Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
Lessons In Industrial Instrumentation-3.pdf
Скачиваний:
5
Добавлен:
25.06.2023
Размер:
2.69 Mб
Скачать

386

CHAPTER 5. AC ELECTRICITY

5.5.3Phasor expressions of impedance

The ultimate purpose of phasors is to simplify AC circuit analysis, so this is what we will explore now. Consider the problem of defining electrical opposition to current in an AC circuit. In DC (direct-current) circuits, resistance (R) is defined by Ohm’s Law as being the ratio between voltage (V ) and current (I):

R = VI

There are some electrical components, though, which do not obey Ohm’s Law. Capacitors and inductors are two outstanding examples. The fundamental reason why these two components do not follow Ohm’s Law is because they do not dissipate energy like resistances do. Rather than dissipate energy (in the form of heat and/or light), capacitors and inductors store and release energy from and to the circuit in which they are connected. The contrast between resistors and these components is remarkably similar to the contrast between friction and inertia in mechanical systems. Whether pushing a flat-bottom box across a floor or pushing a heavy wheeled cart across a floor, work is required to get the object moving. However, the flat-bottom box will immediately stop when you stop pushing it due to the energy loss inherent to friction, while the wheeled cart will continue to coast because it has kinetic energy stored in it. When a resistor disconnected from an electrical source, both voltage and current immediately cease. Capacitors and inductors, however, may store a “charge” of energy when disconnected from a source (capacitors retaining voltage and inductors retaining current).

The relationships between voltage and current for capacitors (C) and inductors (L) are as follows:

I = C

dV

V = L

dI

dt

dt

 

 

Expressed verbally, capacitors pass electric current proportional to how quickly the voltage across them changes over time. Conversely, inductors produce a voltage drop proportional to how quickly current through them changes over time. The symmetry here is beautiful: capacitors, which store energy in an electric field that is proportional to the applied voltage, oppose changes in voltage. Inductors, which store energy in a magnetic field that is proportional to applied current, oppose changes in current. The manner in which a capacitor or an inductor reacts to changes imposed upon it is a direct consequence of the Law of Energy Conservation: since energy can neither appear from nothing nor simply vanish, an exchange of energy must take place in order to alter the amount of energy stored within a capacitor or an inductor. The rate at which a capacitor’s voltage may change is directly related to the rate at which electric charge (current) enters or exits the capacitor. The rate at which an inductor’s current may change is directly related to the amount of electromotive force (voltage) impressed across the inductor.

When either type of component is placed in an AC circuit and subjected to oscillating signals, it will pass a finite amount of alternating current. Even though the mechanism of a capacitor’s or inductor’s opposition to current (called reactance) is fundamentally di erent from that of a resistor (called resistance), just like inertia di ers in its fundamental nature from friction, it is still convenient to express the amount of electrical opposition in a common unit of measurement: the ohm (Ω). To do this, we will have to figure out a way to take the above equations and manipulate them to express each component’s behavior as a ratio of VI .

5.5. PHASORS

387

Let’s start with capacitors. Suppose we impress a 1 volt peak AC voltage across a capacitor, representing that voltage as the exponential ejωt where ω is the angular velocity (frequency) of the signal and t is time:

AC voltage

 

 

 

 

 

source

 

C

 

 

Capacitor

 

 

 

V = ejωt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1 volt peak)

 

 

 

 

 

 

 

 

 

 

 

We will begin by writing the current/voltage relationship for a capacitor, along with the imaginary exponential function for the impressed voltage:

I = C

dV

V = ejωt

dt

Substituting ejωt for V in the capacitor formula, we see we must apply the calculus function of di erentiation to the time-based voltage function:

d

ejωt

I = C dt

Fortunately for us, di erentiation is a very simple20 process with exponential functions:

I = jωCejωt

Remember that our goal here is to solve for the ratio of voltage over current for a capacitor. So far all we have is a function for current (I) in terms of time (t). If we take this function for current and divide that into our original function for voltage, however, we see that ratio simplify quite nicely:

 

 

V

=

 

ejωt

 

 

 

 

I

jωCejωt

 

V

 

 

1

 

 

1

 

=

 

= −j

 

I

jωC

ωC

Note21 how the exponential term completely drops out of the equation, leaving us with a clean ratio strictly in terms of capacitance (C), angular velocity (ω), and of course j.

20Recall from calculus that the derivative of the function ex with respect to x is simply ex. That is, the value of an exponential function’s slope is equal to the value of the original exponential function! If the exponent contains any constants multiplied by the independent variable, those constants become multiplying coe cients after di erentiation. Thus, the derivative of ekx with respect to x is simply kekx. Likewise, the derivative of ejωt with respect to t is jωejωt.

21Note also one of the interesting properties of the imaginary operator: 1j = −j. The proof of this is quite simple:

1j = jj2 = j1 = −j.

2πf C
1

388

CHAPTER 5. AC ELECTRICITY

Next, will will apply this same analysis to inductors. Recall that voltage across an inductor and current through an inductor are related as follows:

V = L dIdt

If we describe the AC current22 through an inductor using the familiar imaginary exponential expression I = ejωt (representing a 1 amp peak AC current at frequency ω), we may substitute this expression for current into the inductor’s characteristic equation to solve for the inductor’s voltage

as a function of time:

 

V = L

dI

 

 

I = ejωt

dt

 

 

 

V = L

d

ejωt

 

 

 

dt

V = jωLejωt

Now that we have the inductor’s voltage expressed as a time-based function, we may include the original current function and calculate the ratio of V over I:

V

=

 

jωLejωt

I

 

ejωt

 

V

= jωL

 

 

I

 

 

 

 

 

In summary, we may express the impedance (voltage-to-current ratio) of capacitors and inductors by the following equations:

 

1

 

 

1

ZL = jωL

ZC =

 

or

− j

 

jωC

ωC

Most students familiar with electronics from an algebraic perspective (rather than calculus) find the expressions XL = 2πf L and XC = easier to grasp. Just remember that angular velocity

(ω) is really “shorthand” notation for 2πf , so these familiar expressions may be alternatively written as XL = ωL and XC = ωC1 .

22Note that we begin this analysis with an exponential expression of the current waveform rather than the voltage waveform as we did at the beginning of the capacitor analysis. It is possible to begin with voltage as a function of time and use calculus to determine current through the inductor, but unfortunately that would necessitate integration rather than di erentiation. Di erentiation is a simpler process, which is why this approach was chosen. If ejωt = L dIdt

R R ejωt

then ejωt dt = L dI. Integrating both sides of the equation yields ejωt dt = L dI. Solving for I yields jωL plus

a constant of integration representing a DC component of current that may or may not be zero depending on where the impressed voltage sinusoid begins in time. Solving for Z = V /I finally gives the result we’re looking for: jωL. Ugly, no?

5.5. PHASORS

389

Furthermore, recall that reactance (X) is a scalar quantity, having magnitude but no direction. Impedance (Z), on the other hand, possesses both magnitude and direction (phase), which is why the imaginary operator j must appear in the impedance expressions to make them complete. The impedance o ered by pure inductors and capacitors alike are nothing more than their reactance values (X) scaled along the imaginary (j) axis (phase-shifted 90o).

These di erent representations of opposition to electrical current are shown here for components exhibiting 50 ohms, resistances (R) and reactances (X) shown as scalar quantities near the component symbols, and impedances (Z) as phasor quantities on the complex plane:

 

+imag

 

+imag

 

+imag

 

R = 50 W

 

XC = 50 W

 

XL = 50 W

 

ZR = 50 W Ð 0o

 

 

 

ZL = 50 W Ð 90o

 

 

 

 

ZL = 0 + j50 W

 

ZR = 50 + j0 W

 

 

 

 

-real

+real

-real

+real

-real

+real

 

 

 

ZC = 50 W Ð -90o

 

 

 

 

 

ZC = 0 - j50 W

 

 

 

-imag

 

-imag

 

-imag