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This section delves into the relationship between work, electrical charge, and potential difference (voltage). You will learn the significance of the coulomb as the unit of charge, how to calculate electric charge in a circuit, and define electric current in amperes. We explore the formula for electrical work (W = Vq) and discuss the efficiency of electrical systems, emphasizing that work output will always be less than work input. By engaging with these concepts, you'll be equipped to solve related problems and understand the practical effects of electrical work.
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Chapter 2 Section 3 Work in Electrical Systems
Objectives • Explain the relationship between work, charge and potential difference (voltage). • Calculate electrical charge in a circuit. • Define units of electric current. • Solve problems in electrical work. • Identify effects of electrical work. • Explain efficiency in terms of work in and work out in electrical systems.
Coulombs • The unit of electric charge is the coulomb (C). • The fundamental charge on an electron is 1.6 X 10-19 C. • 1 C = 1/ 1.6 X 10-19 = 6.25 X 1018 electrons or 6,250,000,000,000,000,000 electrons.
Work in electrical systems • The electric field is given by E = F / q (Sec 1.3). Thus F=Eq. • Potential difference (voltage) is given by V=Ed (Sec 1.3) • W = Fd = Eqd • Thus W = Vq
Work - cont • Thus Work = potential difference X charge moved • W=Vq • The work needed to move one coulomb of charge though a potential difference of one volt is one joule. • 1J = 1 V .C • 1V = 1 J/C
Electric Charge & Current • Electric current is the rate at which charge flows through a circuit. • Current = charge / time; I = q / t • Current is measured in amperes or amps (A) • 1 amp = 1 coulomb per second; 1A = 1 C/s
Efficiency • Efficiency = work out / work in; Wout / Win • % Efficiency = (Wout / Win) x 100% • Efficiency can never be equal to or greater than 100% • Work out will always be less than work in.
Summary • Electric work is done when a charge is moved through a potential difference. • W = Vq • 1 joule = 1 volt X 1 coulomb; 1J = 1V.C • Current is the rate at which charge flows. • 1 Amp = 1 coulomb per second; 1A=1C/s. • % Efficiency = (Wout / Win) x 100%