Part 1: AC Voltages and Current: A Simple AC Circuit

The goal of this chapter was to introduce the concept of an AC waveform and discuss some of the details associated with them. To conclude the chapter, this section presents a simple circuit involving and AC voltage source.

When dealing with circuits that that only contain resistances, things are straight forward even if the voltages in the circuit are AC. This is because resistors, for the most part, are not sensitive to frequency and they simply act like resistors.

Let’s start off by looking at the simple AC circuit shown below in Figure 1. This circuit should be very familiar to you by now and you recognize that the current flow though the resistor is governed by the voltage and Ohm's law.

Figure 1. A simple AC circuit.

Since resistors do not have any sensitivity to frequency, it is a simple matter to plot the voltage across the resistor and the current through it versus time. This is shown in the following figure for a 20Vp-p, 100Hz sine wave and a 2,000Ω resistor.

Figure 2. A plot of the voltage across and current through the resistor.

The voltage at any point in time is given by V = Vp * sin(ωt) where Vp is the peak voltage and ω is the angular frequency in radians per second. In this case Vp is 10V. Knowing that ω = 2πf and the frequency is 100Hz, we can rewrite this as: V = 10 * sin(2*π*100*t). With this equation we can find the voltage for any point in time by substituting the time value into “t” in the preceding equation and produce the plot shown in Figure 2.

If we wanted to find the peak current, we would simply use Ohm's law and the peak voltage value to find the answer. In this case, V=I*R so 10 = I * 2,000 or I = 0.005A.

We can also find the amount of power dissipated in the resistor by using the RMS voltage. The RMS voltage is found by Vrms = Vpeak/√2. In this case, it tells is that the RMS voltage is 7.071V. Using the relationship P = (V^2)/R from Figure 1 of the section titled Power, we find that the power dissipated in the resistor is P = (7.071^2)/2000 or P = 0.025W.