Alternating Current

Introduction

Outline

• Power and Alternating Currents
  • A short review
  • Alternating current
• Practice with the Concepts
• Discussion Points

Power and Alternating Currents

We return now to the idea of power, which we already discussed in terms of doing work in a gravitational field over time. Now we look at the application of the concept of power to the work accomplished by electrical current.

A short review...

• Power = Energy used/time: P = E/t
• An expenditure of electrical energy ΔE = QV, so electrical power is P = QV/t
• Since Q/t is current, we can also write this relationship as P = IV
• Since V = IR, we can also write this as P = I IR = I²R
• Or, since I = R/V, we can write this as P = R/V R = R²/V

The "right" relationship is whichever one applies in a given situation. You should remember the basic concept, which is contained in the first two points: power is energy/unit time, and electrical energy is given by QV. You can derive all the other relationships from the definition of V = IR.

We can also flip this relationship around: Work is the amount of energy expended by exerting power for some period of time: Energy = Power * time. So electrical energy is measured in units of power (watts or kilowatts) and time (hours).

When Thomas Edison first put electrical circuits into public use and sold his incandescent lightbulbs, his circuits used direct current, the same kind of current produced by batteries. Direct current could both light at the lightbulbs and drive motors using storage batteries as a source or for backup during generator interruptions. However, transmission of high voltage from direct current sources encountered significant resistance. Either the current source had to be located within a mile or two of the target load, which meant building direct current generators every few miles, or the current had to be sent with initially high voltages to reach distant targets with enough voltage left to do any work. The problem with this latter solution was that there were at the time no transformers capable of converting the high voltage to a usable voltage for customers close to the generator. The solution lay with in using alternating current as a distribution system for electrical power.

Alternating current

But Edison was no mathematician, and while he could perform thousands of experiments, he lacked the discipline and vision to do the analysis required to develop an alternating current solution. That solution was proposed Nicola Tesla, and when Edison decided not to work with Tesla, Tesla went to work for Westinghouse. The result was "the war of currents", in which both sides used news stories and advertisements to portray the solution of the other company as dangerous, costly, and irresponsible. For nearly a decade around 1890, Edison's General Electric Company and the Westinghouse Company competed for contracts to produce electric power stations throughout the northeastern United States. In 1890, Westinghouse won a contract to build a generator at Niagara Falls using Tesla's patented alternating current system. General Electric eventually capitulated, acquiring a rival alternating current company (Thomson-Houston), so that it could enter the alternating current industry. The result is our centralized power system, which transports electricity at low costs across great distances, employing power transformers to step down the extremely high voltage of the carrier lines to the voltage used by households and businesses.

So it behooves the physics student to know a little bit about how household circuits work and what alternating current is. An electric generator, a spinning magnet turned by an external source of energy such as a waterfall creates induced current in a set of wire loops (we will spend more time on this concept in Chapter 20). The current cycles from all amount to a high positive amount back down to zero, and then runs in the opposite direction to a maximum negative current amount before returning to zero. The amplitude or current amount predicted by our old mathematical friend, the simple harmonic oscillator. In this particular situation, maximum amplitude is the peak current I₀, frequency is oscillations per second, and angular frequency ω is equal (as before) to 2πf. Then instantaneous current I is then a simple harmonic function of f and t:

I = I_O sin ωt = I₀ sin 2πf t

We can use our relationships for power and resistance to write this in a number of ways, including

P = I²R = I_O² sin² ωt

Notice that while I₀ can be positive or negative at any moment, I₀² is always positive. The average power will be one half of the extremes: I²_average (usually written I²) = ½ I₀². By convention, the square root of these two values is the "root mean square value:

I = I_rms = √ (½ I₀²) = I₀ √(½) = I₀/√(2) = 0.707 I₀

If we assume that the resistance wire is constant the wire, we can also write the or relationship in terms of the voltage: P = ½ V₀²/R = I_rmsV_rms. These equivalences will allow us to analyze alternating current in different situations.

Practice with the Concepts

Discussion Points

• Electric power is transferred over large distances at very high voltages. How can the high voltage reduce power losses in the transmission lines?
• Is current used up in a resistor?

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