A Series Circuit

A good way to begin this chapter is by applying what you have already learned. Therefore, first a series RLC circuit shown in Figure 13.1 will be discussed. It consists of two resistors of 10 ohms and 4 ohms connected to an inductor with a reactance of 8 ohms and a capacitor with a reactance of 20 ohms. To illustrate the technique that will be used throughout this chapter, the circuit first will be divided into two complex impedances as shown in Figure 13.2

The impedance of the series combination of R₁ and L is called Z₁. It can be written:

(13–1)

or

The impedance of the series combination of R₂ and C, is called Z₂. It can be written:

(13–2)

or

jX_L is a positive quantity because the vector X_L leads the vector R by 90 degrees on the impedance phasor diagram shown in Figure 13.3 jX_C is a negative quantity because the vector X_C lags the vector R on the same diagram.

The two series impedances, Z₁ and Z₂, can be treated as if they were series resistances. The total impedance of the circuit is:

(13–3)

Substituting the expressions for Z₁ and Z₂,

A Parallel Circuit

Next, a parallel circuit composed of two resistive-reactance branches as shown in Figure 13.4 will be solved. The RL branch has a resistance of 3 ohms and an inductive reactance of 4 ohms. The RC branch has a resistance of 6 ohms and capacitive reactance of 10 ohms. The same technique outlined in the previous example will be used. First divide the circuit into two complex impedances, Z₁ and Z₂, as shown in Figure 13.5. Then write the complex number representation of the impedance for each branch. Using equation 13–1 the resistive-inductive branch, Z₁, can be written as:

The impedance of the resistiveresistive-capacitivendash;capacitive branch, Z₂, can be written:

The total impedance of the circuit is the parallel combination of Z₁ and Z₂. Recall from earlier chapters that when two resistors are connected in parallel, the total resistance can be calculated using the product–over–sum equation:

(13–4)

The total impedance for a circuit can be calculated using a similar equation:

(13–5)

Substituting the rectangular impedance for Z₁and Z₂, this equation becomes:

Once the impedance of the circuits has been converted to this form, addition, subtraction, multiplication, and division can be used to reduce the circuits to a single complex number representing the value of the total impedance. Because the j operator is included, a discussion of these arithmetic operators will be useful before you continue with the complete problem solution.

Math Operations Involving J-operators

You already know how to perform math operations using real numbers. That is basic arithmetic. These same operations can be performed using imaginary numbers; addition, subtraction, multiplication, and division are performed with j–terms as you would add, subtract, multiply, and divide using algebraic techniques. Simply treat the term j as you would the variable X in an algebraic expression and make sure that you recognize the positive and negative quantities. For example, addition and subtraction using imaginary numbers are performed like this:

Here are some additional examples:

Similarly, multiplication and division are performed like this:

Here are some additional examples:

And since one over j equals minus j,

Once division or multiplication has been performed, terms containing a power of j can be simplified using the following basic identities which were introduced in Chapter 12:

For example, j²8 may be simplified:

Addition

To add two rectangular form complex numbers, simply add the two real parts and add the two imaginary parts. The result is expressed as another complex number called the complex sum which has both real and imaginary parts.

(13–6)

For example, suppose the two complex numbers 4 + j2 and 6 + j1 must be added:

Thus, the sum of 4 + j2 and 6 + j1 equals 10 + j3.

As another example,

Subtraction

Subtraction using rectangular form complex numbers is very similar to addition. To subtract two complex numbers, first find the difference between the real parts, then find the difference between the imaginary parts. Remember, this must be the algebraic difference. The result is a complex number with real and imaginary parts.

(13–7)

For example, 4 + j2 is subtracted from 8 + j8.

Effectively, the subtraction process is the same as any algebraic subtraction, you change the sign of the number that is being subtracted and then perform an addition.

As a second example,

Written in a little different form may make it easier to understand.

2–j3 is subtracted from 10–j7 by changingthe sign of 2 –j3 and adding

Multiplication

The multiplication of two rectangular form complex numbers may be performed by using basic binomial algebraic methods. That is, the complex numbers are in the form (R₁ + jX₁) and (R₂ + jX₂). When you multiply one by the other the result is obtained by multiplying each term in the first number times each term in the second number and adding the resulting products:

(13–8)

You can then group like terms. At this time, terms 2 and 3 both have j–prefixes and can be grouped:

Since j² = −1 in term 4, the expression simplifies to:

(13–9)

As an example the product of 9 + j3 and 4–j2

Combining like terms results in:

Since the j² = −1, the result simplifies to:

Finally, combining real terms:

Notice that the product reduces to the real plus imaginary complex number form. It is important to remember to carry along the sign of the reactance terms and apply the rules for the product of signed terms.

Division

The division of two rectangular form complex numbers may also be performed by using algebraic methods. The form is

(13–10)

This process, however, becomes quite involved since the division of a real number by an imaginary number is not possible. Because of this, the denominator must be converted first to a real number by a process called rationalization of the fraction. To do this, both the numerator and denominator must be multiplied by what is called the conjugate of the denominator. The conjugate of a complex number is simply the number with the j-term having an opposite sign. For example, the conjugate of 2 + j2 is 2 – j2. Similarly, the conjugate of 6 – j4 is 6 + j4.

Following this rule to divide 3 + j2 by 2 + j2, you must multiply both the numerator and denominator by the conjugate of the denominator 2 – j2:

In doing so, you do not change the value of the fraction since (2 – j2) divided by (2 – j2) times the fraction is like multiplying the fraction times one.

Multiplying numerators and denominators:

Combining j–terms in the numerator and denominator, the fraction becomes:

Since j² = −1, the expression can be reduced as follows:

Notice that all imaginary terms have been eliminated from the denominator. Recall that this is the purpose of rationalization. Now, the fraction can be split up and the division performed:

The final result is again a complex number with real and imaginary parts. Again, it must be remembered to follow all of the rules that pertain to multiplication and division of signed terms.

Multiplication

To multiply in polar form, one need only multiply the magnitudes and add the angles:

For example,

additional examples:

The algebraic rules for signs apply to the addition of the angles. The polar magnitude is always positive. If in some solution it were to come out negative it means that the angle should be changed by 180 degrees.

Division

Division of polar complex numbers is similar. Simply divide the numerator’s magnitude by the denominator’s magnitude and subtract the denominator’s angle from the numerator’s angle.

Additional examples show how the algebraic convention of signs apply to the subtraction of the angles. The polar magnitude is always positive with the same provision as in multiplication.

Simplifying Impedance for a Series Circuit

Suppose the impedance of a series RLC circuit can be written as:

Combining the j–terms gives:

The resultant impedance of this circuit can be written simply as 80 + j40 in rectangular form. Notice that the j40 term indicates that the total reactance of the circuit is an inductive 40 ohms. The components of this series RLC circuit are shown on the phasor diagram shown in Figure 13.6. The various phasors are drawn with magnitudes to scale and positioned in phase.

The polar form of this impedance can be found using the conversion equation 13–13

Substituting equation 13–13 for the values X and R gives:

Thus, the impedance of 80 + j40 in rectangular form equals 89.4 at 27 degrees in polar form. This polar form corresponds to a vector whose magnitude is the value of the impedance and whose angle is the phase angle of the circuit. This is shown in Figure 13.7. It is the same answer that would be obtained by drawing the vector to scale and using the Pythagorean theorem. Both methods yield the same result.

Since this circuit was simple enough to solve by right triangle phasor analysis, it should also be possible to use complex numbers to arrive at valid solutions of even more complex circuits.

Series Circuit with Two Complex Impedances

In the circuit of Figures 13.1 and 13.2, the total impedance is the series combination of two complex impedances. This was defined previously as,

Combining real terms and imaginary terms, the total impedance of this circuit can be expressed as:

Thus, the total impedance of the circuit can be expressed in rectangular form as 14 – j12.

This can be converted to polar form which indicates the magnitude of the impedance and the phase angle of the circuit:

The total impedance is 18.4 ohms at a negative angle of 40.6 degrees.

A Parallel RLC Circuit

In the circuit of Figure 13.4, the total impedance is the parallel combination of two complex impedances. By applying equation 13–5, the total impedance of this circuit was determined as,

Performing the addition operation in the denominator first results in

The denominator is simplified to 9 – j6. Now, all remaining operations to be performed are either multiplication or division. Since multiplication and division operations are most easily performed in polar form, the easiest way to solve for the total impedance is to convert all rectangular terms to polar form. Using equation 13–13, the 3 + j4 term becomes in polar form. The calculations are:

Similar calculations,

result in a polar form of for the 6 – j10 term. The angle is negative because the j–term is negative.

The 9 – j6 term in polar form is arrived at as follows:

The angle, again, is negative since the j–term is negative. Thus, 9 – j6 is equal to in polar form.

Substituting these equivalent polar–form impedances for their corresponding rectangular–form impedances in the original equation results in the following equation for Z_T

Now the total impedance can be determined simply by the multiplication and division of polar–form terms. Performing the multiplication operation in the numerator first:

Now, performing the division:

Therefore, the total impedance of the circuit is equal to 5.4 ohms at a phase angle of 28 degrees.

Solving for Impedance of RLC Circuit

Now these phasor algebra techniques will be applied to solve for the total impedance of the series–parallel ac circuit shown in Figure 13.8. First divide the circuit into several complex impedances as you have seen done previously to simplify its solution as illustrated in Figure 13.9. Let R₁ and L₁ be impedance Z₁, R₂ and C be impedance Z₂, and R₃ and L₂be impedance Z₃.

Expressed in rectangular form these impedances are:

Looking at the circuit simply in terms of these three impedances, the total impedance of the circuit can be written as Z₁ in series with the parallel combination of Z₂ and Z₃. That is,

(13–14)

If the complex impedances are substituted Z_T is:

Rectangular form addition can be performed in the denominator. This makes Z_T:

Before the first and second term can be added in rectangular form, the second term must have its rectangular components converted to polar form so the multiplication and division can be performed.

The 3 – j4 term is in polar form. The calculations are as follows:

The angles are rounded to two significant figures. The angle is negative because the j term is negative.

The 1 + j2 term is in polar form calculated as follows:

Similarly, the 4 – j2 term is in polar form.

Substituting these polar–form impedances for the rectangular–form impedances which they equal, the total impedance equation for the complex RLC circuit now is:

Performing the multiplication in the numerator results in,

Performing the division results in the total impedance of

To perform the indicated addition, first the second term of the impedance in polar form should be converted to its equivalent rectangular form. Equation 13–15, discussed in previous chapters, is used for this purpose.

Thus, the rectangular form of equals 2 + j1.5.

Substituting this rectangular form into the total impedance equation, and adding Z_Tbecomes,

This is the total impedance of this series–parallel circuit expressed in rectangular form. Note as shown in Figure 13.10 that this impedance indicates that the circuit would appear to a power supply as a 4–ohm resistor in series with an inductor possessing an inductive reactance of 7.5 ohms at the applied frequency. Obviously, since reactances will change with changes in frequency, at a different frequency the circuit may appear to be either an RL or RC circuit.

The total impedance of the circuit can also be converted to its polar form to yield the magnitude of the impedance and the phase angle. It is shown as follows:

This form is usually the most–desirable form since it not only gives the magnitude of the circuit’s impedance, but also its phase angle.

Solution of RLC Circuit

These phasor algebra techniques can now be used to solve for the impedance, total current and voltage drops in the series RLC circuit illustrated in Figure 13.11. The circuit contains 15 ohms resistance, 30 ohms capacitive reactance, 50 ohms of inductive reactance, and a second capacitive reactance of 40 ohms. The applied voltage, E_A, is 50 volts. If the reactances were not given, you would have to use the frequency of the applied voltage to calculate the capacitive and inductive reactances. Writing in j-operator notation, the reactance of C₁ is –j30 ohms, the reactance of the inductor in j50 ohms, and the reactance of C₂ is –j40 ohms.

Impedance

The total impedance of the circuit can be written:

The ohms are understood since only impedance terms are being considered. By

addition of j terms the total impedance of the circuit reduces to

Converting this to polar form results in the total impedance equal to . The calculations are as follows:

Since the j–term is negative, the angle, θ, is negative.

Total Current

The total current of any ac circuit is equal to the applied voltage divided by the total impedance of the circuit:

(13–16)

Using the applied voltage as a reference at zero degrees, it can be expressed as 50 volts at 0 degrees. If it seems unusual that the applied voltage is written in this form, recall that in reality, current and voltages have phase relationships, not resistance, reactance, or impedance. The total current in the circuit is calculated using equation 13–16

Carrying out the indicated division,

Thus, the total current equals .

Voltage Drops

Recall that according to Ohm’s law the voltage drop across any resistor is equal to the current through it times the value of the resistor:

(13–17)

Also recall that the voltage drop across any reactance is equal to the current through it times the value of the reactance:

(13–18)

Since the circuit is a series circuit, the current through all components is the same as the total current, I_T. Thus, to determine the voltage drop across any component in the circuit, simply multiply the total current, , times the resistance, R, or reactance, X, of the component in the circuit.

In order to perform multiplication of two complex numbers, both must be expressed in the same form, either rectangular or polar. The current is expressed in polar form, therefore, to multiply conveniently by the polar form of the current, you should convert all resistances and reactances in the circuit to polar form.

Pure Resistance Polar Form

This is easily accomplished in terms of individual resistances or reactances. For example, R₁, the resistance term of 15 ohms in the circuit of Figure 13.11, can be written in polar form as simply which means the impedance is all resistive with no reactance. This might be more clearly understood when written in rectangular form. In the form R + jX, R = 15 ohms and X = 0 because it is all resistive. The polar form can be obtained by use of the conversion equation:

Pure Reactance Polar Form

The polar form of the reactance of C₁, −j30, is . Its rectangular form is really 0 – j30 which is a vector of a magnitude of 30 plotted on the −j axis. The polar form can also be verified using the conversion equation. Similarly, the reactance of the inductor, j50, can be written in polar form as , and the reactance of C₂, −j40, can be written in polar form as . In summary, the polar forms are:

Notice in Figure 13.12 that these angles of zero degrees, 90 degrees, and −90 degrees denote the phase relationships of the resistances and reactances just like the joperator did for the rectangular forms.

Now that the resistances and reactances of the circuit are written in polar form, they can be multiplied easily by the polar-form current to obtain the voltage drops in the circuit. The voltage across the resistor, ER, equals the current times the resistance:

The voltage across C₁, E_C1, equals the current times the capacitive reactance of C1:

The voltage across the inductor, E_L, equals the current times the inductive reactance:

The voltage across C₂, E_C2, equals the current

times the capacitive reactance of C₂:

Phasor Diagram

Since the applied voltage was used as a reference at 0 degrees, and the total current and voltage drops in the circuit were derived using the applied voltage as reference, all of these angles expressed in polar form in thesecalculations are measured with reference to the applied voltage as shown in Figure 13.13

The phase angle can be determined by plotting the total current in reference to the applied voltage as shown on Figure 13.14

The voltage across the resistor E_T equals 30 volts at 53 degrees. The total current, I_T, is 2 amperes at 53 degrees. It is in phase with E_R. Knowing that the angle between the total current and applied voltage in a circuit is the phase angle, you can see that it is 53 degrees for this circuit.

Now the additional voltage phasors can be added. The voltage across C₁ is 60 volts at −37 degrees as shown in Figure 13.15. Note that this lags E_Rby 90 degrees because it lags the circuit current by 90 degrees.

The voltage across the inductor E_L is 100 volts at 143 degrees. E_L leads I_Tby 90 degrees and E_C1 by 180 degrees, the same relationship as when voltage and impedance phasor diagrams were plotted in previous lessons. The voltage across C₂ is 80 volts at −37 degrees, in phase with E_C1. Since E_C1 and E_C2 are in phase, they are both shown in the same direction, with the vector length of one added to the other. Thus, the total capacitive voltage E_C is 140 volts at −37 degrees.

Using this method provides the phase of every voltage or current in the circuit with respect to the reference quantity used, in this example, the applied voltage E_A.

With the same procedures the values of voltage, current and impedance can be solved for even the most complex RLC circuit.

1. Perform the following operations involving j-operator terms:

2. Perform the following operations involving rectangular form complex numbers; show your work.

3. Perform the indicated operations shown below involving polar form complex numbers.

4. Given the following RLC circuits, express the impedance of the circuit in both polar and rectangular form.

5. Express the resistance or reactance of the following passive elements in rectangular (j-operator) form and then in polar form.

6. Determine the values indicated below for the RLC circuit shown. Express all answers in polar form. Use E_A = for reference.

Solution

1. Perform the indicated operations involving j-operator terms:

2. Perform the indicated operations involving rectangular form complex numbers.

3. Perform the indicated operations shown below involving polar form complex numbers.

4. Given the following RLC circuit, express the impedance of the circuit in both polar and rectangular form.

5. Given the following RLC circuit, express the impedance of the circuit in both polar and rectangular form.

6. Given the following RLC circuit, express the impedance of the circuit in both polar and rectangular form.

7. Given the following RLC circuit, express the impedance of the circuit in both polar and rectangular form.

8. Express the resistance or reactance of the following passive elements in rectangular (joperator) form and then in polar form.

9. Determine the following values for the RLC circuit shown. Express all answers in polar form. Use E_A as reference at 0°.

10. Determine the following values for the RC circuit shown. Express all answers in p o l a r form. Use E_A reference at 0°.

1. Perform the indicated operations involving j-operator terms; simplify.

2. Perform the indicated operations involving rectangular form complex numbers; simplify

3. Perform the indicated operations shown below involving polar form complex numbers.

4. Given the following RLC circuits, express the impedance of the circuit in both polar and rectangular form.

5. Determine the values indicated for the following circuits. Express all answers in polar form. Use EA at 0₀ as reference.