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7.12 Alternating current versus direct current  (Page 2/6)

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To summarize, when dealing with AC, Ohm’s law and the equations for power are completely analogous to those for DC, but rms and average values are used for AC. Thus, for AC, Ohm’s law is written

I rms = V rms R . size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {R} } } {}

The various expressions for AC power P ave size 12{P rSub { size 8{"ave"} } } {} are

P ave = I rms V rms , size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {}
P ave = V rms 2 R , size 12{P rSub { size 8{"ave"} } = { {V rSub { size 8{"rms"} } rSup { size 8{2} } } over {R} } } {}

and

P ave = I rms 2 R . size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R} {}

Why use ac for power distribution?

Most large power-distribution systems are AC. Moreover, the power is transmitted at much higher voltages than the 120-V AC (240 V in most parts of the world) we use in homes and on the job. Economies of scale make it cheaper to build a few very large electric power-generation plants than to build numerous small ones. This necessitates sending power long distances, and it is obviously important that energy losses en route be minimized. High voltages can be transmitted with much smaller power losses than low voltages, as we shall see. (See [link] .) For safety reasons, the voltage at the user is reduced to familiar values. The crucial factor is that it is much easier to increase and decrease AC voltages than DC, so AC is used in most large power distribution systems.

Photograph of transformers installed in transmission lines.
Power is distributed over large distances at high voltage to reduce power loss in the transmission lines. The voltages generated at the power plant are stepped up by passive devices called transformers to 330,000 volts (or more in some places worldwide). At the point of use, the transformers reduce the voltage transmitted for safe residential and commercial use. (Credit: GeorgHH, Wikimedia Commons)

Power losses are less for high-voltage transmission

(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of 1 . 00 Ω size 12{1 "." "00" %OMEGA } {} ? (c) What percentage of the power is lost in the transmission lines?

Strategy

We are given P ave = 100 MW size 12{P rSub { size 8{"ave"} } ="100"`"MW"} {} , V rms = 200 kV size 12{V rSub { size 8{"rms"} } ="200"`"kV"} {} , and the resistance of the lines is R = 1 . 00 Ω size 12{R=1 "." "00"` %OMEGA } {} . Using these givens, we can find the current flowing (from P = IV size 12{P = ital "IV"} {} ) and then the power dissipated in the lines ( P = I 2 R size 12{P = I rSup { size 8{2} } R} {} ), and we take the ratio to the total power transmitted.

Solution

To find the current, we rearrange the relationship P ave = I rms V rms size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {} and substitute known values. This gives

I rms = P ave V rms = 100 × 10 6 W 200 × 10 3 V = 500 A . size 12{I rSub { size 8{"rms"} } = { {P rSub { size 8{"ave"} } } over {V rSub { size 8{"rms"} } } } = { {"100 " times " 10" rSup { size 8{6} } " W"} over {"200 " times " 10" rSup { size 8{3} } " V"} } =" 500 A"} {}

Solution

Knowing the current and given the resistance of the lines, the power dissipated in them is found from P ave = I rms 2 R size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R} {} . Substituting the known values gives

P ave = I rms 2 R = ( 500 A ) 2 ( 1 . 00 Ω ) = 250 kW . size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R = \( "500 A" \) rSup { size 8{2} } \( 1 "." "00 " %OMEGA \) =" 250 kW"} {}

Solution

The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:

% loss= 250 kW 100 MW × 100 = 0 . 250 % . size 12{%" loss=" { {"250"" kW"} over {"100"" MW"} } ´"100"=0 "." "250 %"} {}

Discussion

One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.

It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated with common static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are often more harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in New York City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocating the use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltage transmission.

Section summary

  • Ohm’s law for AC is I rms = V rms R size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {R} } } {} .
  • Expressions for the average power of an AC circuit are P ave = I rms V rms , P ave = V rms 2 R , and P ave = I rms 2 R , analogous to the expressions for DC circuits.

Conceptual questions

Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries.

Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity?

You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain.

Problem exercises

Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power?

2.50 ms

A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined?

In this problem, you will verify statements made at the end of the power losses for [link] . (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a 1 . 00 - Ω size 12{1 "." "00"- %OMEGA } {} transmission line. (c) What percent loss does this represent?

(a) 4.00 kA

(b) 16.0 MW

(c) 16.0%

A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs 9.00 cents /kW ⋅ h size 12{9 "." "00"" cents/kw" cdot h} {} ?

What is the average power consumption of a 120-V AC microwave oven that draws 10.0 A?

1.20 kW

What is the average current through a 500-W room heater that operates on 120-V AC power?

Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC?

(a) 4.0

(b) 0.50

(c) 4.0

Nichrome wire is used in some radiative heaters. Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC.
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Source:  OpenStax, Concepts of physics. OpenStax CNX. Aug 25, 2015 Download for free at https://legacy.cnx.org/content/col11738/1.5
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