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# Alternating Current versus Direct Current

Module by: OpenStax College. E-mail the author

Summary:

• Explain the differences and similarities between AC and DC current.
• Calculate rms voltage, current, and average power.
• Explain why AC current is used for power transmission.

## Alternating Current

Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. Figure 1 shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world.

Figure 2 shows a schematic of a simple circuit with an AC voltage source. The voltage between the terminals fluctuates as shown, with the AC voltage given by

V=V0sin 2πft,V=V0sin 2πft, size 12{V = V rSub { size 8{0} } "sin"" 2"π ital "ft"} {}
(1)

where VV size 12{V} {} is the voltage at time tt size 12{t} {}, V0V0 size 12{V rSub { size 8{0} } } {} is the peak voltage, and ff size 12{f} {} is the frequency in hertz. For this simple resistance circuit, I=V/RI=V/R size 12{I = ital "V/R"} {}, and so the AC current is

I=I0 sin 2πft,I=I0 sin 2πft, size 12{I = I rSub { size 8{0} } " sin 2"π ital "ft"} {}
(2)

where II size 12{I} {} is the current at time tt size 12{t} {}, and I0=V0/RI0=V0/R size 12{I rSub { size 8{0} } = V rSub { size 8{0} } ital "/R"} {} is the peak current. For this example, the voltage and current are said to be in phase, as seen in Figure 1(b).

Current in the resistor alternates back and forth just like the driving voltage, since I=V/RI=V/R size 12{I = ital "V/R"} {}. If the resistor is a fluorescent light bulb, for example, it brightens and dims 120 times per second as the current repeatedly goes through zero. A 120-Hz flicker is too rapid for your eyes to detect, but if you wave your hand back and forth between your face and a fluorescent light, you will see a stroboscopic effect evidencing AC. The fact that the light output fluctuates means that the power is fluctuating. The power supplied is P=IVP=IV size 12{P = ital "IV"} {}. Using the expressions for II size 12{I} {} and VV size 12{V} {} above, we see that the time dependence of power is P=I0V0sin2 2πftP=I0V0sin2 2πft size 12{P= I rSub { size 8{0} } V rSub { size 8{0} } "sin" rSup { size 8{2} } " 2"π ital "ft"} {}, as shown in Figure 3.

### Making Connections: Take-Home Experiment—AC/DC Lights:

Wave your hand back and forth between your face and a fluorescent light bulb. Do you observe the same thing with the headlights on your car? Explain what you observe. Warning: Do not look directly at very bright light.

We are most often concerned with average power rather than its fluctuations—that 60-W light bulb in your desk lamp has an average power consumption of 60 W, for example. As illustrated in Figure 3, the average power PavePave size 12{P rSub { size 8{"ave"} } } {} is

Pave=12I0V0.Pave=12I0V0. size 12{P rSub { size 8{"ave"} } = { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } } {}
(3)

This is evident from the graph, since the areas above and below the (1/2)I0V0(1/2)I0V0 size 12{ $$1/2$$ I rSub { size 8{0} } V rSub { size 8{0} } } {} line are equal, but it can also be proven using trigonometric identities. Similarly, we define an average or rms current IrmsIrms size 12{I rSub { size 8{"rms"} } } {} and average or rms voltage VrmsVrms size 12{V rSub { size 8{"rms"} } } {} to be, respectively,

I rms = I 0 2 I rms = I 0 2 size 12{I rSub { size 8{"rms "} } = { {I rSub { size 8{0} } } over { sqrt {2} } } } {}
(4)

and

Vrms =V02.Vrms =V02. size 12{V rSub { size 8{"rms "} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {}
(5)

where rms stands for root mean square, a particular kind of average. In general, to obtain a root mean square, the particular quantity is squared, its mean (or average) is found, and the square root is taken. This is useful for AC, since the average value is zero. Now,

Pave=IrmsVrms,Pave=IrmsVrms, size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {}
(6)

which gives

Pave=I02V02=12I0V0,Pave=I02V02=12I0V0, size 12{P rSub { size 8{"ave"} } = { {I rSub { size 8{0} } } over { sqrt {2} } } cdot { {V rSub { size 8{0} } } over { sqrt {2} } } = { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } } {}
(7)

as stated above. It is standard practice to quote IrmsIrms size 12{I rSub { size 8{"rms"} } } {}, VrmsVrms size 12{V rSub { size 8{"rms"} } } {}, and PavePave size 12{P rSub { size 8{"ave"} } } {} rather than the peak values. For example, most household electricity is 120 V AC, which means that VrmsVrms size 12{V rSub { size 8{"rms"} } } {} is 120 V. The common 10-A circuit breaker will interrupt a sustained IrmsIrms size 12{I rSub { size 8{"rms"} } } {} greater than 10 A. Your 1.0-kW microwave oven consumes Pave=1.0 kWPave=1.0 kW size 12{P rSub { size 8{"ave"} } =1 "." 0"kW"} {}, and so on. You can think of these rms and average values as the equivalent DC values for a simple resistive circuit.

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

Irms=VrmsR.Irms=VrmsR. size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {R} } } {}
(8)

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

Pave=IrmsVrms,Pave=IrmsVrms, size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {}
(9)
Pave=Vrms2R,Pave=Vrms2R, size 12{P rSub { size 8{"ave"} } = { {V rSub { size 8{"rms"} } rSup { size 8{2} } } over {R} } } {}
(10)

and

Pave=Irms2R.Pave=Irms2R. size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R} {}
(11)

### Example 1: Peak Voltage and Power for AC

(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?

Strategy

We are told that VrmsVrms size 12{V rSub { size 8{"rms"} } } {} is 120 V and PavePave size 12{P rSub { size 8{"ave"} } } {} is 60.0 W. We can use Vrms =V02Vrms =V02 size 12{V rSub { size 8{"rms "} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {} to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.

Solution for (a)

Solving the equation Vrms =V02Vrms =V02 size 12{V rSub { size 8{"rms "} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {} for the peak voltage V0V0 size 12{V rSub { size 8{0} } } {} and substituting the known value for VrmsVrms size 12{V rSub { size 8{"rms"} } } {} gives

V0=2Vrms= 1.414(120 V)= 170 V.V0=2Vrms= 1.414(120 V)= 170 V. size 12{V rSub { size 8{0} } = sqrt {2} V rSub { size 8{"rms"} } =" 1" "." "414" $$"120"" V"$$ =" 170 V"} {}
(12)

Discussion for (a)

This means that the AC voltage swings from 170 V to –170 V–170 V and back 60 times every second. An equivalent DC voltage is a constant 120 V.

Solution for (b)

Peak power is peak current times peak voltage. Thus,

P0=I0V0= 212I0V0= 2Pave.P0=I0V0= 212I0V0= 2Pave. size 12{P rSub { size 8{0} } = I rSub { size 8{0} } V rSub { size 8{0} } =" 2" left ( { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } right ) =" 2"P rSub { size 8{"ave"} } } {}
(13)

We know the average power is 60.0 W, and so

P0= 2(60.0 W)= 120 W.P0= 2(60.0 W)= 120 W. size 12{P rSub { size 8{0} } =" 2" $$"60" "." "0 W"$$ =" 120 W"} {}
(14)

Discussion

So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.

## 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 Figure 4.) 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.

### Example 2: 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 1.00Ω1.00Ω size 12{1 "." "00" %OMEGA } {}? (c) What percentage of the power is lost in the transmission lines?

Strategy

We are given Pave=100 MWPave=100 MW size 12{P rSub { size 8{"ave"} } ="100""MW"} {}, Vrms=200 kVVrms=200 kV size 12{V rSub { size 8{"rms"} } ="200""kV"} {}, and the resistance of the lines is R=1.00ΩR=1.00Ω size 12{R=1 "." "00" %OMEGA } {}. Using these givens, we can find the current flowing (from P=IVP=IV size 12{P = ital "IV"} {}) and then the power dissipated in the lines (P=I2RP=I2R 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 Pave=IrmsVrmsPave=IrmsVrms size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {} and substitute known values. This gives

Irms=PaveVrms=100 × 106 W200 × 103 V= 500 A.Irms=PaveVrms=100 × 106 W200 × 103 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"} {}
(15)

Solution

Knowing the current and given the resistance of the lines, the power dissipated in them is found from Pave=Irms2RPave=Irms2R size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } rSup { size 8{2} } R} {}. Substituting the known values gives

Pave=Irms2R=(500 A)2(1.00 Ω)= 250 kW.Pave=Irms2R=(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"} {}
(16)

Solution

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

% loss=250 kW100 MW×100=0.250 %.% loss=250 kW100 MW×100=0.250 %. size 12{%" loss=" { {"250"" kW"} over {"100"" MW"} } ´"100"=0 "." "250 %"} {}
(17)

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.

### PhET Explorations: Generator:

Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light.

## Section Summary

• Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant.
• The voltage source of an alternating current (AC) system puts out V=V0sin 2πftV=V0sin 2πft size 12{V = V rSub { size 8{0} } "sin2"π ital "ft"} {}, where VV size 12{V} {} is the voltage at time tt size 12{t} {}, V0V0 size 12{V rSub { size 8{0} } } {} is the peak voltage, and ff size 12{f} {} is the frequency in hertz.
• In a simple circuit, I=V/RI=V/R size 12{I = ital "V/R"} {} and AC current is I=I0sin 2πftI=I0sin 2πft size 12{I = I rSub { size 8{0} } "sin2"π ital "ft"} {}, where II size 12{I} {} is the current at time tt size 12{t} {}, and I0=V0/RI0=V0/R size 12{I rSub { size 8{0} } = V rSub { size 8{0} } ital "/R"} {} is the peak current.
• The average AC power is Pave=12I0V0Pave=12I0V0 size 12{P rSub { size 8{"ave"} } = { {1} over {2} } I rSub { size 8{0} } V rSub { size 8{0} } } {}.
• Average (rms) current IrmsIrms size 12{I rSub { size 8{"rms"} } } {} and average (rms) voltage VrmsVrms size 12{V rSub { size 8{"rms"} } } {} are Irms=I02Irms=I02 size 12{I rSub { size 8{"rms"} } = { {I rSub { size 8{0} } } over { sqrt {2} } } } {} and Vrms=V02Vrms=V02 size 12{V rSub { size 8{"rms"} } = { {V rSub { size 8{0} } } over { sqrt {2} } } } {}, where rms stands for root mean square.
• Thus, Pave=IrmsVrmsPave=IrmsVrms size 12{P rSub { size 8{"ave"} } = I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } } {}.
• Ohm’s law for AC is Irms=VrmsRIrms=VrmsR size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {R} } } {}.
• Expressions for the average power of an AC circuit are Pave= Irms VrmsPave= Irms Vrms, Pave = Vrms2RPave = Vrms2R, and Pave= Irms2RPave= Irms2R, analogous to the expressions for DC circuits.

## Conceptual Questions

### Exercise 1

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.

### Exercise 2

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

### Exercise 3

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

### Exercise 1

(a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb’s operating temperature is 2700ºC2700ºC, what is its resistance at 2600ºC2600ºC?

### Exercise 2

Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage?

480 V

### Exercise 3

A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current?

### Exercise 4

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

### Exercise 5

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?

### Exercise 6

In this problem, you will verify statements made at the end of the power losses for Example 2. (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 -Ω1.00 -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%

### Exercise 7

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/kWh9.00 cents/kWh size 12{9 "." "00"" cents/kw" cdot h} {}?

### Exercise 8

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

2.40 kW

### Exercise 9

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

### Exercise 10

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

### Exercise 11

Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of 5.00 mm25.00 mm2 size 12{5 "." "00"" mm" rSup { size 8{2} } } {}, is needed if the operating temperature is 500º C500º C size 12{"500"°C} {}? (c) What power will it draw when first switched on?

### Exercise 12

Find the time after t=0t=0 size 12{t=0} {} when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) V0/2V0/2 size 12{V rSub { size 8{0} } /2} {} (b) V0V0 size 12{V rSub { size 8{0} } } {} (c) 0.

(a) 1.39 ms

(b) 4.17 ms

(c) 8.33 ms

### Exercise 13

(a) At what two times in the first period following t=0t=0 size 12{t=0} {} does the instantaneous voltage in 60-Hz AC equal VrmsVrms size 12{V rSub { size 8{"rms"} } } {}? (b) VrmsVrms size 12{-V rSub { size 8{"rms"} } } {}?

## Glossary

direct current:
(DC) the flow of electric charge in only one direction
alternating current:
(AC) the flow of electric charge that periodically reverses direction
AC voltage:
voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz
AC current:
current that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz
rms current:
the root mean square of the current, I rms = I 0 / 2 I rms = I 0 / 2 size 12{I rSub { size 8{"rms "} } = I rSub { size 8{0} } / sqrt {2} } {} , where I0 is the peak current, in an AC system
rms voltage:
the root mean square of the voltage, V rms = V 0 / 2 V rms = V 0 / 2 size 12{V rSub { size 8{"rms "} } = V rSub { size 8{0} } / sqrt {2} } {} , where V0 is the peak voltage, in an AC system

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