/*! 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 88 Two capacitors are connected in ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of \(2.0 \mu \mathrm{F}\) and the other a capacitance of \(4.0 \mu \mathrm{F}\). These two capacitors together store \(5.4 \times 10^{-5} \mathrm{C}\) of charge. What is the voltage of the battery?

Short Answer

Expert verified
The voltage of the battery is 9.0 volts.

Step by step solution

01

Understand the Problem

We have two capacitors connected in parallel with capacitances of \(2.0 \mu \mathrm{F}\) and \(4.0 \mu \mathrm{F}\). Together, these capacitors store a total charge of \(5.4 \times 10^{-5} \mathrm{C}\). We need to find the voltage \(V\) across the battery.
02

Apply Capacitance Formula for Parallel Connection

For capacitors in parallel, the total capacitance \(C_{total}\) is the sum of individual capacitances: \(C_{total} = C_1 + C_2 = 2.0 \mu \mathrm{F} + 4.0 \mu \mathrm{F} = 6.0 \mu \mathrm{F}\).
03

Convert Units for Capacitance

The total capacitance in microfarads should be converted to farads for calculation: \(6.0 \mu \mathrm{F} = 6.0 \times 10^{-6} \mathrm{F}\).
04

Use Charge Equation

The charge stored on a capacitor is given by \(Q = C \times V\). Here, \(Q = 5.4 \times 10^{-5} \mathrm{C}\) and \(C = 6.0 \times 10^{-6} \mathrm{F}\).
05

Solve for Voltage

Rearrange the equation \(Q = C \times V\) to solve for \(V\): \(V = \frac{Q}{C} = \frac{5.4 \times 10^{-5} \mathrm{C}}{6.0 \times 10^{-6} \mathrm{F}}\).
06

Calculate the Voltage

Perform the division: \(V = \frac{5.4 \times 10^{-5}}{6.0 \times 10^{-6}} = 9.0\) volts.

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 91Ó°ÊÓ!

Key Concepts

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

Capacitance
Capacitance is a property of a capacitor that indicates its ability to store charge. Capacitors are electronic components that store and release electrical energy in a circuit.
They do this by holding electric charge on conductive plates separated by an insulating material, or dielectric. The basic unit of capacitance is the farad (F), but since the farad is a very large unit, capacitance is often measured in microfarads ( \(\mu F\), where \(1 \mu F = 10^{-6} F\)).

The capacitance of a capacitor depends on several factors:
  • Surface area of the plates: Larger plates have a higher capacitance.
  • Distance between plates: Closer plates result in higher capacitance.
  • Dielectric material between the plates: Different materials affect the capacitance based on their dielectric constant.
In the context of capacitors in parallel, the total capacitance is simply the sum of the individual capacitances.
This is because adding capacitors in parallel effectively increases the plate area, thus increasing total capacitance in the circuit.
Charge and Voltage Relationship
The relationship between charge, voltage, and capacitance in a capacitor is defined by the equation \(Q = C \times V\), where:
  • \(Q\) is the charge stored, measured in coulombs (C).
  • \(C\) is the capacitance, measured in farads (F).
  • \(V\) is the voltage across the capacitor, measured in volts (V).
This equation indicates how much charge a capacitor will store depending on its capacitance and the voltage applied to it. It shows that for a given capacitance, the charge stored by the capacitor is directly proportional to the voltage across it.

In a parallel circuit, multiple capacitors receive the same voltage from the battery. Therefore, even though each capacitor stores its charge depending on its individual capacitance, the total charge stored is the sum of the charges on all the capacitors. By understanding this relationship, one can calculate the voltage across a given set of capacitors if the total charge and capacitance are known, which is crucial for determining the functioning of electronic circuits.
Microfarads to Farads Conversion
Understanding the conversion between microfarads and farads is vital when calculating any electrical values involving capacitance in scientific notations.
Since the farad is a large unit, electronic components and circuits often deal with values in microfarads (\(\mu F\)).
Microfarads can be converted to farads using the equation:
  • \(1 \mu F = 10^{-6} F\)
To convert several microfarads to farads, multiply the microfarad value by \(10^{-6}\).
For instance, 6.0 microfarads is converted to farads by calculating \(6.0 \times 10^{-6} F\). This conversion allows us to properly use the capacitance value in standard equations, such as those involving charge and voltage, ensuring the units are consistent throughout the calculation.

Proper understanding and execution of these conversions are essential in academic and professional settings, facilitating easier and accurate computations in electronics-related problems.

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

There are approximately 110 million TVs in the United States. Each uses, on average, 75 W of power and is turned on for 6.0 hours a day. If electrical energy costs \(\$ 0.12\) per \(\mathrm{kWh}\), how much money is spent every day in keeping the TVs turned on?

Two scales on a voltmeter measure voltages up to 20.0 and \(30.0 \mathrm{~V}\), respectively. The resistance connected in series with the galvanometer is \(1680 \Omega\) for the \(20.0-\mathrm{V}\) scale and \(2930 \Omega\) for the 30.0 - \(V\) scale. Determine the coil resistance and the full-scale current of the galvanometer that is used in the voltmeter.

The current in a \(47-\Omega\) resistor is 0.12 A. This resistor is in series with a \(28-\Omega\) resistor, and the series combination is connected across a battery. What is the battery voltage?

The average power used by a stereo speaker is \(55 \mathrm{~W}\). Assuming that the speaker can be treated as a \(4.0-\Omega\) resistance, find the peak value of the ac voltage applied to the speaker.

The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired temperature. Suppose that a 52-gal \(\left(1.00 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\right)\) unit starts with cold water at \(11^{\circ} \mathrm{C}\) and delivers hot water at \(53^{\circ} \mathrm{C}\). The unit is electric and utilizes a resistance heater \((120 \mathrm{~V} \mathrm{ac}, 3.0 \Omega)\) to heat the water. Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.

See all solutions

Recommended explanations on Physics Textbooks

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.