## 1. Series Resistor-Inductor Circuits

In the previous section, we explored what would happen in simple resistor-only and inductor-only AC circuits. Now we will mix the two components together in series form and investigate the effects.

Take this circuit as an example to work with: (Figure below)

Series resistor inductor circuit: Current lags applied voltage by 0o to 90o.

The resistor will offer 5 Ω of resistance to AC current regardless of frequency, while the inductor will offer 3.7699 Ω of reactance to AC current at 60 Hz. Because the resistor’s resistance is a real number (5 Ω ∠ 0o, or 5 + j0 Ω), and the inductor’s reactance is an imaginary number (3.7699 Ω ∠ 90o, or 0 + j3.7699 Ω), the combined effect of the two components will be an opposition to current equal to the complex sum of the two numbers. This combined opposition will be a vector combination of resistance and reactance. In order to express this opposition succinctly, we need a more comprehensive term for opposition to current than either resistance or reactance alone. This term is called impedance, its symbol is Z, and it is also expressed in the unit of ohms, just like resistance and reactance. In the above example, the total circuit impedance is:

Impedance is related to voltage and current just as you might expect, in a manner similar to resistance in Ohm’s Law:

In fact, this is a far more comprehensive form of Ohm’s Law than what was taught in DC electronics (E=IR), just as impedance is a far more comprehensive expression of opposition to the flow of electrons than resistance is. Any resistance and any reactance, separately or in combination (series/parallel), can be and should be represented as a single impedance in an AC circuit.

To calculate current in the above circuit, we first need to give a phase angle reference for the voltage source, which is generally assumed to be zero. (The phase angles of resistive and inductive impedance are always 0o and +90o, respectively, regardless of the given phase angles for voltage or current).

As with the purely inductive circuit, the current wave lags behind the voltage wave (of the source), although this time the lag is not as great: only 37.016o as opposed to a full 90o as was the case in the purely inductive circuit. (Figure below)

Current lags voltage in a series L-R circuit.

For the resistor and the inductor, the phase relationships between voltage and current haven’t changed. Voltage across the resistor is in phase (0o shift) with the current through it; and the voltage across the inductor is +90o out of phase with the current going through it. We can verify this mathematically:

The voltage across the resistor has the exact same phase angle as the current through it, telling us that E and I are in phase (for the resistor only).

The voltage across the inductor has a phase angle of 52.984o, while the current through the inductor has a phase angle of -37.016o, a difference of exactly 90o between the two. This tells us that E and I are still 90o out of phase (for the inductor only).

We can also mathematically prove that these complex values add together to make the total voltage, just as Kirchhoff’s Voltage Law would predict:

Let’s check the validity of our calculations with SPICE: (Figure below)

Spice circuit: R-L.

ac r-l circuit
v1 1 0 ac 10 sin
r1 1 2 5
l1 2 0 10m
.ac lin 1 60 60
.print ac v(1,2) v(2,0) i(v1)
.print ac vp(1,2) vp(2,0) ip(v1)
.end
+
freq          v(1,2)      v(2)        i(v1)
6.000E+01     7.985E+00   6.020E+00   1.597E+00

freq          vp(1,2)     vp(2)       ip(v1)
6.000E+01    -3.702E+01   5.298E+01   1.430E+02
+

Note that just as with DC circuits, SPICE outputs current figures as though they were negative (180o out of phase) with the supply voltage. Instead of a phase angle of -37.016o, we get a current phase angle of 143o (-37o + 180o). This is merely an idiosyncrasy of SPICE and does not represent anything significant in the circuit simulation itself. Note how both the resistor and inductor voltage phase readings match our calculations (-37.02o and 52.98o, respectively), just as we expected them to.

With all these figures to keep track of for even such a simple circuit as this, it would be beneficial for us to use the "table" method. Applying a table to this simple series resistor-inductor circuit would proceed as such. First, draw up a table for E/I/Z figures and insert all component values in these terms (in other words, don’t insert actual resistance or inductance values in Ohms and Henrys, respectively, into the table; rather, convert them into complex figures of impedance and write those in):

Although it isn’t necessary, I find it helpful to write both the rectangular and polar forms of each quantity in the table. If you are using a calculator that has the ability to perform complex arithmetic without the need for conversion between rectangular and polar forms, then this extra documentation is completely unnecessary. However, if you are forced to perform complex arithmetic "longhand" (addition and subtraction in rectangular form, and multiplication and division in polar form), writing each quantity in both forms will be useful indeed.

Now that our "given" figures are inserted into their respective locations in the table, we can proceed just as with DC: determine the total impedance from the individual impedances. Since this is a series circuit, we know that opposition to electron flow (resistance or impedance) adds to form the total opposition:

Now that we know total voltage and total impedance, we can apply Ohm’s Law (I=E/Z) to determine total current:

Just as with DC, the total current in a series AC circuit is shared equally by all components. This is still true because in a series circuit there is only a single path for electrons to flow, therefore the rate of their flow must uniform throughout. Consequently, we can transfer the figures for current into the columns for the resistor and inductor alike:

Now all that’s left to figure is the voltage drop across the resistor and inductor, respectively. This is done through the use of Ohm’s Law (E=IZ), applied vertically in each column of the table:

And with that, our table is complete. The exact same rules we applied in the analysis of DC circuits apply to AC circuits as well, with the caveat that all quantities must be represented and calculated in complex rather than scalar form. So long as phase shift is properly represented in our calculations, there is no fundamental difference in how we approach basic AC circuit analysis versus DC.

Now is a good time to review the relationship between these calculated figures and readings given by actual instrument measurements of voltage and current. The figures here that directly relate to real-life measurements are those in polar notation, not rectangular! In other words, if you were to connect a voltmeter across the resistor in this circuit, it would indicate 7.9847 volts, not 6.3756 (real rectangular) or 4.8071 (imaginary rectangular) volts. To describe this in graphical terms, measurement instruments simply tell you how long the vector is for that particular quantity (voltage or current).

Rectangular notation, while convenient for arithmetical addition and subtraction, is a more abstract form of notation than polar in relation to real-world measurements. As I stated before, I will indicate both polar and rectangular forms of each quantity in my AC circuit tables simply for convenience of mathematical calculation. This is not absolutely necessary, but may be helpful for those following along without the benefit of an advanced calculator. If we were to restrict ourselves to the use of only one form of notation, the best choice would be polar, because it is the only one that can be directly correlated to real measurements.

Impedance (Z) of a series R-L circuit may be calculated, given the resistance ® and the inductive reactance (XL). Since E=IR, E=IXL, and E=IZ, resistance, reactance, and impedance are proportional to voltage, respectively. Thus, the voltage phasor diagram can be replaced by a similar impedance diagram. (Figure below)

Series: R-L circuit Impedance phasor diagram.

+

Example:

Given: A 40 Ω resistor in series with a 79.58 millihenry inductor. Find the impedance at 60 hertz.

XL = 2πfL
XL = 2π·60·79.58×10-3
XL = 30 Ω
Z = R + jXL
Z = 40 + j30
|Z| = sqrt(402 + 302) = 50 Ω
∠Z = arctangent(30/40) = 36.87o
Z = 40 + j30 = 50∠36.87o
+
• REVIEW:

• Impedance is the total measure of opposition to electric current and is the complex (vector) sum of ("real") resistance and ("imaginary") reactance. It is symbolized by the letter "Z" and measured in ohms, just like resistance ® and reactance (X).

• Impedances (Z) are managed just like resistances ® in series circuit analysis: series impedances add to form the total impedance. Just be sure to perform all calculations in complex (not scalar) form! ZTotal = Z1 + Z2 + . . . Zn

• A purely resistive impedance will always have a phase angle of exactly 0o (ZR = R Ω ∠ 0o).

• A purely inductive impedance will always have a phase angle of exactly +90o (ZL = XL Ω ∠ 90o).

• Ohm’s Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I

• When resistors and inductors are mixed together in circuits, the total impedance will have a phase angle somewhere between 0o and +90o. The circuit current will have a phase angle somewhere between 0o and -90o.

• Series AC circuits exhibit the same fundamental properties as series DC circuits: current is uniform throughout the circuit, voltage drops add to form the total voltage, and impedances add to form the total impedance.

## 2. Series Resistor-Capacitor Circuits

In the last section, we learned what would happen in simple resistor-only and capacitor-only AC circuits. Now we will combine the two components together in series form and investigate the effects. (Figure below)

Series capacitor circuit: voltage lags current by 0o to 90o.

The resistor will offer 5 Ω of resistance to AC current regardless of frequency, while the capacitor will offer 26.5258 Ω of reactance to AC current at 60 Hz. Because the resistor’s resistance is a real number (5 Ω ∠ 0o, or 5 + j0 Ω), and the capacitor’s reactance is an imaginary number (26.5258 Ω ∠ -90o, or 0 - j26.5258 Ω), the combined effect of the two components will be an opposition to current equal to the complex sum of the two numbers. The term for this complex opposition to current is impedance, its symbol is Z, and it is also expressed in the unit of ohms, just like resistance and reactance. In the above example, the total circuit impedance is:

Impedance is related to voltage and current just as you might expect, in a manner similar to resistance in Ohm’s Law:

In fact, this is a far more comprehensive form of Ohm’s Law than what was taught in DC electronics (E=IR), just as impedance is a far more comprehensive expression of opposition to the flow of electrons than simple resistance is. Any resistance and any reactance, separately or in combination (series/parallel), can be and should be represented as a single impedance.

To calculate current in the above circuit, we first need to give a phase angle reference for the voltage source, which is generally assumed to be zero. (The phase angles of resistive and capacitive impedance are always 0o and -90o, respectively, regardless of the given phase angles for voltage or current).

As with the purely capacitive circuit, the current wave is leading the voltage wave (of the source), although this time the difference is 79.325o instead of a full 90o. (Figure below)

Voltage lags current (current leads voltage)in a series R-C circuit.

As we learned in the AC inductance chapter, the "table" method of organizing circuit quantities is a very useful tool for AC analysis just as it is for DC analysis. Let’s place out known figures for this series circuit into a table and continue the analysis using this tool:

Current in a series circuit is shared equally by all components, so the figures placed in the "Total" column for current can be distributed to all other columns as well:

Continuing with our analysis, we can apply Ohm’s Law (E=IR) vertically to determine voltage across the resistor and capacitor:

Notice how the voltage across the resistor has the exact same phase angle as the current through it, telling us that E and I are in phase (for the resistor only). The voltage across the capacitor has a phase angle of -10.675o, exactly 90o less than the phase angle of the circuit current. This tells us that the capacitor’s voltage and current are still 90o out of phase with each other.

Let’s check our calculations with SPICE: (Figure below)

Spice circuit: R-C.

+
ac r-c circuit
v1 1 0 ac 10 sin
r1 1 2 5
c1 2 0 100u
.ac lin 1 60 60
.print ac v(1,2) v(2,0) i(v1)
.print ac vp(1,2) vp(2,0) ip(v1)
.end
+
freq          v(1,2)      v(2)        i(v1)
6.000E+01     1.852E+00   9.827E+00   3.705E-01

freq          vp(1,2)     vp(2)       ip(v1)
6.000E+01     7.933E+01  -1.067E+01  -1.007E+02
+

Once again, SPICE confusingly prints the current phase angle at a value equal to the real phase angle plus 180o (or minus 180o). However, its a simple matter to correct this figure and check to see if our work is correct. In this case, the -100.7o output by SPICE for current phase angle equates to a positive 79.3o, which does correspond to our previously calculated figure of 79.325o.

Again, it must be emphasized that the calculated figures corresponding to real-life voltage and current measurements are those in polar form, not rectangular form! For example, if we were to actually build this series resistor-capacitor circuit and measure voltage across the resistor, our voltmeter would indicate 1.8523 volts, not 343.11 millivolts (real rectangular) or 1.8203 volts (imaginary rectangular). Real instruments connected to real circuits provide indications corresponding to the vector length (magnitude) of the calculated figures. While the rectangular form of complex number notation is useful for performing addition and subtraction, it is a more abstract form of notation than polar, which alone has direct correspondence to true measurements.

Impedance (Z) of a series R-C circuit may be calculated, given the resistance ® and the capacitive reactance (XC). Since E=IR, E=IXC, and E=IZ, resistance, reactance, and impedance are proportional to voltage, respectively. Thus, the voltage phasor diagram can be replaced by a similar impedance diagram. (Figure below)

Series: R-C circuit Impedance phasor diagram.

+

Example:

Given: A 40 Ω resistor in series with a 88.42 microfarad capacitor. Find the impedance at 60 hertz.

XC = 1/(2πfC)
XC = 1/(2π·60·88.42×10-6)
XC = 30 Ω
Z = R - jXC
Z = 40 - j30
|Z| = sqrt(402 +  (-30)2) = 50 Ω
∠Z = arctangent(-30/40) = -36.87o
Z = 40 - j30 = 50∠-36.87o
+
• REVIEW:

• Impedance is the total measure of opposition to electric current and is the complex (vector) sum of ("real") resistance and ("imaginary") reactance.

• Impedances (Z) are managed just like resistances ® in series circuit analysis: series impedances add to form the total impedance. Just be sure to perform all calculations in complex (not scalar) form! ZTotal = Z1 + Z2 + . . . Zn

• Please note that impedances always add in series, regardless of what type of components comprise the impedances. That is, resistive impedance, inductive impedance, and capacitive impedance are to be treated the same way mathematically.

• A purely resistive impedance will always have a phase angle of exactly 0o (ZR = R Ω ∠ 0o).

• A purely capacitive impedance will always have a phase angle of exactly -90o (ZC = XC Ω ∠ -90o).

• Ohm’s Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I

• When resistors and capacitors are mixed together in circuits, the total impedance will have a phase angle somewhere between 0o and -90o.

• Series AC circuits exhibit the same fundamental properties as series DC circuits: current is uniform throughout the circuit, voltage drops add to form the total voltage, and impedances add to form the total impedance.

## 3. Series R, L, and C Circuits

### 3.1. LIEC (todo)

Let’s take the following example circuit and analyze it: (Figure below)

Example series R, L, and C circuit.

The first step is to determine the reactances (in ohms) for the inductor and the capacitor.

The next step is to express all resistances and reactances in a mathematically common form: impedance. (Figure below) Remember that an inductive reactance translates into a positive imaginary impedance (or an impedance at +90o), while a capacitive reactance translates into a negative imaginary impedance (impedance at -90o). Resistance, of course, is still regarded as a purely "real" impedance (polar angle of 0o):

+
(((02079.png)))

Example series R, L, and C circuit with component values replaced by impedances.

Now, with all quantities of opposition to electric current expressed in a common, complex number format (as impedances, and not as resistances or reactances), they can be handled in the same way as plain resistances in a DC circuit. This is an ideal time to draw up an analysis table for this circuit and insert all the "given" figures (total voltage, and the impedances of the resistor, inductor, and capacitor).

Unless otherwise specified, the source voltage will be our reference for phase shift, and so will be written at an angle of 0o. Remember that there is no such thing as an "absolute" angle of phase shift for a voltage or current, since its always a quantity relative to another waveform. Phase angles for impedance, however (like those of the resistor, inductor, and capacitor), are known absolutely, because the phase relationships between voltage and current at each component are absolutely defined.

Notice that I’m assuming a perfectly reactive inductor and capacitor, with impedance phase angles of exactly +90 and -90o, respectively. Although real components won’t be perfect in this regard, they should be fairly close. For simplicity, I’ll assume perfectly reactive inductors and capacitors from now on in my example calculations except where noted otherwise.

Since the above example circuit is a series circuit, we know that the total circuit impedance is equal to the sum of the individuals, so:

Inserting this figure for total impedance into our table:

We can now apply Ohm’s Law (I=E/R) vertically in the "Total" column to find total current for this series circuit:

Being a series circuit, current must be equal through all components. Thus, we can take the figure obtained for total current and distribute it to each of the other columns:

Now we’re prepared to apply Ohm’s Law (E=IZ) to each of the individual component columns in the table, to determine voltage drops:

Notice something strange here: although our supply voltage is only 120 volts, the voltage across the capacitor is 137.46 volts! How can this be? The answer lies in the interaction between the inductive and capacitive reactances. Expressed as impedances, we can see that the inductor opposes current in a manner precisely opposite that of the capacitor. Expressed in rectangular form, the inductor’s impedance has a positive imaginary term and the capacitor has a negative imaginary term. When these two contrary impedances are added (in series), they tend to cancel each other out! Although they’re still added together to produce a sum, that sum is actually less than either of the individual (capacitive or inductive) impedances alone. It is analogous to adding together a positive and a negative (scalar) number: the sum is a quantity less than either one’s individual absolute value.

If the total impedance in a series circuit with both inductive and capacitive elements is less than the impedance of either element separately, then the total current in that circuit must be greater than what it would be with only the inductive or only the capacitive elements there. With this abnormally high current through each of the components, voltages greater than the source voltage may be obtained across some of the individual components! Further consequences of inductors' and capacitors' opposite reactances in the same circuit will be explored in the next chapter.

Once you’ve mastered the technique of reducing all component values to impedances (Z), analyzing any AC circuit is only about as difficult as analyzing any DC circuit, except that the quantities dealt with are vector instead of scalar. With the exception of equations dealing with power (P), equations in AC circuits are the same as those in DC circuits, using impedances (Z) instead of resistances ®. Ohm’s Law (E=IZ) still holds true, and so do Kirchhoff’s Voltage and Current Laws.

To demonstrate Kirchhoff’s Voltage Law in an AC circuit, we can look at the answers we derived for component voltage drops in the last circuit. KVL tells us that the algebraic sum of the voltage drops across the resistor, inductor, and capacitor should equal the applied voltage from the source. Even though this may not look like it is true at first sight, a bit of complex number addition proves otherwise:

Aside from a bit of rounding error, the sum of these voltage drops does equal 120 volts. Performed on a calculator (preserving all digits), the answer you will receive should be exactly 120 + j0 volts.

We can also use SPICE to verify our figures for this circuit: (Figure below)

Example series R, L, and C SPICE circuit.

ac r-l-c circuit
v1 1 0 ac 120 sin
r1 1 2 250
l1 2 3 650m
c1 3 0 1.5u
.ac lin 1 60 60
.print ac v(1,2) v(2,3) v(3,0) i(v1)
.print ac vp(1,2) vp(2,3) vp(3,0) ip(v1)
.end
freq          v(1,2)      v(2,3)      v(3)        i(v1)
6.000E+01     1.943E+01   1.905E+01   1.375E+02   7.773E-02

freq          vp(1,2)     vp(2,3)     vp(3)       ip(v1)
6.000E+01     8.068E+01   1.707E+02  -9.320E+00  -9.932E+01

The SPICE simulation shows our hand-calculated results to be accurate.

As you can see, there is little difference between AC circuit analysis and DC circuit analysis, except that all quantities of voltage, current, and resistance (actually, impedance) must be handled in complex rather than scalar form so as to account for phase angle. This is good, since it means all you’ve learned about DC electric circuits applies to what you’re learning here. The only exception to this consistency is the calculation of power, which is so unique that it deserves a chapter devoted to that subject alone.

• REVIEW:

• Impedances of any kind add in series: ZTotal = Z1 + Z2 + . . . Zn

• Although impedances add in series, the total impedance for a circuit containing both inductance and capacitance may be less than one or more of the individual impedances, because series inductive and capacitive impedances tend to cancel each other out. This may lead to voltage drops across components exceeding the supply voltage!

• All rules and laws of DC circuits apply to AC circuits, so long as values are expressed in complex form rather than scalar. The only exception to this principle is the calculation of power, which is very different for AC.

### 3.2. Reactance, Impedance in Series RLC Circuits (Navy,todo)

To explain the various properties that exist within ac circuits, the series RLC circuit will be used. Figure 1 is the schematic diagram of the series RLC circuit. The symbol shown in Figure 1 that is marked E is the general symbol used to indicate an ac voltage source.

Figure 1. Series RLC circuit.

#### 3.2.1. Reactance

The effect of inductive reactance is to cause the current to lag the voltage, while that of capacitive reactance is to cause the current to lead the voltage. Therefore, since inductive reactance and capacitive reactance are exactly opposite in their effects, what will be the result when the two are combined? It is not hard to see that the net effect is a tendency to cancel each other, with the combined effect then equal to the difference between their values. This resultant is called REACTANCE; it is represented by the symbol X; and expressed by the equation X = XL - XC or X = XC - XL. Thus, if a circuit contains 50 ohms of inductive reactance and 25 ohms of capacitive reactance in series, the net reactance, or X, is 50 ohms - 25 ohms, or 25 ohms of inductive reactance.

For a practical example, suppose you have a circuit containing an inductor of 100 wH in series with a capacitor of .001 μF, and operating at a frequency of 4 MHz. What is the value of net reactance, or X?

Now assume you have a circuit containing a 100 - μH inductor in series with a .0002 μF capacitor, and operating at a frequency of 1 MHz. What is the value of the resultant reactance in this case?

You will notice that in this case the inductive reactance is smaller than the capacitive reactance and is therefore subtracted from the capacitive reactance.

These two examples serve to illustrate an important point: when capacitive and inductive reactance are combined in series, the smaller is always subtracted from the larger and the resultant reactance always takes the characteristics of the larger.

O13. What is the formula for determining total reactance in a series circuit where the values of XC and XL are known?

O14. What is the total amount of reactance (X) in a series circuit which contains an XL of 20 ohms and an XC of 50 ohms? (Indicate whether X is capacitive or inductive)

#### 3.2.2. Impedance

From your study of inductance and capacitance you know how inductive reactance and capacitive reactance act to oppose the flow of current in an ac circuit. However, there is another factor, the resistance, which also opposes the flow of the current. Since in practice ac circuits containing reactance also contain resistance, the two combine to oppose the flow of current. This combined opposition by the resistance and the reactance is called the IMPEDANCE, and is represented by the symbol Z.

Since the values of resistance and reactance are both given in ohms, it might at first seem possible to determine the value of the impedance by simply adding them together. It cannot be done so easily, however. You know that in an ac circuit which contains only resistance, the current and the voltage will be in step (that is, in phase), and will reach their maximum values at the same instant. You also know that in an ac circuit containing only reactance the current will either lead or lag the voltage by one-quarter of a cycle or 90 degrees. Therefore, the voltage in a purely reactive circuit will differ in phase by 90 degrees from that in a purely resistive circuit and for this reason reactance and resistance are rot combined by simply adding them.

When reactance and resistance are combined, the value of the impedance will be greater than either. It is also true that the current will not be in step with the voltage nor will it differ in phase by exactly 90 degrees from the voltage, but it will be somewhere between the in-step and the 90-degree out-of-step conditions. The larger the reactance compared with the resistance, the more nearly the phase difference will approach 90° . The larger the resistance compared to the reactance, the more nearly the phase difference will approach zero degrees.

If the value of resistance and reactance cannot simply be added together to find the impedance, or Z, how is it determined? Because the current through a resistor is in step with the voltage across it and the current in a reactance differs by 90 degrees from the voltage across it, the two are at right angles to each other. They can therefore be combined by means of the same method used in the construction of a right- angle triangle.

Assume you want to find the impedance of a series combination of 8 ohms resistance and 5 ohms inductive reactance. Start by drawing a horizontal line, R, representing 8 ohms resistance, as the base of the triangle. Then, since the effect of the reactance is always at right angles, or 90 degrees, to that of the resistance, draw the line XL representing 5 ohms inductive reactance, as the altitude of the triangle. This is shown in Figure 2. Now, complete the hypotenuse (longest side) of the triangle. Then, the hypotenuse represents the impedance of the circuit.

Figure 2. Vector diagram showing relationship of resistance, inductive reactance, and impedance in a series circuit.

One of the properties of a right triangle is:

Now suppose you apply this equation to check your results in the example given above.

When you have a capacitive reactance to deal with instead of inductive reactance as in the previous example, it is customary to draw the line representing the capacitive reactance in a downward direction. This is shown in Figure 3. The line is drawn downward for capacitive reactance to indicate that it acts in a direction opposite to inductive reactance which is drawn upward. In a series circuit containing capacitive reactance the equation for finding the impedance becomes:

Figure 3. Vector diagram showing relationship of resistance, capacitive reactance, and impedance in a series circuit.

In many series circuits you will find resistance combined with both inductive reactance and capacitive reactance. Since you know that the value of the reactance, X, is equal to the difference between the values of the inductive reactance, XL, and the capacitive reactance, XC, the equation for the impedance in a Series circuit containing R, XL, and XC then becomes:

In Figure 4 you will see the method which may be used to determine the impedance in a series circuit consisting of resistance, inductance, and capacitance.

Figure 4. Vector diagram showing relationship of resistance, reactance (capacitive and inductive), and impedance in a series circuit.

Assume that 10 ohms inductive reactance and 20 ohms capacitive reactance are connected in series with 40 ohms resistance. Let the horizontal line represent the resistance R. The line drawn upward from the end of R, represents the inductive reactance, X,. Represent the capacitive reactance by a line drawn downward at right angles from the same end of R. The resultant of X, and XC is found by subtracting X; from Xc. This resultant represents the value of X.

Thus:

The line, Z, will then represent the resultant of R and X. The value of Z can be calculated as follows:

O17. What is the value of Z in a series ac circuit where XL = 6 ohms, XC = 3 ohms, and R = 4 ohms?

#### 3.2.3. Series RLC Circuit Example

The principles and formulas that have been presented in this chapter are used in all ac circuits. The examples given have been series circuits.

This section of the chapter will not present any new material, but will be an example of using all the principles presented so far. You should follow each example problem step by step to see how each formula used depends upon the information determined in earlier steps. When an example calls for solving for square root, you can practice using the square-root table by looking up the values given.

The example series RLC circuit shown in Figure 5 will be used to solve for XL, XC, X, Z, IT, true power, reactive power, apparent power, and power factor.

The values solved for will be rounded off to the nearest whole number.

First solve for XL and XC.

Figure 5. Example series RLC circuit

Now solve for X

Use the value of X to solve for Z.

This value of Z can be used to solve for total current (IT).

Since current is equal in all parts of a series circuit, the value of IT can be used to solve for the various values of power.

The power factor can now be found using either apparent power and true power or resistance and impedance. The mathematics in this example is easier if you use impedance and resistance.

### 3.3. Ohms Law for AC

In general, Ohm’s law cannot be applied to alternating-current circuits since it does not consider the reactance which is always present in such circuits. However, by a modification of Ohm’s law which does take into consideration the effect of reactance we obtain a general law which is applicable to ac circuits. Because the impedance, Z, represents the combined opposition of all the reactances and resistances, this general law for ac is,

this general modification applies to alternating current flowing in any circuit, and any one of the values may be found from the equation if the others are known.

For example, suppose a series circuit contains an inductor having 5 ohms resistance and 25 ohms inductive reactance in series with a capacitor having 15 ohms capacitive reactance. If the voltage is 50 volts, what is the current? This circuit can be drawn as shown in Figure 6.

Figure 6. Series LC circuit.

Now suppose the circuit contains an inductor having 5 ohms resistance and 15 ohms inductive reactance in series with a capacitor having 10 ohms capacitive reactance. If the current is 5 amperes, what is the voltage?