems becomes a lot easier. You can quickly decide how changing configurations affects overall performance. Whether you're turning a complicated network into a simpler one or determining the real-world effects in practical applications, knowing how to calculate equivalent capacitance opens up many engineering and physics possibilities.

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Most popular questions from this chapter

A 50 -pF capacitor is immersed in silicone oil, which has a dielectric constant of \(2.6 .\) When the capacitor is connected to a 24 - \(V\) battery, \((\) a) what will be the charge on the capacitor? (b) How much energy is stored in the capacitor?

A uniform electric field of \(10 \mathrm{kV} / \mathrm{m}\) points vertically upward. How far apart are the equipotential planes that differ by \(100 \mathrm{~V} ?\)

In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of \(20 \mathrm{MV}\). What is their kinetic energy if they started from rest? Give your answer in (a) \(\mathrm{eV},\) (b) \(\mathrm{keV},\) (c) \(\mathrm{MeV}\), (d) \(\mathrm{GeV},\) and (e) joules.

(a) At one-third the original distance from a positive point charge, by what factor is the electric potential changed: \((1) 1 / 3,(2) 3,(3) 1 / 9,\) or (4) \(9 ?\) Why? (b) How far from \(a+1.0-\mu C\) charge is a point with an electric potential value of \(10 \mathrm{kV} ?\) (c) How much of a change in potential would occur if the point were moved to three times that distance?

(a) The equipotential surfaces in the neighborhood of a positive point charge are spheres. Which sphere is associated with the higher electric potential: (1) the smaller one, (2) the larger one, or (3) they are associated with the same potential? (b) Calculate the amount of work (in electron-volts) it would take to move an electron from \(12.6 \mathrm{~m}\) to \(14.3 \mathrm{~m}\) away from \(\mathrm{a}+3.50-\mu \mathrm{C}\) point charge.

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