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.
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=I02⋅V02=12I0V0,Pave=I02⋅V02=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)(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.