As shown in Table 2.1, the expressions for the voltages and currents of AC circuits containing resistors, capacitors, and inductors involve differentials and integrals with respect to time. Expressions for V (t) and I(t) of AC circuits can be obtained directly by solving the first-order and second-order ordinary differential equations that govern their behavior. The differential equation solutions for RC and RLC circuits subjected to step and sinusoidal inputs are presented later in Chapter 4. At this point, a working knowledge of AC circuits can be gained through some elementary considerations.
When capacitors and inductors are exposed to time-varying voltages in AC circuits, they each create a reactance to the voltage. Reactance plus resistance equals impedance. Symbolically, X + R = Z. Often an RLC component is described by its impedance because it encompasses both re- sistance and reactance. Impedance typically is considered generalized resis- tance [2]. For DC circuit analysis, impedance is resistance because there is no reactance. For this case, Z = R.
Voltages and currents in AC circuits usually do not vary simultaneously in the same manner. An increase in voltage with time, for example, can be followed by a corresponding increase in current at some later time. Such changes of voltage and current in time are characterized best by using com- plex number notation. This notation is described in more detail in Chapter 9.
Assume that the voltage, V (t), and the current, I(t), are represented by the complex numbers Voeiφ and Ioeiφ, respectively. Here eiφ is given by
Euler’s formula,
eiφ= cos φ + i sin φ, (2.35)
because i is used for the current. Throughout this text, i symbolizes the imaginary number. The real voltage and real current are obtained by multi- plying each by the complex number representation eiωtand then taking the
real part, Re, of the resulting number. That is,
V (t) = Re(V eiωt) = Re(V ) cos ωt − Im(V ) sin ωt (2.36) and
I(t) = Re(Ieiωt) = Re(I) cos ωt − Im(I) sin ωt, (2.37) in which ω is the frequency in rad/s. The frequency in cycles/s is f , where 2πf = ω.
Expressions for capacitive and inductive reactances can be derived using the voltage and current expressions given in Equations 2.36 and 2.37 [2]. For a capacitor, I(t) = CdV (t)/dt. Differentiating Equation 2.36 with respect to time yields the current across the capacitor,
I(t) = −VoCω sin ωt = Re[Vo
ei/ωC
1/iωC]. (2.38)
The denominator of the real component is the capacitive reactance,
XC= 1/iωC. (2.39)
In other words, V (t) = I(t)XC.
For the inductor, V (t) = LdI(t)/dt. Differentiating Equation 2.37 with respect to time yields the voltage across the inductor,
V (t) = −IoLω sin ωt = Re[iωLIoei/ωC]. (2.40)
The numerator of the real component is the inductive reactance,
XC= i/ωL. (2.41)
Simply put, V (t) = I(t)XL.
Because the resistances of capacitors and inductors are effectively zero, their impedances equal the reactances. Further, the resistor has no reac- tance, so its impedance is its resistance. Thus, ZR = R, ZC = 1/iωC, and
ZL= iωL.
Ohm’s law is still valid for AC circuits. It now can be written as V = ZI. Also the rules for adding resistances apply to adding impedances, where for impedances in series
ZT =
X
Zi, (2.42)
and for impedances in parallel
ZT = 1/
X ( 1
Zi
Example Problem 2.5
Statement: An electrical circuit loop is comprised of a resistor, R, a voltage source, Eo, and a 1-µF capacitor, C, that can be added into the loop by a switch. When the
switch is closed, all three electrical components are in series. When the switch is open, the loop has only the resistor and the voltage source in series. Eo= 3 V and R = 2 Ω.
Determine the expression for the current as a function of time, I(t), immediately after the switch is closed.
Solution: Applying Kirchhoff’s second law to the loop gives RI(t) + 1
C Z
I(t)dt = Eo.
This equation can be differentiated with respect to time to yield RdI(t) dt + I(t) C = dEo dt = 0,
because Eo is a constant 3 V. This equation can be integrated to obtain the result
I(t) = C1e−t/RC, where C1 is a constant. Now at t = 0 s, current flows through the
loop and equals Eo/R. This implies that C1= 3/2 = 1.5 A. Thus, for the given values
of R and C, I(t) = 1.5e−t/(2 × 10−6)
. This means that the current in the loop becomes almost zero in approximately 10 µs after the switch is closed.
FIGURE 2.12
2.7
*Equivalent Circuits
Th´evenin’s equivalent circuit theorem states that any two-terminal network of linear impedances, such as resistors and inductors, and voltage sources can be replaced by an equivalent circuit consisting of an ideal voltage source, ET h, in series with an impedance, ZT h. This circuit is shown in the
top of Figure 2.12.
Norton’s equivalent circuit theorem states that any two-terminal network of linear impedances and current sources can be replaced by an equivalent circuit consisting of an ideal current source, IT h, in parallel with
an impedance, ZT h. This circuit is illustrated in the bottom of Figure 2.12.
The voltage of the Th´evenin equivalent circuit is the current of the Nor- ton equivalent circuit times the equivalent impedance. Obtaining the equiv- alent impedance sometimes can be tedious, but it is very useful in under- standing circuits, especially the more complex ones.
The Th´evenin equivalent voltage and equivalent impedance can be determined by examining the open-circuit voltage and the short-circuit cur- rent. The Th´evenin equivalent voltage, ET h, is the open-circuit voltage,
which is the potential difference that exists between the circuit’s two ter- minals when nothing is connected to the circuit. This would be the voltage measured using an ideal voltmeter. Simply put, ET h= Eoc, where the sub-
script oc denotes open circuit. The Th´evenin equivalent impedance, ZT h, is Eth divided by the short-circuit current, Isc, where the subscript
sc denotes short circuit. The short-circuit current is the current that would pass through an ideal ammeter connected across the circuit’s two terminals. An example diagram showing an actual circuit and its Th´evenin equiv- alent is presented in Figure 2.13. In this figure, ZT his represented by RT h
because the only impedances in the circuit are resistances. Rmdenotes the
meter’s resistance, which would be infinite for an ideal voltmeter and zero for an ideal ammeter.
The Th´evenin equivalents can be found for the actual circuit. Kirchhoff’s voltage law implies
Ei = I(R1+ R2). (2.44)
Also, the voltage measured by the ideal voltmeter, Em, using Ohm’s law
and noting that R2<< Rm, is
Em= Eth= IR2. (2.45)
Combining Equations 2.44 and 2.45 yields the open-circuit or Th´evenin equivalent voltage,
ET h= Ei
R2
R1+ R2
FIGURE 2.13
A circuit and its Th´evenin equivalent.
Further, the short-circuit current would be
Isc= Ei/R1. (2.47)
So, the Th´evenin equivalent resistance is RT h≡ ET h Isc = R1R2 R1+ R2 . (2.48)
The resulting Th´evenin equivalent circuit is shown in the bottom of Figure 2.13.
An alternative approach to determining the Th´evenin impedance is to replace all voltage sources in the circuit by their internal impedances and then find the circuit’s output impedance. Usually the voltage sources’ in- ternal impedances are negligible and can be assumed to be equal to zero, effectively replacing all voltage sources by short circuits. For the circuit shown in Figure 2.13, this approach would lead to having the resistances R1
and R2 in parallel to ground, leading directly to Equation 2.48.
This alternative approach can be applied also when determining the Th´evenin equivalent resistance, which is the output impedance, of the
Wheatstone bridge circuit shown in Figure 2.10. Assuming a negligible in- ternal impedance for the voltage source Ei, RT his equivalent to the parallel
combination of R1and R2in series with the parallel combination of R3and
R4. That is, RT h= R1R2 R1+ R2 + R3R4 R3+ R4. (2.49) Example Problem 2.6
Statement: For the circuit shown in the top of Figure 2.13, determine the Th´evenin equivalent resistance and the Th´evenin equivalent voltage, assuming that Vs= 20 V,
R1 = 6 Ω, R2= 3 Ω, and Rm= 3 MΩ.
Solution: Because Rm>> R2, the Th´evenin equivalent voltage is given by Equation
2.46 and the Th´evenin equivalent resistance by Equation 2.48. Substitution of the given values for Vs= Ei, R1, and R2into these equations yields Eth= (20)
h 3 6+3 i = 6.67 V and Rth= (6)(3)6+3 = 2 Ω.
2.8
*Meters
FIGURE 2.14Voltage and current meters.
All voltage and current meters can be represented by Th´evenin and Norton equivalent circuits, as shown in Figure 2.14. These meters are char- acterized by their input impedances. An ideal voltmeter has an infinite input impedance such that no current flows through it. An ideal amme- ter has zero input impedance such that all the connected circuit’s current flows through it. The actual devices differ from their ideal counterparts only in that the actual impedances are neither zero nor infinite, but finite.
A voltmeter is attached in parallel to the point of interest in the circuit. An ammeter is attached in series with the point of interest in the circuit. A good voltmeter has a very high input impedance, typically greater than 1 MΩ. Because of this, a good voltmeter connected to a circuit draws negligible current from the circuit and, therefore, has no additional voltage difference present between the voltmeter’s terminals. Likewise, because a good amme- ter has a very low input impedance, typically less than 1 Ω, almost all of the attached circuit’s current flows through the ammeter.
Resistance measurements typically are made using an ohmmeter. The resistance actually is determined by passing a known current through the test leads of a meter and the unknown resistance and then measuring the total voltage difference across them. This is called the two-wire method. This approach is valid provided that the unknown resistance is much larger than the resistances of the test leads. In practice, this problem is circum- vented by using a multimeter and the four-wire method. This method requires the use of two additional test leads. Two of the leads carry a known current through the unknown resistance and then back to the meter, while the other two leads measure the resulting voltage drop across the unknown resistance. The meter determines the resistance by Ohm’s law and then displays it.
2.9
*Impedance Matching and Loading Error
When the output of one electronic component is connected to the input of another, the output signal may be altered, depending upon the component impedances. Each measurement circumstance requires a certain relation be- tween the output component’s output impedance and the input component’s input impedance to avoid signal alteration. If this impedance relation is not maintained, then the output component’s signal will be altered upon connec- tion to the input component. A common example of impedance mismatch is when an audio amplifier is connected to a speaker with a high input impedance. This leads to a significant reduction in the power transmitted to the speaker, which results in a low volume from the speaker.
A loading error can be introduced whenever one circuit is attached to another. Loading error, eload, is defined in terms of the difference be-
tween the true output impedance, Rtrue, the impedance that would be mea-
sured across the circuit’s output terminals by an ideal voltmeter, and the impedance measured by an actual voltmeter, Rmeas. Expressed on a per-
centage basis, the loading error is eload= 100 Rtrue− Rmeas Rtrue . (2.50)
FIGURE 2.15
Voltage circuit (top) and current circuit (bottom) illustrating loading error. Loading errors that occur when measuring voltages, resistances, or cur- rent can be avoided by following two simple rules. These rules, summarized at the end of this section, can be derived by considering two circuits, one in which an actual voltage source circuit is connected to an actual voltmeter, and the other in which an actual current source circuit is connected to an actual ammeter. These circuits are shown in Figure 2.15.
For the voltage circuit, Kirchhoff’s voltage law applied around the outer circuit loop gives
Vm= Vs− IoRout. (2.51)
Kirchhoff’s current law applied at node A yields
Io= IA= Vm/Rin, (2.52)
where all of the current flows through the voltmeter’s Rin. Substituting
Equation 2.52 into Equation 2.51 results in
Vm= Vs 1 1 +Rout Rin = Vs R in Rin+ Rout . (2.53)
that Rtrue= Rout and Rmeas = (RinRout)/(Rin+ Rout), the loading error becomes eload,V = Rout Rin+ Rout . (2.54)
For the current circuit, Kirchhoff’s current law applied at node B yields
Is= IB+ Io. (2.55)
Kirchhoff’s voltage law applied around the circuit loop containing Rin and
Rout gives
ImRin= IBRout. (2.56)
Substituting Equation 2.56 into Equation 2.55 results in
Im= Is 1 1 +Rin Rout = Is Rout Rin+ Rout . (2.57)
When Rin << Rout, Im = Is. Noting for the current measurement case
that Rtrue = Rin and Rmeas = (RinRout)/(Rin+ Rout), the loading error
becomes eload,I = Rin Rin+ Rout . (2.58)
Loading errors can be avoided between two circuits by connecting them via a buffer that has near-infinite input and near-zero output impedances. This is one of the many uses of operational amplifiers. These are presented in Chapter 3.
Example Problem 2.7
Statement: Determine the minimum input impedance, Rmin, of a voltage measure-
ment circuit that would have less than 0.5 % loading error when connected to a circuit having an output impedance of 50 Ω.
Solution: Direct application of Equation 2.54 implies 0.5
100=
50 Ω 50 Ω + Rmin
.
Solving for the minimum input impedance gives Rmin= 9950 Ω, or approximately 10
kΩ. This condition can be met by using a unity-gain operational amplifier in the non- inverting configuration at the input of the voltage-measurement circuit (see Chapter 3).
The impedance relation for optimum power transmission between an output source and an input circuit can be determined [8]. For the voltage circuit in Figure 2.15, noting that the power received, Pin, equals Vin2/Rin,
Equation 2.53 becomes Pin= Vs2 Rin (Rin+ Rout)2 . (2.59)
Differentiating Equation 2.59 with respect to Rin, setting the result equal
to zero and solving for Rin gives
Rin= Rout. (2.60)
Substitution of Equation 2.60 into the derivative equation shows that this condition ensures a maximum transmission of power. Equation 2.60 rep- resents true impedance matching, where the two impedances have the same value.
Example Problem 2.8
Statement: Determine the power that is transmitted, Pt, between two connected
circuits if the output circuit impedance is 6.0 Ω, the input circuit impedance is 4.0 Ω, and the source voltage is 12 V.
Solution: Substitution of the given values into Equation 2.53 gives Vm= 12
h
4 6+4
i = 4.8 V. Now, the power transmitted is given by Pt = Vin2/Rin = 4.82/4 = 5.8 W,
with the correct number of significant figures.
Impedance matching also is critical when an output circuit that generates waveforms is connected by a cable to a receiving circuit. In this situation, the high-frequency components of the output circuit can reflect back from the receiving circuit. This essentially produces an input wave to the receiving circuit that is different from that intended. When a cable with characteristic impedance, Rcable, is connected to a receiving circuit of load impedance,
Rin, and these impedances are matched, then the the input wave will not
be reflected. The reflected wave amplitude, Ar, is related to the incident
wave amplitude, Ai, by Ar= Ai Rcable− Rin Rcable+ Rin . (2.61)
When Rcable< Rin, the reflected wave is inverted. When Rcable> Rin, the
reflected wave is not inverted [2].
The rules for impedance matching and for loading error minimization, as specified by Equations 2.53, 2.57, 2.60, and 2.61, are as follows:
• Rule 1 − loading error minimization: When measuring a voltage, the input impedance of the measuring device must be much greater than the equivalent circuit’s output impedance.
• Rule 2 − loading error minimization: When measuring a current, the input impedance of the measuring device must be much less than the equivalent circuit’s output impedance.
• Rule 3 − impedance matching: When transmitting power to a load, the output impedance of the transmission circuit must equal the input impedance of the load for maximum power transmission.
• Rule 4 − impedance matching: When transmitting signals having high frequency through a cable, the cable impedance must equal the load impedance of the receiving circuit.
2.10
*Electrical Noise
Electrical noise is defined as anything that obscures a signal [2]. Noise is characterized by its amplitude distribution, frequency spectrum, and the physical mechanism responsible for its generation. Noise can be subdivided into intrinsic noise and interference noise. Intrinsic noise is random and primarily the result of thermally induced molecular motion in any resistive element (Johnson noise), current fluctuations in a material (shot noise), and local property variations in a material (1/f or pink noise). The first two are intrinsic and cannot be eliminated. The latter can be reduced through quality control of the material that is used.
Noise caused by another signal is called interference noise. Interference noise depends on the amplitude and frequency of the noise source. Common noise sources include AC-line power (50 Hz to 60 Hz), overhead fluorescent lighting (100 Hz to 120 Hz), and sources of radio-frequency (RF) and elec- tromagnetically induced (EMI) interference, such as televisions, radios, and high-voltage transformers.
The causes of electrical interference include local electric fields, magnetic fields, and ground loops. These noticeably affect analog voltage signals with amplitudes less than one volt. A surface at another electric potential that is near a signal-carrying wire will establish an undesirable capacitance between the surface and the wire. A local magnetic field near a signal-carrying wire will induce an additional current in the wire. A current flowing through one ground point in a circuit will generate a signal in another part of the circuit that is connected to a different ground point.
Most interference noise can be attenuated to acceptable levels by proper shielding, filtering, and amplification. For example, signal wires can be shielded by a layer of conductor that is separated from the signal wire by an insulator. The electric potential of the shield can be driven at the same po- tential as the signal through the use of operational amplifiers and feedback,
thereby obviating any undesirable capacitance [5]. Pairs of insulated wires carrying similar signals can be twisted together to produce signals with the same mode of noise. These signals subsequently can be conditioned using common-mode rejection techniques. Use of a single electrical ground point for a circuit almost always will minimize ground-loop effects. Signal ampli- fication and filtering also can be used. In the end though, it is better to eliminate the sources of noise than to try to cover them up.
The magnitude of the noise is characterized through the signal-to-noise ratio (SNR). This is defined as
SNR ≡ 10 log10 V2 s V2 n , (2.62)
where Vsand Vndenote the source and noise voltages, respectively. The volt-
age values usually are rms values (see Chapter 9). Also, a center frequency and range of frequencies are specified when the SNR is given.
2.11
Problem Topic Summary
Topic Review Problems Homework Problems
Basics 1, 2, 4, 6, 7, 13, 14, 15 8, 9, 10 20, 21, 22, 23, 24, 25 Circuits 3, 5, 8, 9, 10, 11, 12 4, 5, 7, 8, 11, 12 16, 17, 18, 19 Systems 8, 9, 10, 11 1, 2, 3, 6 Op Amps 23 13, 14 TABLE 2.2
Chapter 2 Problem Summary
2.12
Review Problems
1. Three 11.9 µF capacitors are placed in series in an electrical circuit. Compute the total capacitance in µF to one decimal place.
2. Which of the following combination of units is equivalent to 1 J? (a) 1 C·A·W, (b) 1 W·s/C, (c) 1 N/C, (d) 1 C·V.
FIGURE 2.16 Electrical circuit.
3. For the electrical circuit depicted in Figure 2, given R1 = 160 Ω, R3 =
68 Ω, I1 = 0.9 A, I3 = 0.2 A, and R2= R4, find the voltage potential,
Quantity Famous Person
current James Joule
charge Charles Coulomb
electric field work Georg Ohm electric potential James Watt
resistance Andre Ampere
power Michael Faraday
inductance Joseph Henry
capacitance Alessandro Volta TABLE 2.3
Famous people and electric quantities.
4. The ends of a wire 1.17 m in length are suspended securely between two insulating plates. The diameter of the wire is 0.000 05 m. Given that the electric resistivity of the wire is 1.673 × 10−6 Ω·m at 20.00◦C and
that its coefficient of thermal expansion is 56.56 × 10−5/◦C, compute
the internal resistance in the wire at 24.8 ◦C to the nearest whole ohm.
5. A wire with the same material properties given in the previous problem is used as the R1 arm of a Wheatstone bridge. The bridge is designed to
be used in deflection method mode and to act as a transducer in a system used to determine the ambient temperature in the laboratory. The length of the copper wire is fixed at 1.00 m and the diameter is 0.0500 mm. R2= R3= R4= 154 Ω and Ei = 10.0 V. For a temperature of 25.8◦C,
compute the output voltage, Eo, in volts to the nearest hundredth.
6. Which of the following effects would most likely not result from routing an AC signal across an inductor? (a) A change in the frequency of the