/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Three capacitors having capacita... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 61

Three capacitors having capacitances of \(8.4,8.4,\) and 4.2\(\mu \mathrm{F}\) are connected in series across a \(36-\mathrm{V}\) potential difference. (a) What is the charge on the \(4.2-\mu F\) capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

Short Answer

Expert verified
(a) 100.8 渭C; (b) 1.814 mJ; (c) 14.4 V; (d) 2.088 mJ.

Step by step solution

01

Find Equivalent Capacitance

For capacitors in series, the equivalent capacitance \(C_{eq}\) is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] where \(C_1 = 8.4\, \mu F\), \(C_2 = 8.4\, \mu F\), and \(C_3 = 4.2\, \mu F\). Calculating this, we get: \[ \frac{1}{C_{eq}} = \frac{1}{8.4} + \frac{1}{8.4} + \frac{1}{4.2} = 0.3571\, \mu F^{-1} \] Thus, \(C_{eq} \approx 2.8\, \mu F\).
02

Calculate Charge on Each Capacitor

The charge \(Q\) on capacitors in series is the same. Using \(Q = C_{eq} \cdot V\), where \(V = 36\, V\): \[ Q = 2.8\, \mu F \times 36\, V = 100.8\, \mu C \] Thus, the charge on the \(4.2\, \mu F\) capacitor is also \(100.8\, \mu C\).
03

Calculate Total Energy in Series Configuration

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). For the series, the energy is: \[ E_{total} = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times 2.8\, \mu F \times (36\, V)^2 = 1.814\, mJ \]
04

Find Voltage Across Each Capacitor in Parallel

When disconnected and reconnected in parallel without discharging, the voltage across each capacitor is found using total charge \(Q_{total} = Q \times 3\) and total capacitance. \(Q_{total} = 100.8\, \mu C \times 3 = 302.4\, \mu C\), and \(C_{total} = C_1 + C_2 + C_3 = 21\, \mu F\). Thus, voltage \(V'\) is: \[ V' = \frac{Q_{total}}{C_{total}} = \frac{302.4\, \mu C}{21\, \mu F} \approx 14.4\, V \]
05

Calculate Total Energy in Parallel Configuration

In the parallel combination, the energy stored in each capacitor is \(E = \frac{1}{2} C V'^2\). Total energy \(E'\) is: \[ E' = 3 \times \frac{1}{2} C_1 (V')^2 = 3 \times \frac{1}{2} \times 8.4\, \mu F \times (14.4 \ V)^2 = 2.088\, mJ \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Equivalent Capacitance
When capacitors are connected in series, the equivalent capacitance of the system can be found using the reciprocal formula: \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\]This formula shows that the total capacitance for capacitors in series is always less than the smallest capacitance in the series.
  • It sums the reciprocals of the individual capacitances.
  • The calculation ensures that capacitance diminishes because, in a series, the same charge must pass through each capacitor.
For instance, if you have capacitances of 8.4, 8.4, and 4.2 microfarads, the equivalent capacitance would be approximately 2.8 microfarads. Calculating the equivalent capacitance in series is crucial because it allows you to explore how capacitors share charge and energy across a circuit. Understanding this lays a foundation for working with complex circuits involving multiple capacitors.
Charge on Capacitors
In a series circuit, the charge \(Q\) across each capacitor is the same. This is because the current flowing through the circuit must pass through each capacitor, rendering each charge carry equal amounts of electron flow. The total charge in this case is calculated by:\[Q = C_{eq} \cdot V\]where \(C_{eq}\) is the equivalent capacitance, and \(V\) is the applied potential difference across the series. For a 36 V potential difference, using our calculated \(C_{eq} = 2.8 \mu F\), the charge \(Q\) on each capacitor is approximately 100.8 渭C.
  • Always remember, in series circuits, though the charge is identical on all capacitors, the potential difference (voltage) can vary.
  • This characteristic is fundamental when designing circuits that require specific charge distributions.
It is critical to grasp this concept as it influences how capacitors behave in different circuit configurations.
Energy Stored in Capacitors
Capacitors store energy in the electric field between their plates. The energy \(E\) stored in a capacitor, when connected in series, can be determined using:\[E = \frac{1}{2} C_{eq} V^2\]where \(V\) is the voltage. For the total series configuration, with a voltage of 36 V, the energy stored would be approximately 1.814 millijoules.
  • Energy storage is a vital feature in applications like power supply systems where capacitors smooth out voltage fluctuations.
  • Understanding energy calculation helps in ensuring the safe and efficient design of circuits.
Analyzing stored energy assists in efficiently managing power within electronic systems.
Voltage Across Capacitors in Parallel
When capacitors that were originally in series are reconfigured to be in parallel, without discharging, they will share a common voltage \(V'\) across them. This is because the capacitors are essentially connected to the same two points in a circuit, ensuring identical electrical potential across each. Calculate the new voltage across each capacitor in the parallel setup by using:\[V' = \frac{Q_{total}}{C_{total}}\]Here, \(Q_{total}\) is the pre-calculated total charge, and \(C_{total}\) is the sum of the individual capacitances now in parallel. For example, with a total charge of 302.4 碌C and a total capacitance of 21 碌F, the voltage \(V'\) comes out to be 14.4 V.
  • This uniform voltage facilitates equal energy distribution across each capacitor.
  • In parallel, capacitors can supply energy or smooth voltage efficiently, which is beneficial in many electronic applications.
Understanding this characteristic is important for tailoring capacitive circuits for specific needs and functionalities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 \(\mathrm{mm}^{2}\) , and the separation between the plates is 0.700 \(\mathrm{mm}\) before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circulatry can detect a change in capacitance of 0.250 \(\mathrm{pF}\) , how far must the key be depressed before the circuitry detects its depression?

B10 Potential in Human Cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is 5.0 \(\mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{\boldsymbol{E}}\) in the wall-between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) Atypical cell in the human body has a volume of\(10^{-16} \mathrm{m}^{3} .\) Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of \(5.4 .\) Repeat parts (a) and (b) in this case.

A cylindrical air capacitor of length 15.0 \(\mathrm{m}\) stores \(3.20 \times 10^{-9} \mathrm{J}\) of energy when the potential difference between the two conductors is 4.00 \(\mathrm{V}\) . (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

Capacitance of the Earth. Consider a spherical capacitor with one conductor being a solid conducting sphere of radius \(R\) and the other conductor being at infinity. (a) Use Eq. (24.1) and what you know about the potential at the surface of a conducting sphere with charge \(Q\) to derive an expression for the capacitance of the charged sphere. (b) Use your result in part (a) to calculate the capacitance of the earth. The earth is a good conductor and has a radius of 6380 \(\mathrm{km} .\) Compare your results to the capacitance of typical capacitors used in electronic circuits, which ranges from 10 \(\mathrm{pF}\) to 100 \(\mathrm{pF} .\)

A parallel-plate vacuum capacitor has 8.38 \(\mathrm{J}\) of energy stored in it. The separation between the plates is 2.30 \(\mathrm{mm}\) . If the separation is decreased to 1.15 \(\mathrm{mm}\) , what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Turning Points in Physics

Read Explanation

Oscillations

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.