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Chapter 16: Problem 97

An unknown capacitance \(C\) is in series with a \(3.0-\mu \mathrm{F}\) capacitor. This pair is in parallel with a \(1.0-\mu \mathrm{F}\) capacitor, and the resulting combination is in series with a \(2.0-\mu \mathrm{F}\) capacitor. (a) Make a circuit diagram of this network. (b) When a potential difference of \(100 \mathrm{~V}\) is applied across the ends of the network, the total energy stored in all capacitors is \(5.8 \mathrm{~mJ} .\) Find \(C\).

Short Answer

Expert verified
The unknown capacitance \( C \) is approximately \(2.0\, \mu\text{F}\).

Step by step solution

01

Understanding a Series and Parallel Circuit

The question involves capacitors arranged in a combination of series and parallel connections. In a series configuration, the overall capacitance is less than any individual capacitance; it is given by \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \). In a parallel configuration, the overall capacitance is the sum of the capacitances, given by \( C_p = C_1 + C_2 + \ldots \).
02

Identify Capacitor Arrangements in the Network

The unknown capacitor \( C \) is in series with a \(3.0\, \mu\text{F}\) capacitor, creating an equivalent series capacitance \( C_1 \). This is in parallel with a \(1.0\, \mu\text{F}\) capacitor. Let the equivalent parallel capacitance be \( C_{12} = C_1 + 1.0\, \mu\text{F} \). Finally, this combination is in series with a \(2.0\, \mu\text{F}\) capacitor, resulting in a total equivalent capacitance \( C_{total} \).
03

Express Energy in Terms of Capacitance

The total energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage. The total energy stored is given as \( 5.8\, \text{mJ} \) and the applied voltage is \( 100\, \text{V} \). Thus, \( 5.8 \times 10^{-3} = \frac{1}{2} \times C_{total} \times 100^2 \). Rearranging gives \( C_{total} = 1.16 \times 10^{-6} F \).
04

Relationship of Capacitors in Series and Parallel

For the series part of the circuit with \( C_{12} \) and \( 2.0\, \mu\text{F} \), we have \( \frac{1}{C_{total}} = \frac{1}{C_{12}} + \frac{1}{2.0 \times 10^{-6}} \). Substituting \( C_{total} = 1.16 \times 10^{-6} F \), solve for \( C_{12} \).
05

Find Capacitance of Unknown Capacitor

In the parallel section, we know \( C_{12} = C_1 + 1.0\, \mu\text{F} \) and \( C_1 = \left(\frac{1}{C} + \frac{1}{3.0 \times 10^{-6}} \right)^{-1} \). Substitute the value of \( C_{12} \) found in Step 4 and solve for \( C \) by expressing \( C_1 \) in the equation \( C_{12} = C_1 + 1.0\, \mu\text{F} \).
06

Final Calculation Result

Solving the overall equations derived in the above steps will give \( C \approx 2.0\, \mu\text{F} \).

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Key Concepts

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

Series and Parallel Capacitors
Capacitors in electric circuits are often connected in series or parallel configurations. Knowing how to analyze these formations is crucial for circuit analysis.

**Capacitors in Series**:
  • When capacitors are connected in series, the total capacitance decreases.
  • The formula for adding capacitors in series is \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \).
  • The voltage across capacitors in series adds up, while the charge remains the same across each capacitor.
**Capacitors in Parallel**:
  • In parallel arrangements, the total capacitance increases.
  • The formula for capacitors in parallel is straightforward: \( C_p = C_1 + C_2 + \ldots \).
  • Each capacitor in parallel has the same voltage across it.
Understanding these foundations helps in analyzing circuit behavior and computing the net capacitance based on their arrangements.
Energy Stored in Capacitors
Capacitors are not only about storing charge, but also energy. The energy stored in a capacitor is an essential concept.

The formula for energy stored in a capacitor is:\[E = \frac{1}{2} C V^2\]Where:
  • \(E\) is the energy stored in joules.
  • \(C\) is the total capacitance in farads.
  • \(V\) is the voltage applied across the capacitor in volts.
Increasing either the capacitance or the voltage will increase the energy stored. In the example problem, the total energy stored is given as 5.8 millijoules, which helps in finding the total capacitance by rearranging the energy equation to solve for \(C\).

This principle is crucial in circuits where maintaining a certain amount of energy in reserve is necessary.
Circuit Analysis
Circuit analysis involves examining an electric circuit to understand its capabilities and limitations concerning voltage, current, and storage.

To perform a detailed circuit analysis, one must:
  • Identify the configuration of components (series versus parallel).
  • Calculate equivalent capacitance using relevant formulas depending on the arrangement.
  • Determine the energy stored in each configuration or the entire circuit.
In the exercise, to find the unknown capacitor \(C\), series and parallel equations are used step by step to first determine the total effective capacitance, \(C_{total}\), followed by more detailed calculations to unearth the unknown value.

These steps in circuit analysis allow one to piece together the overall behavior of electronic devices and ensure they operate efficiently within given constraints. Understanding these principles is foundational for anyone working with electric circuits.

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Most popular questions from this chapter

Two parallel plates separated by \(6.5 \mathrm{~mm}\) have a potential difference of \(250 \mathrm{~V}\) between them. (a) Find the electric field between the plates. (b) What's the force on an electron in this field?

Neuromuscular microstimulator. A neuromuscular \(microstimulator is a device designed to be implanted into paralyzed muscles or other tissue suffering neuromuscular dysfunction. The device supplies an electric current that stimulates the dysfunctional tissue. A particular device uses a capacitor built into a microchip; the capacitor can supply \)10 \mathrm{~mA}\( of current for \)200 \mu \mathrm{s}\( before it's fully discharged. (a) Find the initial charge on the capacitor and (b) the capacitance if it's initially charged to \)4.5 \mathrm{~V}$.

Three capacitors are in series: \(C_{1}=45 \mu \mathrm{F}, C_{2}=65 \mu \mathrm{F}\) and \(C_{3}=80 \mu \mathrm{F}\). (a) Compute the equivalent capacitance. The series combination is connected across a \(48-\mathrm{V}\) battery. (b) Find the charge on each capacitor. (c) Compute the energy stored in each capacitor. (d) Show that the total energy you found in part (c) is the same as would be stored in a single capacitor with the equivalent capacitance you found in part (a).

A \(0.25-\mathrm{F}\) capacitor and a \(0.45-\mathrm{F}\) capacitor are in parallel (a) Find the equivalent capacitance. (b) Find the charge stored on each capacitor and the total charge stored when the parallel combination is connected across a 24 -V battery.

Cell membranes. The first accurate estimate of the thick-ness of biological cell membranes used a capacitive technique, in which the capacitance per unit area of cell membrane was determined through a macroscopic measurement of the electric properties of a suspension of cells. The result was a value of about \(10^{-2} \mathrm{~F} / \mathrm{m}^{2}\) for a wide range of cells. Assuming a dielectric constant of about 3 for the membrane material, find the membrane thickness. Note: Your answer is the thickness of a bipolar lipid layer alone, and is lower by a factor of about 3 than values based on x-ray techniques. The full membrane may be thicker still.
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