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 3.19 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 3.20 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
3.38 V=V0sin2πft,
where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz. For this simple resistance circuit, I=V/R, and so the AC current is
3.39 I=I0sin 2πft,
where I is the current at time t, and I0=V0/R is the peak current. For this example, the voltage and current are said to be in phase, as seen in Figure 3.19(b).
Current in the resistor alternates back and forth just like the driving voltage, since I=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=IV. Using the expressions for I and V above, we see that the time dependence of power is P=I0V0sin22πft, as shown in Figure 3.21.
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 3.21, the average power Pave is
3.40 Pave=12I0V0.
This is evident from the graph, since the areas above and below the (1/2)I0V0 line are equal, but it can also be proven using trigonometric identities. Similarly, we define an average or rms current Irms and average or rms voltage Vrms to be,
3.41 Irms=I0√2
and
3.42 Vrms=V0√2,
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,
3.43 Pave=IrmsVrms,
which gives
3.44 Pave=I0√2⋅V0√2=12I0V0,
as stated above. It is standard practice to quote Irms, Vrms, and Pave rather than the peak values. For example, most household electricity is 120 V AC, which means that Vrms is 120 V. The common 10-A circuit breaker will interrupt a sustained Irms greater than 10 A. Your 1.0-kW microwave oven consumes Pave=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
3.45 Irms=VrmsR.
The various expressions for AC power Pave are
3.46 Pave=IrmsVrms,
3.47 Pave=V2rmsR,
and
3.48 Pave=I2rmsR.
Example 3.9 Peak Voltage and Power for Alternating Current
(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 Vrms is 120 V and Pave is 60.0 W. We can use Vrms=V0√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=V0√2 for the peak voltage V0 and substituting the known value for Vrms gives
3.49 V0=√2Vrms=1.414(120 V)=170 V.
Discussion for (a)
This means that the AC voltage swings from 170 V to –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,
3.50 P0=I0V0=2(12I0V0)=2Pave.
We know the average power is 60.0 W, and so
3.51 P0=2(60.0 W)=120 W.
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.