/*! 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 2 Electric current is the fundamen... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 2

Electric current is the fundamental electrical dimension in SI; its unit is the ampere (A). Determine units for the following quantities, as combinations of fundamental SI units. (a) Electric power; (b) Electric charge; (c) Electric potential difference; (d) Electric resistance; (e) Electric capacitance.

Short Answer

Expert verified
(a) Power: W = J/s, (b) Charge: C = A鈰卻, (c) Voltage: V = J/C, (d) Resistance: 惟 = V/A, (e) Capacitance: F = C/V.

Step by step solution

01

Identifying the formula for Electric Power

Electric power is defined as the rate at which electrical energy is transferred by an electric circuit. The formula is given by \( P = IV \), where \( P \) is power, \( I \) is current, and \( V \) is electric potential difference (voltage). The SI unit for power is the watt (W), which is defined as 1 joule per second \(( ext{J/s})\). Using the definition of voltage \( V = ext{Joules per coulomb (J/C)} \) and current \( I = ext{coulombs per second (C/s)} \), we can write:\[ P = I imes V = (C/s) \times (J/C) = J/s = W \]
02

Identifying the formula for Electric Charge

Electric charge is the fundamental property of matter that exhibits electrostatic attraction or repulsion in the presence of other matter. The unit of electric charge in the SI system is the coulomb (C). Charge \( Q \) is defined by \( Q = I imes t \), where \( I \) is current and \( t \) is time. Since \( I \) is in amperes \(( ext{A})\) and \( t \) is in seconds \(( ext{s})\), the unit for charge is:\[ Q = I imes t = ext{A} imes ext{s} = ext{C} \]
03

Determining the unit for Electric Potential Difference

Electric potential difference or voltage \( V \) is defined as the work done per unit charge to move a charge between two points. It is given by \( V = W/Q \), where \( W \) is work done in joules \(( ext{J})\) and \( Q \) is charge in coulombs \(( ext{C})\). The unit for voltage is the volt (V), which is defined as:\[ V = ext{J/C} \]
04

Determining the unit for Electric Resistance

Electric resistance \( R \) is the opposition to the flow of electric current, defined by Ohm's Law \( V = IR \). Rearranging this gives \( R = V/I \). With voltage \( V \) in volts \(( ext{V})\) and current \( I \) in amperes \(( ext{A})\), resistance is measured in ohms \(( ext{惟})\):\[ R = ext{V/A} \] which, simplifying with \[ ext{V} = ext{J/C} \text{ and } ext{A} = ext{C/s} \] results in:\[ R = ext{(J/C)}/( ext{C/s}) = ext{J/s (A}^2) = ext{惟} \]
05

Determining the unit for Electric Capacitance

Electric capacitance \( C \) is the ability of a system to store electric charge per unit voltage, defined by \( C = Q/V \). With charge \( Q \) in coulombs \(( ext{C})\) and voltage \( V \) in volts \(( ext{V})\), capacitance is measured in farads \(( ext{F})\):\[ C = ext{C/V} \] which simplifies with \[ ext{V} = ext{J/C} \] to yield:\[ C = ext{C}^2/ ext{J} = ext{F} \]

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.

Electric Power
Electric power is a measure of how much electrical energy is transferred in a circuit over time. Imagine it as the speed at which energy is used or produced within an electrical device. Whenever you turn on a light bulb or a television, they consume electric power. The formula to calculate electric power is given by \( P = IV \), where \( P \) represents power, \( I \) is the electric current, and \( V \) is the electric potential difference.
The SI unit for electric power is the watt (W), which is equivalent to one joule per second (J/s). So, when we calculate electric power, we are essentially determining how many joules of energy are being transferred every second.
  • A wattage of 60 W means 60 joules of energy are transferred every second.
  • Power consumption in homes is often measured in kilowatt-hours (kWh); 1 kWh means 1000 watts used over 1 hour.
Electric Charge
Electric charge is a fundamental property of matter that allows objects to exert electric force on each other. It is what causes static electricity and sparks. Electric charge comes in two types: positive and negative, commonly represented by protons and electrons.
In terms of SI units, electric charge is measured in coulombs (C). The formula to determine electric charge is \( Q = I \times t \), where \( Q \) is charge, \( I \) is electric current in amperes (A), and \( t \) is time in seconds (s). Thus, a current of 1 A flowing for 1 second will transfer a charge of 1 coulomb.
  • A common battery might store a charge of several thousand coulombs.
  • Lightning can involve the transfer of several hundred million coulombs.
Electric Potential Difference
Electric potential difference, or voltage, is the measure of the energy change experienced by a charge moving between two points in an electric field. Think of it as an electric pressure that pushes charges through a circuit.
The SI unit for electric potential difference is the volt (V). Voltage is calculated using the formula \( V = W/Q \), where \( V \) is voltage, \( W \) is work done in joules (J), and \( Q \) is charge in coulombs (C). Thus, 1 volt equals 1 joule per coulomb.
  • A car battery commonly has a voltage of 12 volts, meaning it transfers 12 joules of energy per coulomb of charge it moves.
  • Voltage helps in overcoming the resistance of electrical components to maintain current flow.
Electric Resistance
Electric resistance is the measure of how much a material opposes the flow of electric current. It's like the friction that electrons experience as they move through a wire. Devices such as resistors are used to control current flow in a circuit.
Resistance is measured in ohms (惟), using Ohm's Law which states \( V = IR \), or rearranged, \( R = V/I \). Here, \( V \) represents voltage in volts, and \( I \) represents current in amperes. A higher resistance means more opposition to current flow.
  • Resistance in wires leads to heat generation; that is how electric heaters work.
  • Materials with low resistance, like copper, are preferred for wiring to reduce energy loss.
Electric Capacitance
Electric capacitance is the ability of a body to store an electric charge. You can think of capacitors, which are devices that store and release electrical energy, much like a battery. They are essential in maintaining voltage stability in circuits.
Capacitance is measured in farads (F). It is determined through the formula \( C = Q/V \), where \( C \) is capacitance, \( Q \) is charge in coulombs, and \( V \) is voltage in volts. This means that one farad equals 1 coulomb per volt.
  • Capacitors help in filtering out voltage fluctuations in power supplies.
  • They can temporarily store energy to provide power during short drops in supply.

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

Consider the following proposal for a decimal calendar. The fundamental unit is the decimal year \((\mathrm{Yr}),\) equal to the number of conventional (SI) seconds required for the earth to complete a circuit of the sun. Other units are defined in the table below. Develop, where possible, factors for converting decimal calendar units to conventional calendar units. Discuss pros and cons of the proposal $$\begin{array}{c|c|c} \hline \text { Decimal Calendar Unit } & \text { Symbol } & \text { Definition } \\ \hline \text { Second } & \mathrm{Sc} & 10^{-6} \mathrm{Yr} \\ \text { Minute } & \mathrm{Mn} & 10^{-5} \mathrm{Yr} \\ \text { Hour } & \mathrm{Hr} & 10^{-4} \mathrm{Yr} \\ \text { Day } & \mathrm{Dy} & 10^{-3} \mathrm{Yr} \\ \text { Week } & \mathrm{Wk} & 10^{-2} \mathrm{Yr} \\ \text { Month } & \mathrm{Mo} & 10^{-1} \mathrm{Yr} \\ \text { Year } & \mathrm{Yr} & \\ \hline \end{array}$$

What is the value of \(g_{c}\) and what are its units in a system in which the second, the foot, and the pound mass are defined as in Sec. \(1.2,\) and the unit of force is the poundal, defined as the force required to give \(1\left(\mathrm{lb}_{\mathrm{m}}\right)\) an acceleration of \(1(\mathrm{ft})(\mathrm{s})^{-2} ?\)

The reading on a mercury manometer at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) (open to the atmosphere at one end) is \(56.38 \mathrm{cm} .\) The local acceleration of gravity is \(9.832 \mathrm{m} \mathrm{s}^{-2}\). Atmospheric pressureis \(101.78 \mathrm{kPa}\). What is the absolute pressurein kPa being measured? The density of mercury at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) is \(13.534 \mathrm{g} \mathrm{cm}^{-3}\)

Verify that the SI unit of kinetic and potential energy is the joule

The turbines in a hydroelectric plant are fed by water falling from a 50 -m height. Assuming \(91 \%\) efficiency for conversion of potential to electrical energy, and \(8 \%\) loss of the resulting power in transmission, what is the mass flowrate of water required to power a \(200 \mathrm{W}\) light bulb?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

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.