course index

Recall from last lecture:

Summary of the properties of circuit elements.
ResistorCapacitorInductor
units ohm, W = V / A farad, F = C / V henry, H = V s / A
symbol R C L
relation V = I R Q = C V emf = -L (DI / Dt)
power dissipated P = I V = I² R = V² / R 0 0
energy stored 0 PEC = C V² / 2 PEL = L I² / 2
time constant 0 t = RC t = L/R

Ch. 21: Alternating Current Circuits and Electromagnetic Waves

This is the final chapter in our discussion of electricity and magnetism. We will begin the chapter with an examination of ac circuits; simple circuits consisting of resistors, capacitors, and inductors connected to a source of ac voltage, like the generator of the last chapter. AC circuits are important, because our electrical power grid is based on delivering power via ac currents.

The latter half of the chapter introduces the concepts and equations related to electromagnetic waves. Electromagnetic waves span the spectrum from micro waves to radio waves to infrared radiation to visible light to ultraviolet light to X-rays and gamma rays.

21.1 Resistors in an ac Circuit

An ac circuit is a circuit powered by an ac generator. We discussed ac generators in chapter 20, and for the purposes here, they can be considered as a voltage source that varies sinusoidally with time (called an ac voltage source):

Dv = DVm sin(wt) = DVm sin(2p f t)
where Vm is the maximum voltage that appears, and f is the frequency in hertz. In this chapter, I'll denote voltages and currents that vary with time by lower case letters, v and i. Upper case letters, like Vm, represent constant voltage or current values.

Consider a circuit consisting of a resistor and an ac voltage source (sketch, showing symbol for ac voltage source). What is the current through the resistor? At any instant, the voltage across the resistor is v, and the current is i = v/R. The result is as shown in Figure 21.2:

i = v/R = (Vm / R) sin( 2p f t) = Im sin( 2p f t)
When plotted as a function of time, we see that the current is always proportional to the voltage, that is, it varies sinusoidally, and remains in phase with the voltage. When the voltage is positive the current is positive; when the voltage reaches its peak, the current reaches its peak; when the voltage is zero, the current is zero.

The average value of the current over one cycle is zero. (The same is true of the voltage.) This does NOT mean that the power dissipated in the resistor is zero. The current is zero because it is positive for half a cycle, then negative for the other half. But during each half, the power dissipated in the resistor is positive; a current moving through a resistor dissipates power, no matter which way it is moving.

It is useful to have an expression for the (average) power dissipated. Start with the expression for power:

P = i² R = Im² R sin²( 2p f t)
Next, we must average this expression over one cycle. Since Im and R are constants, the only thing to average is the sin² function, and the average of sin² over one cycle is 1/2. Thus:
P = ½ Im² R
The average current squared we call Ir² = <i²>av = ½ Im².
Ir = Im / sqrt(2) = 0.707 Im
The square root of this current, Ir, we call the root mean square (rms) current. The rms current is equal to the amount of direct current that will dissipate the same energy as the alternating current:
P = Ir² R

Similarly, we define the rms voltage as:

Vr = Vm / Sqrt(2) = 0.707 Vm
The 120V from our electrical outlets is the rms voltage. The peak voltage is about 170V.

In terms of rms voltage and current, we can write the ac equivalent of Ohm's law:

Vr = Ir R

Example: P21.2

An ac voltage source has an output of Dv = 150 sin 377t. Find (a) the rms voltage output, (b) the frequency of the source, and (c) the voltage at t = (1/120)s. (d) Find the maximum current in the circuit when the generator is connected to a 50.0W resistor.

(a) Use the relation between rms and peak voltage. The peak voltage is the value when the sine function is 1.
Vr = Vm / sqrt(2) = 0.707(150V) = 106V.

(b) The operand of the sine function equals 2p f t, therefore,
f = 377/2p = 60.0 Hz.

(c) At t=(1/120)s,
Dv = 150 sin (377/120) = 0.V
Remember that the operand of the sine function is in radians, so set your calculator to radians before calculating.

(d) When connected to a 50.0W resistor, the maximum current is related to the maximum voltage via Ohm's law:
Im = Vm / R = 150V / 50.0W = 3.00A.

21.2 Capacitors in an ac Circuit

Next, we want to understand how a capacitor affects an ac circuit, so let's begin with the simplest circuit, just a capacitor and ac voltage source (sketch). Consider what happens as the voltage goes through a cycle (Figure 21.5). First, the capacitor is uncharged, so as the voltage increases, current flows easily to begin charging the capacitor. Then, as the voltage increases, more charge is added, but at a slower rate, so the current decreases. When the voltage reaches its peak value, current momentarily stops flowing. As the voltage decreases, the capacitor begins to discharge, and the current becomes negative. And you get the idea.

The capacitor limits the current in the current. In addition, the current and voltage are out of phase; we say that the voltage lags the current, since the current peaks a quarter of a cycle (90°) before the voltage peaks.

We summarize the effect of the capacitor by the capacitive reactance:

XC = 1 / (2p f C)
The reactance has units of ohms, just like resistance. In terms of the reactance, the rms voltage across the capacitor is related to the rms current through the capacitor by:
VC = Ir XC

21.3 Inductors in an ac Circuit

And finally, consider what happens when an inductor is put in a circuit with an ac voltage source. Similar arguments apply as for the capacitor, but this time we find that the current leads the voltage by a quarter cycle (90°), and that the rms current is related to the rms voltage across the solenoid by:

VL = Ir XL
The quantity XL is called the inductive reactance, given by:
XL = 2p f L

Example: P21.15

A 2.40mF capacitor is connected across an alternating voltage with an rms value of 9.00V. The rms current in the capacitor is 25.0mA. (a) What is the source frequency? (b) If the capacitor is replaced by an ideal coil with an inductance of 0.160H, what is the rms current in the coil?

(a) From the rms voltage and current, we can determine the capacitive reactance, and this is related to the capacitance and the frequency of the source.
XC = Vr / Ir = 9.00V / 25.0mA = 360W.
f = 1 / (2p C XC) = 1 / (2p(2.40mF)(360W)) = 184Hz.

At this frequency, the inductive reactance is
XL = 2p f L = 2p (184Hz)(0.160H) = 185W.
The rms current can be found from the rms voltage and reactance:
Ir = Vr / XL = 9.00V / 185W = 48.6mA.

Summary of the properties of ac circuit elements.
ResistorCapacitorInductor
units ohm, W = V / A farad, F = C / V henry, H = V s / A
symbol R C L
reactance XR = R XC = 1 / (2p f C) XL = 2p f L
phase 0° -90° 90°
© Robert Harr 2000