16
CHAPTER
Alternating-Current Circuit Analysis
WHEN YOU SEE AN AC CIRCUIT THAT CONTAINS COILS AND/OR CAPACITORS, YOU CAN ENVISION A complex-number half-plane, either RX (resistance-reactance) or GB (conductance-susceptance). The RX half-plane applies to series circuit analysis. The GB half-plane applies to parallel circuit analysis.
Complex Impedances in Series
In any situation where resistors, coils, and capacitors are connected in series, each component has an impedance that we can represent as a vector in the RX half-plane. The vectors for resistors remain constant regardless of the frequency. But the vectors for coils and capacitors vary as the frequency increases or decreases.
Pure Reactances
Pure inductive reactance (XL) and capacitive reactance (XC) simply add together when we connect coils and capacitors in series. That is,
X = XL + XC
In the RX half-plane, their vectors add, but because these vectors point in exactly opposite directions—inductive reactance upward and capacitive reactance downward, as shown in Fig. 16-1—the resultant sum vector inevitably points either straight up or straight down, unless the reactances are equal and opposite, in which case their sum equals the zero vector.
16-1 We can represent pure inductance and pure capacitance as reactance vectors that point straight up and down.
Problem 16-1
Suppose that we connect a coil and a capacitor in series with jXL = j200 and jXC = −j150. What’s the net reactance?
Solution
We add the values to get
jX = j200 + (−j150) = j(200 − 150) = j50
Suppose that we connect a coil and capacitor in series with jXL = j30 and jXC = −j110. What’s the net reactance?
Solution
Again, we add the values, obtaining
jX = j30 + (−j110) = j(30 − 110) = −j80
Problem 16-3
Suppose that we connect a coil of L = 5.00 μH and a capacitor of C = 200 pF in series, and then drive an AC signal through the combination at f = 4.00 MHz. What’s the net reactance?
Solution
First, we calculate the reactance of the inductor at 4.00 MHz to get
Next, we calculate the reactance of the capacitor at 4.00 MHz (noting that 200 pF = 0.000200 μF) to get
Finally, we add the reactances and round off to three significant figures, obtaining
jX = j125.664 + (−j198.943) = −j73.3
Beware of Cumulative Rounding Errors!
Do you wonder why we carried through some extra digits during the course of the preceding calculations, rounding off only at the very end of the process? The extra digits in the intermediate steps reduce our risk of ending up with a so-called cumulative rounding error. In situations where repeated rounding introduces many small errors, the final calculation can end up a full digit or two off, even after rounding to the appropriate number of significant figures. Let’s use this precaution, when necessary, in all of our future calculations.
Problem 16-4
What’s the net reactance of the above-described combination at f =10.0 MHz?
Solution
First, we calculate the reactance of the inductor at 10.0 MHz to get
Next, we calculate the reactance of the capacitor at 10.00 MHz to get
Finally, we add the reactances and round off, obtaining
jX = j314.16 + (−j79.58) = j235
Adding Impedance Vectors
Whenever the resistance in a series circuit reaches values that are significant compared with the reactance, the impedance vectors no longer point straight up and straight down. Instead, they run off towards the “northeast” (for the inductive part of the circuit) and “southeast” (for the capacitive part). Figure 16-2 shows an example of this condition.
16-2 When we have resistance along with reactance, the impedance vectors are neither vertical nor horizontal.
When two impedance vectors don’t lie along a single line, we must use vector addition to get the correct net impedance vector. Figure 16-3 shows the geometry of vector addition. We construct a parallelogram, using the two vectors Z1 = R1 + jX1 and Z2 = R2 + jX2 as two adjacent sides of the figure. The diagonal of the parallelogram constitutes the vector representing the net complex impedance. In a parallelogram, pairs of opposite angles always have equal measures. These equalities are indicated by the pairs of single and double arcs in Fig. 16-3.
16-3 Parallelogram method of complex-impedance vector addition.
Formula for Complex Impedances in Series
Consider two complex impedances, Z1 = R1 + jX1 and Z2 = R2 + jX2. If we connect these impedances in series, we can represent the net impedance Z as the vector sum
Z = (R1 + jX1) + (R2 + jX2) = (R1 + R2) + j(X1 + X2)
The resistance and reactance components add separately. Remember that inductive reactances are positive imaginary while capacitive reactances are negative imaginary!
Series RLC Circuits
With an inductance, capacitance, and resistance in series (Fig. 16-4), you can imagine the resistance R as belonging entirely to the coil, if you want to take advantage of the above formulas. Then you have only two vectors to add (instead of three), when you calculate the impedance of the series RLC circuit. Mathematically, the situation works out as follows:
Z = (R + jXL) + (0 + jXC) = R + j(XL + XC)
16-4 A series resistanceinductance-capacitance (RLC) circuit.
Again, remember that XC is never positive! So, although these general formulas contain addition symbols exclusively, you must add in a negative value (the equivalent of subtraction) when you include a capacitive reactance.
Problem 16-5
Suppose that we connect a resistor, a coil, and a capacitor in series with R = 50 ohms, XL = 22 ohms, and XC = −33 ohms. What’s the net impedance Z?
We can consider the resistor as part of the coil, obtaining 50 + j22 and 0 − j33. Adding these gives the resistance component of 50 + 0 = 50, and the reactive component of j22 − j33 = −j11. Therefore, Z = 50 − j11.
Problem 16-6
Consider a resistor, a coil, and a capacitor in series with R = 600 ohms, XL = 444 ohms, and XC = −444 ohms. What’s the net impedance, Z?
Solution
Again, we can imagine the resistor as part of the inductor. Then the complex impedance vectors are 600 + j444 and 0 − j444. Adding these, the resistance component equals 600 + 0 = 600, and the reactive component equals j444 − j444 = j0. We have a net impedance Z = 600 + j0, a purely resistive impedance.
Series Resonance
When a series-connected RLC circuit has zero net reactance at a certain frequency, we say that the circuit exhibits series resonance at that frequency.
Problem 16-7
Consider a resistor, a coil, and a capacitor connected in series. The resistor has a value of 330 ohms, the capacitance equals 220 pF, and the inductance equals 100 μH. We operate the circuit at 7.15 MHz. What’s the complex impedance?
Solution
First, we calculate the inductive reactance. Remember that
XL = 6.2832fL
and that megahertz and microhenrys go together in the formula. We multiply to obtain
jXL = j(6.2832 × 7.15 × 100) = j4492
Next, we calculate the capacitive reactance using the formula
XC = −1/(6.2832fC)
We can convert 220 pF to microfarads, obtaining C = 0.000220 μF. Then we have
jXC = −j[1/(6.2832 × 7.15 × 0.000220)] = −j101
Now, we can lump the resistance and the inductive reactance together, so one of the impedances becomes 330 + j4492. The other impedance equals 0 − j101. Adding these gives us
Z = 330 + j4492 − j101 = 330 + j4391
We can justify only three significant digits of accuracy here, so we might want to state this result as Z = 330 + j4.39k (remembering that “k” stands for “kilohms”).
Suppose that we connect a resistor, a coil, and a capacitor in series. The resistance equals 50.0 ohms, the inductance equals 10.0 μH, and the capacitance equals 1000 pF. We operate the circuit at 1592 kHz. What’s the complex impedance?
Solution
First, let’s calculate the inductive reactance. Note that 1592 kHz = 1.592 MHz. Plugging in the numbers, we obtain
jXL = j(6.2832 × 1.592 × 10.0) = j100
Next, we calculate the capacitive reactance. Let’s convert picofarads to microfarads, and use megahertz for the frequency. Then we have
When we put the resistance and the inductive reactance together into a single complex number, we get 50.0 + j100. We know that the capacitor’s impedance is 0 − j100. Adding the two complex numbers, we get
Z = 50.0 + j100 − j100 = 50.0 + j0
Our circuit exhibits a pure resistance of 50.0 ohms at 1592 kHz.
Complex Admittances in Parallel
When you see resistors, coils, and capacitors in parallel, remember that each component, whether a resistor, an inductor, or a capacitor, has an admittance that you can represent as a vector in the GB half-plane. The vectors for pure conductances remain constant, even as the frequency changes. But the vectors for the coils and capacitors vary with frequency.
Pure Susceptances
Pure inductive susceptance (BL) and capacitive susceptance (BC) add together when coils and capacitors appear in parallel. That is,
B = BL + BC
Remember that BL is never positive, and BC is never negative; we must invert the “sign scenario” with susceptance values as compared with reactance values.
In the GB half-plane, pure jBL and jBC vectors add. Because such vectors always point in exactly opposite directions—capacitive susceptance upward and inductive susceptance downward—the sum, jB, points either straight up or straight down, as shown in Fig. 16-5, unless the susceptances happen to be equal and opposite, in which case they cancel and the result equals the zero vector.
16-5 We can represent pure capacitance and pure inductance as susceptance vectors that point straight up and down.
Problem 16-9
Consider a coil and capacitor connected in parallel with jBL = −j0.05 and jBC = j0.08. What’s the net susceptance?
We add the values to get
jB = jBL + jBC = −j0.05 + j0.08 = j0.03
Problem 16-10
Suppose that we connect a coil and capacitor in parallel with jBL = −j0.60 and jBC = j0.25. What’s the net susceptance?
Solution
Again, we add the values to obtain
jB = −j0.60 + j0.25 = −j0.35
Problem 16-11
Suppose that we connect a coil of L = 6.00 μH and a capacitor of C = 150 pF in parallel and drive a signal through them at f = 4.00 MHz. What’s the net susceptance?
Solution
First, we calculate the susceptance of the inductor at 4.00 MHz, obtaining
Next, we calculate the susceptance of the capacitor (converting its value to microfarads) at 4.00 MHz, getting
Finally, we add the inductive and capacitive susceptances and round off to three significant figures, ending up with
jB = −j0.00663144 + j0.00376992 = −j0.00286
What’s the net susceptance of the above parallel-connected inductor and capacitor at a frequency of f = 5.31 MHz?
Solution
First, we calculate the susceptance of the inductor at 5.31 MHz to get
jBL = −j[1/(6.2832 × 5.31 × 6.00)] = −j0.00499544
Next, we calculate the susceptance of the capacitor (converting its value to microfarads) at 5.31 MHz, getting
jBC = j(6.2832 × 5.31 × 0.000150) = j0.00500457
Finally, we add the inductive and capacitive susceptances and round off to three significant figures, obtaining
jB = −j0.00499544 + j0.00500 = j0.00
Adding Admittance Vectors
When the conductance is significant in a parallel circuit containing inductance and capacitance, the admittance vectors don’t point straight up and down. Instead, they run off towards the “northeast” (for the capacitive part of the circuit) and “southeast” (for the inductive part), as shown in Fig. 16-6. You’ve seen how vectors add in the RX half-plane. In the GB half-plane, things work in the same way. The net admittance vector equals the sum of the component admittance vectors.
16-6 When we have conductance along with susceptance, admittance vectors are neither vertical nor horizontal.
Formula for Complex Admittances in Parallel
Imagine that we connect two admittances Y1 = G1+ jB1 and Y2 = G2 + jB2 in parallel. We can find the net admittance Y as the complex-number sum
Y = (G1 + jB1) + (G2 + jB2) = (G1 + G2) + j(B1 + B2)
Parallel RLC Circuits
When we connect a coil, capacitor, and resistor in parallel (Fig. 16-7), we can think of the resistance as a conductance in siemens (symbolized S), which equals the reciprocal of the value in ohms. If we consider the conductance as part of the inductor, we have only two complex numbers to add, rather than three, when finding the admittance of a parallel RLC circuit. We can use the formula
Y = (G + jBL) + (0 + jBC) = G + j(BL + BC)
16-7 A parallel resistance-inductance-capacitance (RLC) circuit. Here, G represents conductance (the reciprocal of resistance), so we can just as well call this a conductance-inductance-capacitance (GLC) circuit.
Again, we must remember that BL is never positive! So, although the formulas here have addition symbols in them, we actually subtract a value for the inductive susceptance by adding in a negative number.
Problem 16-13
Suppose that we connect a resistor, a coil, and a capacitor in parallel. The resistor has a conductance of G = 0.10 S. The susceptances are jBL = −j0.010 and jBC = j0.020. What’s the complex admittance of this combination?
Solution
Let’s consider the resistor as part of the coil, so we have two complex admittances in parallel: 0.10 − j0.010 and 0.00 + j0.020. Adding these values “part-by-part” gives us a conductance component of 0.10 + 0.00 = 0.10 and a susceptance component of −j0.010 + j0.020 = j0.010. Therefore, the complex admittance equals 0.10 + j0.010.
Problem 16-14
Consider a resistor, a coil, and a capacitor in parallel. The resistor has a conductance of G = 0.0010 S. The susceptances are jBL = −j0.0022 and jBC = j0.0022. What is the complex admittance of this combination?
Solution
Again, let’s consider the resistor as part of the coil. Then the complex admittances are 0.0010 − j0.0022 and 0.0000 + j0.0022. Adding these, we get a conductance component of 0.0010 + 0.0000 = 0.0010 and a susceptance component of −j0.0022 + j0.0022 = j0.0000. Thus, the admittance equals 0.0010 + j0.0000, a pure conductance.
Parallel Resonance
When a parallel RLC circuit lacks net susceptance at a certain frequency, we have a condition called parallel resonance at that frequency.
Problem 16-15
Suppose that we connect a resistor, a coil, and a capacitor in parallel. The resistor has a value of 100 ohms, the capacitance is 200 pF, and the inductance is 100 μH. We operate the circuit at a frequency of 1.00 MHz. What’s the net complex admittance?
Solution
First, let’s calculate the inductive susceptance. We recall the formula and plug in the numbers. Megahertz and microhenrys go together, so we have
Next, we calculate the capacitive susceptance. We can convert 200 pF to 0.000200 μF, so we have
We can consider the conductance, which equals 1/100 = 0.0100 S, and the inductive susceptance together so that one of the parallel-connected admittances equals 0.0100 − j0.00159155. The other admittance is 0 + j0.00125664. When we add these complex numbers and then round off the susceptance coefficient to three significant figures, we get
Problem 16-16
Consider a resistor, a coil, and a capacitor in parallel. The resistance is 10.00 ohms, the inductance is 10.00 μH, and the capacitance is 1000 pF. The frequency is 1592 kHz. What’s the complex admittance?
Solution
First, let’s calculate the inductive susceptance. We can convert the frequency to megahertz; 1592 kHz = 1.592 MHz. Now we plug in the numbers to get
jBL = −j[1/(6.2832 × 1.592 × 10.00)] = −j0.00999715
Next, we calculate the capacitive susceptance. We can convert 1000 pF to 0.001000 μF, so we obtain
jBC = j(6.2832 × 1.592 × 0.001000) = j0.01000285
Finally, we consider the conductance, which equals 1/10.00 = 0.1000 S, and the inductive susceptance in a single component, so that one of the parallel-connected admittances is 0.1000 − j0.00999715. The other is 0 + j0.01000285. Adding these complex numbers and then rounding off the susceptance coefficient to four significant figures, we obtain
Y = 0.1000 − j0.00999715 + j0.01000285
= 0.1000 − j0.000
Converting Complex Admittance to Complex Impedance
As we’ve seen, the GB half-plane looks like the RX half-plane, although mathematically they differ. We can convert a quantity from a complex admittance G + jB to a complex impedance R + jX using the formulas
R = G/(G2 + B2)
and
X = −B/(G2 + B2)
If we know the complex admittance, we should find the resistance and reactance components individually, using the above formulas. Then we can assemble the two components into the complex impedance R + jX.
Problem 16-17
Suppose that a circuit has an admittance Y = 0.010 − j0.0050. What’s the complex impedance, assuming the frequency never varies?
Solution
In this case, G = 0.010 S and B = −0.0050 S. We determine G2 + B2 as follows:
Knowing this common denominator, we can calculate R and X as
R = G/0.000125 = 0.010/0.000125 = 80 ohms
and
X = −B/0.000125 = 0.0050/0.000125 = 40 ohms
Our circuit has a complex impedance of Z = 80 + j40.
Putting It All Together
When we encounter a parallel circuit containing resistance, inductance, and capacitance, and we want to determine the complex impedance of the combination, we should go through the following steps in order:
• Calculate the conductance G of the resistor.
• Calculate the susceptance BL of the inductor.
• Calculate the susceptance BC of the capacitor.
• Determine the net susceptance B = BL + BC.
• Determine the quantity G2 + B2.
• Compute R in terms of G and B using the appropriate formula.
• Compute X in terms of G and B using the appropriate formula.
• Write down the complex impedance as the sum R + jX.
Suppose that we connect a resistor of 10.0 ohms, a capacitor of 820 pF, and a coil of 10.0 μH in parallel. We operate the circuit at 1.00 MHz. What’s the complex impedance?
Solution
Let’s proceed according to the steps described above, leaving in some extra digits in the susceptance figures and then rounding off to three significant figures at the end of the process, as follows:
• Calculate G = 1/R = 1/10.0 = 0.100.
• Calculate BL = −1/(6.2832fL) = −1/(6.2832 × 1.00 × 10.0) = −0.0159155.
• Calculate BC = 6.2832fC = 6.2832 × 1.00 × 0.000820 = 0.00515222. (We must remember to convert the capacitance from picofarads to microfarads.)
• Calculate B = BL + BC = −0.0159155 + 0.00515222 = −0.0107633.
• Define G2 + B2 = 0.1002 + (−0.0107633)2 = 0.0101158.
• Calculate R = G/0.0101158 = 0.100/0.0101158 = 9.89.
• Calculate X = −B/0.0101158 = 0.0107633/0.0101158 = 1.06.
• The complex impedance equals R + jX = 9.89 + j1.06.
Problem 16-19
Suppose that we connect a resistor of 47.0 ohms, a capacitor of 500 pF, and a coil of 10.0 μH in parallel. What’s their complex impedance at 2.25 MHz?
Solution
We proceed in the same fashion as we did when solving Problem 16-18, leaving in some extra digits in the conductance and susceptance figures until the end, as follows:
• Calculate G = 1/R = 1/47.0 = 0.0212766.
• Calculate BL = −1/(6.2832fL) = −1/(6.2832 × 2.25 × 10.0) = −0.00707354.
• Calculate BC = 6.2832fC = 6.2832 × 2.25 × 0.000500 = 0.0070686. (We convert the capacitance to microfarads.)
• Calculate B = BL + BC = −0.00707354 + 0.0070686 = 0.00000.
• Define G2 + B2 = 0.02127662 + 0.000002 = 0.00045269.
• Calculate R = G/0.00045269 = 0.0212766/0.00045269 = 47.000.
• Calculate X = −B/0.00045269 = 0.00000/0.00045269 = 0.00000.
• The complex impedance is R + jX = 47.000 + j0.00000. When we round both values off to three significant figures, we get 47.0 + j0.00. This complex-number quantity represents a pure resistance equal to the value of the resistor in the circuit.
Reducing Complicated RLC Circuits
Sometimes we’ll see circuits with several resistors, capacitors, and/or coils in series and parallel combinations. We can always reduce such a circuit to an equivalent series or parallel RLC circuit that contains one resistance, one capacitance, and one inductance.
Series Combinations
Resistances in series simply add. Inductances in series also add. Capacitances in series combine in a more complicated way, which you learned earlier. If you don’t remember the formula, it is
C = 1/(1/C1 + 1/C2 + … + 1/Cn)
where C1, C2, …, and Cn represent the individual capacitances, and C represents the net capacitance of the series combination. Figure 16-8A shows an example of a “complicated” series RLC circuit. Figure 16-8B shows the equivalent circuit with one resistance, one capacitance, and one inductance.
16-8 At A, a “complicated” series circuit containing multiple resistances and reactances. At B, the same circuit simplified. Resistances are in ohms; inductances are in microhenrys (μH); capacitances are in picofarads (pF).
Parallel Combinations
Resistances and inductances combine in parallel just as capacitances combine in series. Capacitances in parallel simply add up. Figure 16-9A shows an example of a “complicated” parallel RLC circuit. Figure 16-9B shows the equivalent circuit with one resistance, one capacitance, and one inductance.
16-9 At A, a “complicated” parallel circuit containing multiple resistances and reactances. At B, the same circuit simplified. Resistances are in ohms; inductances are in microhenrys (μH); capacitances are in picofarads (pF).
“Nightmare” Circuits
Imagine an RLC circuit like the one shown in Fig. 16-10. How would you find the net complex impedance at, say, 8.54 MHz? You’ll rarely encounter circuits such as this one in practical applications—or if you do, no one will likely ask you to calculate the net impedance at any particular frequency. But you can rest assured that, given a frequency, a complex impedance does exist, no matter how complicated the circuit might be.
16-10 A series-parallel “nightmare circuit” containing multiple resistances and reactances. Resistances are in ohms; inductances are in microhenrys (μH); capacitances are in picofarads (pF).
A true electronics “geek” could use a computer to work out the theoretical complex impedance of a circuit, such as the one in Fig. 16-10, at a specific frequency. In practice, however, an engineer might take an experimental approach by building the circuit, connecting a signal generator to its input terminals, and measuring the resistance R and the reactance X at the frequency of interest using a lab instrument called an impedance bridge.
Ohm’s Law for Alternating Current
We can state Ohm’s Law for DC circuits as a simple relationship between three variables: the current I (in amperes), the voltage E (in volts), and the resistance R (in ohms). Here are the formulas, in case you don’t recall them:
E = IR
I = E/R
R = E/I
In AC circuits containing no reactance, these same formulas apply, as long as we work with root-mean-square (RMS) voltages and currents. If you need to refresh your memory concerning the meaning of RMS, refer back to Chap. 9.
Purely Resistive Impedances
When the impedance Z in an AC circuit contains no reactance, all of the current and voltage exist through and across a pure resistance R. In that case, we can express Ohm’s Law in terms of the three formulas
E = IZ
I = E/Z
Z = E/I
where Z = R, and the values I and E represent the RMS values for the current and voltage.
Complex Impedances
When you want to determine the relationship between current, voltage, and resistance in an AC circuit that contains reactance along with the resistance, things get quite interesting. Recall the formula for the square of the absolute-value impedance in a series RLC circuit:
Z2 = R2 + X2
This equation tells us that
Z = (R2 + X2)1/2
so Z equals the length of the vector R + jX in the complex impedance plane. This formula applies only for series RLC circuits.
The square of the absolute-value impedance for a parallel RLC circuit, in which the resistance equals R and the reactance equals X, is defined as
Z2 = R2X2/(R2 + X2)
We can calculate Z, the absolute-value impedance, directly as
Z = [R2X2/(R2 + X2)]1/2
The ½ power of a quantity represents the positive square root of that quantity.
Imagine that a series RX circuit (shown by the generic block diagram of Fig. 16-11) has a resistance of R = 50.0 ohms and a capacitive reactance of X = −50.0 ohms. If we apply 100 V RMS AC to this circuit, what’s the RMS current?
16-11 A series circuit containing resistance and reactance. Illustration for Problems 16-20 through 16-23.
Solution
First, let’s calculate the complex impedance using the above formula for series circuits, going to a few extra digits to prevent cumulative rounding errors in later calculations. We have
which we can round off to 70.7 ohms. Now we can calculate the current, using the original “un-rounded” figure for Z, as
I = E/Z = 100 / 70.71068 = 1.414214 A RMS
which we can round off to 1.41 A RMS.
What are the RMS AC voltages across the resistance and the reactance, respectively, in the circuit described in Problem 16-20?
Solution
The Ohm’s Law formulas for DC will work here. We determined the current, going to several extra digits, as I = 1.414214 A RMS, so the voltage drop ER across the resistance is
ER = IR = 1.414214 × 50.0 = 70.7107 V RMS
which rounds off to 70.7 V RMS. The voltage drop EX across the reactance is
EX = IX = 1.41 × (−50.0) = −70.7107 V RMS
which rounds off to −70.7 V RMS. Do you see why we included all the extra digits in our intermediate calculations? If we’d used the rounded figure for the current (1.41 A RMS) in the foregoing calculations, we’d have gotten 70.5 and −70.5 V RMS for the answers we just found. That provides a great example of how cumulative rounding errors can mislead us if we’re not careful!
Signs and Phase with RMS Values
Here’s an important note in regards to the signs (plus or minus) in RMS figures. When we deal with RMS values, minus signs have no meaning. An RMS value can never be negative, so the minus sign in this result constitutes a mathematical artifact. We can consider the voltage across the reactance as 70.7 V RMS, equal in magnitude to the voltage across the resistance. But the phase is different, and that’s what the minus sign tells us!
Note that the voltages across the resistance and the reactance (a capacitive reactance in the above-described case because it’s negative) don’t add up to 100 V RMS, which appears across the whole circuit. This phenomenon occurs because, in an AC circuit containing resistance and reactance, we always observe a difference in phase between the voltage across the resistance and the voltage across the reactance. The voltages across the components add up to the applied voltage vectorially, but not always arithmetically.
Suppose that a series RX circuit (Fig. 16-11) has R = 10.0 ohms and X = 40.0 ohms. The applied voltage is 100 V RMS AC. What’s the current?
Solution
First, we calculate the complex impedance using the above formula for series circuits, obtaining
When we use Ohm’s Law to calculate the current, we get
I = E/Z = 100/41.23106 = 2.425356 A RMS
which rounds off to 2.43 A RMS.
What are the RMS AC voltages across the resistance and the reactance, respectively, in the circuit described in Problem 16-22?
Solution
Knowing the current (and using our “un-rounded” result), we can calculate the voltage across the resistance as
ER = IR = 2.425356 × 10.0 = 24.25356 V RMS
which rounds off to 24.3 V RMS. The voltage across the reactance is
EX = IX = 2.425356 × 40.0 = 97.01424 V RMS
which rounds off to 97.0 V RMS. If we take the arithmetic sum ER + EX, we get 24.25356 + 97.01424 = 121.2678 V RMS, which rounds off to 121.3 V RMS, as the total voltage across R and X. Again, this value comes out different from the actual applied voltage. The simple DC rule does not work here, for the same reason it didn’t work in the scenario of Problem 16-21. The AC voltage across the resistance doesn’t add up arithmetically with the AC voltage across the reactance because the two AC waves differ in phase.
Suppose that a parallel RX circuit (shown by the generic block diagram of Fig. 16-12) has R = 30.0 ohms and X = −20.0 ohms. We supply the circuit with E = 50.0 V RMS. What’s the total current drawn from the AC supply?
16-12 A parallel circuit containing resistance and reactance. Illustration for Problems 16-24 and 16-25.
Solution
First, we find the absolute-value impedance, remembering the formula for parallel circuits and going to a few extra digits to avoid cumulative rounding errors. We get
The total current is therefore
I = E/Z = 50/16.64332 = 3.004208 A RMS
which rounds off to 3.00 A RMS.
What are the RMS currents through the resistance and the reactance, respectively, in the circuit described in Problem 16-24?
Solution
The Ohm’s Law formulas for DC will work here. We can calculate the current through the resistance as
IR = E/R = 50.0/30.0 = 1.67 A RMS
For the current through the reactance, we calculate
IX = E/X = 50.0/(−20.0) = −2.5 A RMS
As before, we can neglect the minus sign when we think in terms of RMS, so we can call this current 2.5 A RMS. Note that if we directly add the current across the resistance to the current across the reactance, we don’t get 3.00 A, the actual total current. This effect takes place for the same reason that AC voltages don’t add arithmetically in AC circuits that contain resistance and reactance. The constituent currents, IR and IX, differ in phase. Vectorially, they add up to 3.00 A RMS, but arithmetically, they don’t.
Quiz
Refer to the text in this chapter if necessary. A good score is 18 correct. Answers are in the back of the book.
1. Suppose that in a series RLC circuit, R = 50 ohms and no net reactance exists. In which direction does the complex-impedance vector point?
(a) Straight up
(b) Straight down
(c) Straight toward the right
(d) Downward and toward the right
2. Suppose that in a parallel RLC circuit, G = 0.05 S and B = −0.05 S. In which direction does the complex-admittance (not the complex-impedance) vector point?
(a) Straight down
(b) Straight toward the right
(c) Upward and toward the right
(d) Downward and toward the right
3. Suppose that in a parallel RLC circuit, R = 10 ohms and jXC = −j10. In which direction does the complex-admittance (not the complex-impedance) vector point?
(a) Straight up
(b) Straight toward the right
(c) Upward and toward the right
(d) Downward and toward the right
4. A vector pointing upward and toward the right in the GB half-plane would indicate
(a) pure conductance.
(b) conductance and inductive susceptance.
(c) conductance and capacitive susceptance.
(d) None of the above
5. A vector pointing upward and toward the left in the RX half-plane would indicate
(a) pure resistance.
(b) resistance and inductive reactance.
(c) resistance and capacitive reactance.
(d) None of the above
6. Suppose that a coil has a reactance of j20 ohms. What’s the susceptance, assuming that the circuit contains nothing else?
(a) j0.050 S
(b) −j0.050 S
(c) j20 S
(d) −j20 S
7. Suppose that a capacitor has a susceptance of j0.040 S. What’s the reactance, assuming that the circuit contains nothing else?
(a) j0.040 ohms
(b) −j0.040 ohms
(c) j25 ohms
(d) −j25 ohms
8. Suppose that we connect a coil and capacitor in series with jXL = j50 and jXC = −j100. What’s the net reactance?
(a) j50
(b) j150
(c) −j50
(d) −j150
9. Suppose that we connect a coil of L = 3.00 μH and a capacitor of C = 100 pF in series, and then drive an AC signal through the combination at a frequency of f = 6.00 MHz. What’s the net reactance?
(a) −j152
(b) −j378
(c) j152
(d) j378
10. Consider a resistor, a coil, and a capacitor in series with R = 10 ohms, XL = 72 ohms, and XC = −83 ohms. What’s the net impedance Z?
(a) 10 + j11
(b) 10 − j11
(c) 82 − j11
(d) −73 − j11
11. Consider a resistor, a coil, and a capacitor connected in series. The resistor has a value of 220.0 ohms, the capacitance equals 500.00 pF, and the inductance equals 44.00 μH. We operate the circuit at a frequency of 5.650 MHz. What’s the complex impedance?
(a) 220.0 + j1506
(b) 220.0 − j1506
(c) 0.000 + j1506
(d) 220.0 + j0
12. Suppose that we connect a resistor, a coil, and a capacitor in series. The resistance equals 75.3 ohms, the inductance equals 8.88 μH, and the capacitance equals 980 pF. We operate the circuit at a frequency of 1340 kHz. What’s the complex impedance?
(a) 75.3 + j0.00
(b) 75.3 + j46.4
(c) 75.3 − j46.4
(d) 0.00 − j75.3
13. Consider a coil and capacitor connected in parallel with jBL = −j0.32 and jBC = j0.20. What’s the net susceptance?
(a) j0.52
(b) −j0.52
(c) j0.12
(d) −j0.12
14. Suppose that we connect a coil of 8.5 μH and a capacitor of 100 pF in parallel and drive a signal through them at 7.10 MHz. What’s the net susceptance?
(a) −j0.0045
(b) j0.0018
(c) −j0.0026
(d) None of the above
15. What’s the net susceptance of the parallel-connected inductor and capacitor described in Question 14 if we double the frequency to 14.2 MHz?
(a) −j0.0090
(b) j0.0036
(c) −j0.0013
(d) None of the above
16. Consider a resistor, a coil, and a capacitor in parallel. The resistance is 7.50 ohms, the inductance is 22.0 μH, and the capacitance is 100 pF. The frequency is 5.33 MHz. What’s the complex admittance?
(a) 0.133 + j0.00199
(b) 0.133 − j0.00199
(c) 7.50 + j503
(d) 7.50 − j503
17. Suppose that a circuit has an admittance of Y = 0.333 + j0.667. What’s the complex impedance, assuming the frequency does not change?
(a) 1.80 − j0.833
(b) 1.80 + j0.833
(c) 0.599 − j1.20
(d) 0.599 + j1.20
18. Suppose that we connect a resistor of 25 ohms, a capacitor of 0.0020 μF, and a coil of 7.7 μH in parallel (not in series!). We operate the circuit at 2.0 MHz. What’s the complex impedance?
(a) 8.1 + j22
(b) 8.1 − j22
(c) 22 + j8.1
(d) 22 − j8.1
19. Suppose that a series RX circuit has a resistance of R = 20 ohms and a capacitive reactance of X = −20 ohms. Suppose that we apply 42 V RMS AC to this circuit. How much current flows?
(a) 0.67 A RMS
(b) 1.5 A RMS
(c) 2.3 A RMS
(d) 3.0 A RMS
20. Suppose that a parallel RX circuit has R = 50 ohms and X = 40 ohms. We supply the circuit with E = 155 V RMS. How much current does the entire circuit draw from the AC source?
(a) 5.0 A RMS
(b) 2.5 A RMS
(c) 400 mA RMS
(d) 200 mA RMS