Voltage and Current Relationships

The voltage across a resistor, E_R, is in phase with the current through it as shown in Figure 7.3. For a capacitor, however, recall from Chapter 6 that the current leads the voltage by 90 degrees, as shown in Figure 7.4. Since the resistor and capacitor are in series with one another, the common factor in both phase relationships is the current.

Phasor diagrams can be drawn for the voltage and current waveforms of Figure 7.3 and Figure 7.4. The voltage E_R, equal to 8 volts, across the resistor is in phase with the current, I, in the series circuit as shown in Figure 7.5. The same series current, I, leads the voltage E_Cacross the capacitor by 90 degrees as shown in Figure 7.6. E_Cis 6 volts.

Comparing the phase relationships of the two voltage drops, it is found that the voltage across the resistor leads the voltage across the capacitor by 90 degrees. Because these two voltages are out of phase, they cannot be added as the voltage drops in a resistive series circuit would be added to obtain the total applied voltage. Mathematically, therefore:

(7–3).

The total voltage must be calculated, therefore, by another type of addition known as vector addition.

Common Phasor Diagram

Since I is the same in both of the phasor diagrams of Figure 7.5 and Figure 7.6, a common phasor diagram can be formed as shown in Figure 7.7. It is now very apparent that E_Rand E_Care 90 degrees out of phase. More specifically, the voltage across the resistor leads the voltage across the capacitor by 90 degrees. In Figure 7.7 E_Rand E_C are two vectors. Their length represents their amplitude and their direction represents their phase.

Parallelogram Method

When two vectors are added, the parallelogram method is used. The sum of the two vectors, as shown in Figure 7.8, is represented by a third vector called the resultant. To add two vectors using the parallelogram method, the parallelogram begun by the two original vectors are merely completed. The two original vectors in Figure 7.8 are the two solid lines with arrows at points E_Rand E_C. Then a line parallel to each vector beginning at the tip of the other and shown as the two dotted lines in Figure 7.8 are drawn.

The first line begins at the tip of E_Cand is parallel to line E_R. The second line begins at the tip of E_Rand is parallel to line E_C. The point where these two dotted lines intersect determines where the tip of the resultant vector will be located. The resultant vector is then drawn from the origin (as are all vectors) to the point of intersection as shown.

If the length and phase of E_Rand E_Care drawn to scale, the length of the resultant vector obtained in this manner will indicate its magnitude. Therefore, using the diagram, the resultant vector′s magnitude and phase relationship to the other two vectors can be measured. The resultant is the vector sum of E_R and E_C.

Voltage Vector Addition

The voltage across the capacitor, E_C, and the voltage across the resistor, E_R, in Figure 7.7 must be added vectorially in a series RC circuit. Using the parallelogram method, the vector sum of E_Rand E_Cequals the total applied voltage, E_A. A diagram to perform this calculation is shown in Figure 7.9. This diagram is the voltage vector diagram for a series RC circuit.

Resistance and Reactance Vector Addition

Recall that in a purely resistive series circuit, the total ohms of resistance is equal to the sum of all individual ohms of resistance.

(7–4)

In a purely reactive (capacitors only) series circuit, the total ohms of reactance is equal to the sum of all individual ohms of reactance.

(7–5).

However, in a series RC circuit, like the one shown in Figure 7.10, there exists a combination of ohms resistance and ohms reactance.

This combination is called impedance, also measured in ohms. Just as voltages were represented by vectors whose length represents their magnitude and whose direction represents their phase, R, X_C, and the impedance Z can be represented as vectors. For a series RC circuit, the relationship of the vectors is shown in Figure 7.11. Both resistance and reactance impede the current; therefore, the total impedance of an RC series circuit is the vector sum of the resistance and reactance in the circuit as shown in Figure 7.11. The length of the Z vector is its magnitude and the direction its phase.

By Ohm′s law,

(7–6).

The current, I, flowing in the series RC circuit of Figure 7.10 is the same through each circuit element:

(7–7).

and the vector E_Rin Figure 7.12 is made equal to IR. The current is the same for the capacitor:

(7–8).

and the E_Cvector of Figure 7.12 is made equal to IX_C. E_Cand E_Rpreviously were plotted at right angles since E_Rleads E_Cby 90 degrees.

Also by Ohm′s law, the total applied voltage is equal to the circuit current, I, times the total impedance:

(7–9).

As a result, IZ is substituted for the applied voltage E_Ain the voltage vector diagram as shown in Figure 7.12. In that diagramml:

(7–10).

(7–11).

Because the total current is the common factor, it can be factored out of equation 7–11.

(7–12).

Equation 7–12 becomes, for a series RC Circuit,

(7–13).

This shows clearly that Z is the vector sum of R and X_C.

The resultant diagram is shown in Figure 7.13.

The parallelogram method is used to show the relationship of the resultant total impedance, Z, to the value of the resistance and capacitance in a series RC circuit. Its magnitude is its length and its direction its phase.

Pythagorean Theorem

The method used to calculate impedance is based upon a mathematical theorem involving right triangles stated centuries ago by a man named Pythagoras. He found that there was a special relationship between the lengths of the three sides of a right triangle, which is a triangle in which the major angle is 90 degrees. He found that the length of the longest side, called the hypotenuse, is equal to the square root of the sum of the square of the length of each of the other two sides. Mathematically, from Figure 7.14.

(7–14).

For example, in Figure 7.14, side A is 3 inches long, and side B is 4 inches long. Side C, hypotenuse, is calculated:

The length of the hypotenuse of this triangle, then, is 5 inches.

Pythagorean Theorem Applied to Impedance Solutions

The Pythagorean theorem can be applied to circuit problems involving resistance and reactance. Figure 7.15 shows the resistive-reactance phasor diagram for a series RC circuit. The total impedance of the circuit, Z, is the vector sum of the resistance, R, and reactance, X_C. If the values of the resistance and reactance are known, the impedance can be calculated using the Pythagorean theorem.

Notice that the length of the reactance vector is the same as the length between the tip of the resistance vector and the tip of the impedance vector. In fact, the vector diagram can be drawn with the reactance vector placed as shown in Figure 7.16. Now, the right triangle of the Pythagorean theorem becomes evident. Applying the Pythagorean theorem, the total impedance (the hypotenuse) can be calculated,

(7–15).

Total impedance is equal to the square root of the resistance squared plus the reactance squared.

Figure 7.17 shows a typical series RC circuit. The impedance can be calculated using equation 7–15. The resistance of this RC circuit is 12 ohms and the capacitive reactance is 16 ohms.

(7–15).

Thus, the total impedance of the circuit is 20 ohms.

Pythagorean Theorem Applied to Voltage Solutions

Similar mathematical calculations using the Pythagorean theorem can be performed to show the relationship between the voltage drops and applied voltage in a series RC circuit. In this case the phasor diagram is drawn as shown in Figure 7.18. Shifting the capacitive voltage vector, as shown in Figure 7.19, equation 7–15 can be rewritten to calculate easily the applied voltage, E_A.

(7–16).

An example circuit is shown in Figure 7.20 which also shows the vector relationship between E_R, E_C, and E_A. In that circuit the voltage across the capacitor is 10 volts, rms; the voltage across the resistor is 20 volts, rms. Using equation 7–16, E_Ais calculated:

Thus, the total applied voltage to the circuit is 22.36 volts(rms).

Calculations of X_C

First, capacitive reactance, XC, will be calculated in a typical series RC circuit shown in Figure 7.21. The value of the resistor is 30 ohms. The value of the capacitor is 4 microfarads. The applied voltage is 100 VAC with a frequency of 995 hertz. To determine the total impedance of this circuit, the value of the capacitive reactance must first be calculated.

Therefore, X_Cequals 40 ohms.

Calculation of Z

The impedance phasor diagram may now be drawn to show the relationship between the values of resistance, reactance, and impedance in the circuit as shown in Figure 7.22. The value of impedance is calculated by using the Pythagorean theorem (equation 7–15). The vectors being plotted are the sides of a right triangle.

Thus, the value of the total impedance of this circuit is 50 ohms. Impedance is defined as the total opposition of a resistive-reactance circuit to the flow of alternating current. It is something that impedes the current.

Calculation of I_T

Recall that in a purely resistive circuit or purely reactive circuit, the total current in the circuit is calculated using Ohm′s law by dividing the applied voltage by the total resistance or reactance. For resistance:

(7–17).

And for reactance:

(7–18).

Similarly, in a resistive-reactive circuit, like the one shown in Figure 7.23, the total current is found using Ohm′s law by dividing the applied voltage by the total impedance of the circuit.

(7–19).

This relationship between voltage, current, and impedance is often referred to as Ohm′s law for ac circuits.

Figure 7.24 is a typical resistive-reactive series circuit, with circuit values as shown. Using equation 7–19, the total current in the circuit can be calculated. That equation requires that Z_Tbe known. As calculated in the previous section, Z_Tfor the circuit is 50 ohms. Therefore,

The total circuit current equals 2 amperes.

Calculations of E_Rand E_C

Voltage drops across the resistor and capacitor are now calculated as they would be in any series circuit. The voltage drop across the resistor is calculated:

The voltage drop across the capacitor is then calculated:

Relationship of E_R, E_C, and E_Ain Series RC Circuits

Notice that the voltage across the capacitor plus the voltage across the resistor add to 140 volts. This is more than the applied voltage and at first, seems to contradict Kirchhoff′s voltage law. Recall that Kirchhoff′s voltage law essentially states that the total voltage applied to any closed circuit path always equals the sum of the voltage drops across the individual parts of the path. Also remember, however, that the voltage across the resistor and the voltage across the capacitor are out of phase by 90 degrees as shown in Figure 7.25. Therefore, they must be added vectorially like this:

The applied voltage in the example circuit is 100 VAC. Thus, the vector sum of the circuit voltage drops equals the applied voltage. This type of calculation can be performed to check the calculated voltage drops in the circuit.

Calculations of Phase Angle in Series RC Circuits

Recall that when originally forming the voltage phasor diagram for a series RC circuit it was determined that the current is the same throughout a series circuit. Therefore, current was used as a reference quantity. This total current is in phase with the voltage across the resistor. Notice in Figure 7.26, however, that the applied voltage and the total current are out of phase. Specifically, the current leads the applied voltage by a certain number of degrees. This angle by which the total applied voltage and total current are out-of-phase is called the phase angle of circuit. The phase angle, by definition, is the number of degrees the total current and the total applied voltage are out of phase. The phase angle is usually denoted by the Greek letter theta, θ.

The phase angle in a series RC circuit can also be recognized as the angle between the voltage across the resistor and applied voltage as shown in Figure 7.27. Figure 7.12 showed that the impedance phasor diagram is proportional to the voltage phasor diagram because

and the fact that I cancels because it is common to each term. Therefore, as shown in Figure 7.27, the voltage across the resistor ER is in phase with the series circuit current. The phase angle in a series RC circuit is also the angle between the resistance vector and the impedance vector as shown in Figure 7.28.

Definition of Tangent Function

The value of the phase angle can be calculated by using a trigonometric function called the tangent function. Recall from Chapter 2 that the tangent, abbreviated tan, of an angle in a right triangle, is equal to the ratio of the length of the side opposite the angle divided by the length of the side adjacent to the angle:

(7–20).

For example, as shown in Figure 7.29, the side of a right triangle opposite from the angle 0 is 10 units long and the side adjacent to the angle 0 is 5 units long. Then,

Thus, the tangent of θ in this right triangle is equal to 2. A calculator or trigonometric function table is then used to determine the angle which has a tangent of 2. The angle which has a tangent of 2 is about 64 degrees.

Relationship of Tangent and Arctangent

Recall from equation 7–20 that the value obtained by dividing the lengths of the sides of a right triangle is called the ratio of the two sides. It will be a ratio that has a value from 0 to infinity. If given the ratio, the angle can be found using a calculator or trigonometric table.

The inverse process of determining the angle when the ratio is known is called finding the arctangent, abbreviated arctan.

(7–21).

This equation reads: θ is an angle of a right triangle whose tangent is the ratio called out in the arctan function.

For example, the arctangent of the ratio of 2 is about 64 degrees:

Said inversely, the tangent of 64 degrees is approximately 2:

Taking a second example, the tangent of 35° = 0.7. Therefore, the arctangent of 0.7 = 35°. Sometimes, arctangent is abbreviated by writing: tan⁻¹(tan to the minus one). Both notations indicate the arctangent function. Therefore, the arctangent of a ratio indicates an angle whose tangent is the ratio.

Application of Tangent Function in Solving for Phase Angles

Using the tangent function to find the phase angle of the RC circuit in Figure 7.24 the side opposite the phase angle in the voltage phasor diagram of Figure 7.27 is the vector E_C. The length of this vector represents the value of the voltage drop across the capacitor. The side adjacent to the phase angle in the voltage phasor diagram is the vector, E_R, whose length represents the value of the voltage drop across the resistor.

The tangent of the phase angle is equal to the ratio of the opposite side divided by the adjacent side,

for the voltage phasor diagram. The arctangent of this ratio yields the value of the phase angle, θ:

In the series RC circuit shown in Figure 7.30, the voltage across the capacitor is 80 volts and the voltage across the resistor is 60 volts. Therefore,

Completing the calculations the arctangent of ER divided by E_Cequals the arctangent of 80 volts divided by 60 volts which equals the arctangent of 1.33. Using your calculator or trigonometric table, you can determine that the tangent function has a value of 1.33 for an angle of about 53 degrees, or the arctangent of 1.33 is 53 degrees. Therefore, the phase angle of this RC series circuit is 53 degrees.

Recall that earlier it was shown that the impedance phasor diagram of a series circuit is proportional to the voltage phasor diagram. Since these two phasor diagrams are proportional, the phase angle also can be determined using an impedance phasor diagram. This will provide an identical value for the phase angle for the RC circuit of Figure 7.24.

The side opposite to the phase angle in the impedance phasor diagram in Figure 7.28 is the reactance vector, X_C. The length of this vector represents the value of the capacitive reactance. The side adjacent to the phase angle in the impedance phasor diagram is the resistance vector, R. Its length represents the value of the resistor. Recall that the tangent ratio is the opposite side divided by adjacent side, therefore,

(7–22).

The arctangent of this ratio yields the value of the phase angle.

(7–23a).

There is one other important point about calculating theta of a series RC circuit. In the series RC circuit of Figure 7.24, the value of the capacitive reactance is 40 ohms and the value of the resistor is 30 ohms. Therefore:

40 divided by 30 results in a ratio of 1.33. The arctangent of 1.33 is 53 degrees. This is the same phase angle derived using the voltage phasor diagram. Thus, it should be remembered that the value of the phase angle in a series RC circuit can be determined from either the voltage or impedance phasor diagram.

Negative Phase Angle in Series RC Circuits

The value of the phase angle in a series RC circuit is considered to be a negative value. That is, the phase angle in the example circuit of Figure 7.24 is actually negative 53 degrees as shown in Figure 7.31. The reason the phase angle is considered to be negative is because, previously, the angle of rotation of phasors in a counter-clockwise direction was designated as positive. Thus, the angle of rotation of a phasor is a clockwise direction, in agreement with the convention, is said to be negative. This is illustrated graphically in Figure 7.32.

Remember that the total current in the series RC circuit was used as the reference, which is at zero degrees. Thus, as shown in Figure 7.33, the negative sign of the phase angle indicates that the applied voltage is rotated 53 degrees clockwise from the current vector direction.

Alternative Series RC Circuit Solution Method

It might be instructive to note that the method of solution for values of impedance, current, voltage, and phase angle is not limited to the order of solutions illustrated in the previous section. Recall also from Chapter 2 that

(2–21).

and

(2–22).

In the impedance phasor diagram of Figure 7.28, the opposite side to the phase angle (θ) is the capacitive reactance vector (X_C). The adjacent side to the phase angle (θ) is the resistive vector (R). The hypotenuse is the impedance vector (Z). Substituting these values into the sine or cosine function equations shown above results in

(7–23b).

or

(7–23c).

Rewriting these equations produces

(7–23d).

or

(7–23e).

Now, an alternative solution method follows using these new equations (7–23d and 7–23e).

If you are given the values of R, C, applied voltage, and frequency in a series RC circuit such as the one shown in Figure 7.21, the values of X_C, phase angle, impedance, circuit current, capacitive, and resistive voltages can be found.

First, determine the capacitive reactance (X_C) using the capacitive reactance equation as done earlier.

Then, using equation 7–23a, the phase angle may be found using the tangent function.

Now, using equation 7–23d

or using equation 7–23e above

Note the values of Z match the value calculated earlier. If current and voltage values are also desired, they can be found using the following calculations.

and

What is illustrated here, then, is a different approach to determining the same values we found previously.

Calculation of Real Power

In the example series RC circuit of Figure 7.30, the power dissipated by the resistor can be calculated by multiplying the voltage across the resistor by the current through the resistor. The current is total current. Recall that previously the current through the resistor, I_R, was calculated to be 2 amperes, and E_Rand E_Cwere calculated to be 60 volts and 80 volts, respectively. Thus,

The real power in the circuit equals 120 watts. It is the power dissipated as heat in the circuit.

Calculation of Reactive Power

The reactive power of the capacitor is calculated as it was in a purely capacitive circuit.

The reactive power in the circuit is 160 VAR. No actual heat is generated unless the capacitor has resistance included in it.

Calculation of Apparent Power

As stated previously, in a series RC circuit like the one in Figure 7.30, there exists a combination of real power and reactive power. This combined power calculated using total current and total voltage circuit values is called apparent power. Since the apparent power is a combination of real and reactive power, it cannot be designated either watts or VAR. Instead, it is measured in units called volt-amperes, abbreviated VA.

Power in a circuit is always calculated by multiplying a voltage times a current.

(7–24).

Equation 7–24 is such an equation and it can be used to calculate total power in any resistive–reactive circuit. In the circuit being used (Figure 7.30) E_A= 100 volts and I_T= 2 amperes. Thus,

Recall that in a purely resistive circuit the total real power is the sum of all the individual real power values:

(7–25).

and in a purely reactive circuit the total reactive power is the sum of all individual reactive power values:

(7–26).

However, in resistive and reactive circuits, the simple sum of the real power and reactive power does not equal the apparent power.

(7–27).

This occurs because the phase relationship between the voltage across and the current through each component are different.

In series RC circuits, the phase relationships of the three power determinations are similar to the voltage phase relationships, as shown in Figure 7.34.

Since each power determination is made by multiplying the voltage shown times the current, the power phasor diagram is proportional to the voltage phasor diagram by a factor of the total current, as shown in Figure 7.34.

Mathematically:

(7–28).

or

(7–29).

The total apparent power is calculated using the Pythagorean theorem.

It is equal to the square root of the real power squared plus the reactive power squared. Thus, the total apparent power in the example circuit of Figure 7.24 and Figure 7.30 as calculated, 200 volt-amperes, should be equal to the square root of 120 watts squared plus 160 VAR squared.

This is the relationship between the apparent power and the real and reactive power. Note in Figure 7.35 that the angle between the real and apparent power is the phase angle. In this example the phase angle is negative 53 degrees.

Summary of Phase Relationships in Parallel RC Circuits

In the circuit of Figure 7.36, as in any parallel circuit, the voltage across all components is the same: E_A=E_R=E_C. However, the algebraic sum of the branch currents does not equal the total current in the circuit.

(7–31).

This occurs because the phase relationships between the voltage and current are different for each component.

Recall that the voltage across a resistor is in phase with the current (Figure 7.37a) and, in a capacitor, the current leads the voltage by 90 degrees (Figure 7.37b).

Since the components in the circuit are in parallel with one another, the common factor in both relationships is the voltage, because the voltage across both components is the same. This voltage is the applied voltage, E_A.

For discussion purposes, the diagram of Figure 7.37b is rotated by 90 degrees to make E_C horizontal, as shown in Figure 7.38a. Because the components in the example circuit are parallel with one another, the common factor in both relationships is the voltage. The voltage across both components is the applied voltage, therefore, E_C is equal to E_A and E_R equals E_A.

Using the voltage as a common reference (instead of current which was used in an analysis of a series RC circuit), the two individual phasor diagrams can be combined into one as shown in Figure 7.38b. However, in order to match common voltage, the E_C vector has been rotated 90 degrees to match the direction of the E_Rand E_Avectors. Also note that the current through the resistor is shown in phase with the applied voltage across it, while holding to the theory that the capacitive current is leading the applied voltage by 90 degrees. Comparing the phase relationships of the two branch currents, the capacitive current leads the resistive current by 90 degrees.

Calculation of I_R and I_C

The individual branch currents in a parallel circuit can be calculated as they were calculated in a purely resistive or purely capacitive circuit. Simply divide the voltage across the branch by the opposition to current in that branch.

In the resistive branch, the opposition to the flow of current is measured in ohms of resistance. Thus, in the circuit, shown in Figure 7.39, the resistive current is determined by dividing the applied voltage by the value of the resistor.

(7–32).

In the capacitive branch, the opposition to the flow of current is measured in ohms of reactance. Therefore the capacitive current is determined by dividing the applied voltage by the reactance of the capacitor.

(7–33).

Calculation of I_T

To obtain the total current, I_T, the currents in a parallel RC circuit cannot simply be added as the branch currents in a parallel resistive circuit would be added. Instead, these currents must be added vectorially.

As shown in Figure 7.40, the length of the capacitive current vector is the same as the length between the tip of the total current vector and the tip of the resistive current vector. Thus, the capacitive current vector may be shifted over to form a right triangle.

Now, the Pythagorean theorem can be used to determine the total current in the circuit. According to the Pythagorean theorem, the total current in a parallel RC circuit is equal to the square root of the resistive current squared plus the capacitive current squared.

(7–34).

The parallel RC circuit shown in Figure 7.41 can be used to illustrate how to use this equation to calculate I_T. In the circuit the resistive current is 5 amperes, and the capacitive current is 12 amperes. Therefore,

The total current is 13 amperes.

Calculation of Z_TUsing Ohm′s Law

Once the total current is known, the total impedance of the circuit can be determined using Ohm′s law for ac circuits:

(7–35).

Rearranging to solve for impedance, the total impedance of the circuit is equal to the applied voltage divided by the total current:

(7–36).

For example, in the circuit in Figure 7.41 the applied voltage is 26 volts. The circuit impedance is calculated:

The circuit′s impedance is 2 ohms.

Calculation of Z_T Using Impedance Vectors

The total impedance can also be determined directly. However, a more complex equation is involved. Note that in Figure 7.42a the total current is equal to the applied voltage divided by the total impedance. The resistive current is equal to the applied voltage divided by the resistance. And the capacitive current is equal to the applied voltage divided by the capacitive reactance. These E over Z, E over R, and E over X quantities can be substituted for the current they equal in the current phasor diagram as shown in Figure 7.42b.

Since the applied voltage is the common factor, it can be factored out. The result, as shown in Figure 7.43, is an impedance phasor diagram relating the reciprocals of resistance, reactance, and impedance. Now, using the Pythagorean theorem, it can be written that:

(7–37).

Therefore,

(7–38).

The reciprocal of the impedance is equal to the square root of the reciprocal of the resistance squared plus the reciprocal of the reactance squared. The impedance, then, is equal to one divided by the square root of the reciprocal of the resistance squared plus the reciprocal of the reactance squared.

Calculations using

and

show that

This method serves to check the impedance value obtained by dividing the applied voltage by the total current. It is described simply to provide an alternate method of calculation of impedance in a parallel RC Circuit.

Calculation of Phase Angle

Recall that the phase angle is the number of degrees of phase difference between the applied voltage and the total current. Also recall that the applied voltage is in phase with the resistive current and I_T is the phasor sum of I_R and I_C. Therefore, the phase angle, theta, is located between the EA value and I_T value on the phasor diagram in Figure 7.44. The phase angle in a parallel RC circuit can also be recognized as the angle between the resistive current and the total current.

The value of the phase angle can be calculated by finding the arctangent of the ratio of the opposite to adjacent sides:

(7–39).

As shown in Figure 7.45, the opposite side′s length represents the value of the capacitive current. The adjacent side′s length represents the value of the resistive current. Thus, the tangent of the phase angle, θ, is equal to the ratio of the capacitive current divided by the resistive current as shown in equation 7–40.

(7–40).

Therefore,

(7–41).

The arctangent of the capacitive current divided by the resistive current equals the value of the phase angle. The phase angle, theta is an angle whose tangent is the ratio of capacitive current divided by resistive current.

For the example of Figure 7.41

Circuit Summary

The parallel RC circuit to be used in this example is shown in Figure 7.46. The value of the resistor is 15 ohms. The value of the capacitor is such that at the applied frequency, it has a capacitive reactance of 20 ohms. The applied voltage is 60 VAC.

Calculation of I_R and I_C

First the resistive branch current, IR, is determined

The capacitive branch current, I_C, is determined in a similar manner:

The current phasor diagram can now be drawn to show the relationships between the resistive and capacitive branch currents and the total circuit current. This diagram is shown in Figure 7.47.

Calculation of I_T

Using equation 7.34, the total current can be calculated.

IR = 4 amperes, and I_C= 3 amperes.

Therefore,

The total current flowing in the circuit is 5 amperes.

Calculation of Z

The total impedance (Z) of the circuit can now be calculated dividing the applied voltage by the total current. Since, E_A=60 volts and I_T=5 amperes, therefore,

(7–42a).

Total impedance is 12 ohms.

Calculation of Phase Angle

The phase angle can be recognized as the angle between the resistive current and the total current on the current phasor diagram in Figure 7.48. The value of I_C=3 amperes, and IR =4 amperes. The phase angle is calculated using equation 7–41:

The phase angle is equal to the arctangent of the ratio of the capacitive current to the resistive current. Thus, the phase angle, θ, is equal to the arctangent of 3 amperes divided by 4 amperes which equals the arctangent 0.75. Using a calculator or trigonometric table, the angle whose arctangent is 0.75 is approximately 37 degrees. Thus, the phase angle of this parallel RC circuit is approximately 37 degrees.

Positive Phase Angle in Parallel RC Circuit

The phase angle is said to be a positive 37 degrees, as shown in Figure 7.49. This is because the applied voltage in the parallel RC circuit was used as a reference at zero degrees. The positive sign is used to indicate that the total current phasor is rotated 37 degrees counter-clockwise from the applied voltage phasor.

In Figure 7.50 it can be seen that in a series RC circuit, the phase angle is negative. But in a parallel circuit, as shown in Figure 7.51, the phase angle is positive. The sign of the phase angle is used simply to indicate in which direction of rotation from the reference that the phase angle is measured, therefore, if you know only that the circuit is an RC circuit, you can readily determine whether the resistor and capacitor are connected in series or parallel by the sign of the phase angle.

Alternative Parallel RC Circuit Solution Method

It might be instructive to note that the method of solution for values of currents, impedance, voltage, and phase angle is not limited to the order of solutions illustrated in the previous section. As illustrated previously in this chapter, recall from Chapter 2 that

(2–21).

and

(2–22).

In the current phasor diagram of Figure 7.44, the opposite side to the phase angle (θ)is the capacitive current vector (I_C). The adjacent side to the phase angle (θ)is the resistive current vector (I_R). The hypotenuse is the total current vector (I_T). Substituting these values into the sine or cosine function equations shown above results in

(7–42b).

or

(7–42c).

Rewriting these equations produces

(7–42d).

or

(7–42e).

Now, an alternative solution method follows using these new equations (7–42d and 7–42e).

If you are given the values of R, C, applied voltage, and frequency in a parallel RC circuit such as the one shown in Figure 7.52, the values of X_C, phase angle, circuit branch currents, total circuit current, and impedance can be determined.

First, determine the capacitive reactance (X_C) using the capacitive reactance equation as done earlier.

Then, using equations 7–32 and 7–33, the individual branch currents may be found using Ohm′s law.

(7–32).

(7–33).

Now the phase angle of the circuit may be found using the arctangent function:

Now, using equation 7–42d above

or using equation 7–42e above

Note that the value of total current matches the value calculated earlier.

If the value of total circuit impedance is desired, it may now be determined using the following relationship:

What is illustrated here, then, is a different approach to determining the same values we found previously.

Calculation of P_R(Real Power)

The real power, P_R, is equal to the voltage across the resistor times the value of the current flowing through it.

Real power is 240 watts.

Calculation of P_C (Reactive Power)

The reactive power, P_C, is equal to the voltage across the capacitor times the value of the current through it.

(7–43).

In the example circuit, E_C=60 volts; I_C=3 amperes. Therefore:

Reactive power is 180 VAR.

Calculation of P_T (Total Apparent Power)

The total apparent power, P_T, is equal to the applied voltage times the total current:

(7–44).

In the example circuit, E_A=60 volts; I_T=5 amperes. Therefore:

Total apparent power is 300 volt-amperes.

Power Phasor Relationships in Parallel RC Circuits

Since each power determination is made by multiplying the current shown times the applied voltage, the power phasor diagram is proportional to the current phasor diagram by a factor of the applied voltage as shown in Figure 7.54.

As in the series RC circuit, the total apparent power is the vector sum of the real and reactive power, as shown in Figure 7.55. That is, the apparent power is equal to the square root of the real power squared plus the reactive power squared.

(7–45).

Using the parallel RC circuit values as shown in Figure 7.53 the apparent power, P_T, can be calculated. In that circuit P_R=240 w; P_C=180 VAR.

Therefore, the total apparent power calculated in this way is 300 volt-amperes, which matches previous calculations and serves as a check that they were correct.

1. Draw the phasor diagrams for the voltage and power circuit values in the circuit below.

    Solution:In a series circuit, current is common and used as the reference. E_Ris plotted in phase with I_T, I_R, and I_C. E_Clags I_Cby 90 degrees. E_Ais the phasor sum of E_Rand E_C. Current is a common factor in the power equations, and the power values are plotted in phase with the voltage values.

2. Draw the phasor diagrams for the voltage and power circuit values in the circuit below.

    Solution:In a parallel circuit, voltage is common and used as the reference circuit value in the phasor diagram. Voltage is a common factor in the power equations, and the power values are plotted in phase with the current values. I_Cleads E_C by 90°. I_Ris in phase with E_R.

3. Draw the Pythagorean theorem circuit value relationships for R, X_C, and Z_Tfor the circuit in question 1.

    Solution:

4. Given the following angles find the tangent of the angle.

    Calculator Solution:Enter the angle in degrees into the calculator and press the tan key. The tangent of the angle will be on the calculator.

    Table Solution:Obtain a table of natural trigonometric functions for angles in decimal degress. Enter the table by locating the angle in the degree column. Scan across to the tangent column and read the tangent value.

5. Given the following arctangents find the angle.

Calculator Solution:Enter the ratio into the calculator so it is displayed. Press the inv and tan keys or the tan⁻¹key and the calculator will display the angle in decimal degrees.

Table Solution:Enter the table of natural trigonometric functions by locating the ratio in the tangent column. Read across to the angle in decimal degress in the degree column.

6. Fill in the blanks.

Solution:Use same calculator or table procedure as for question 4 or 5.

7. In the phasor diagram for a parallel RC circuit shown, calculate the phase angle.

    Solution:

8. Solve for the circuit values specified for the circuit below.

Solution:

9. Solve for the circuit values specified for the circuit below:

Solution:

10. Solve for the circuit values specified for the circuit below:

Solution:Simplify the circuit.

1. Given the following angles find the tangent of the angle.

2. Given the following tangent values, determine the angle.

3. What is the definition of arctangent?

4. In each of the diagrams a., b., c., and d., solve for the value of the hypotenuse of the right triangle.

5. Solve for the phase angle in each of the circuits a., b., c., and d.

6. Draw a phasor diagram for the circuit shown. Show voltage, current, and power phasors.

7. 

8. Solve for the circuit values specified for this circuit.

9. Solve for the circuit values specified for this circuit.

10. Solve for the values specified for this circuit.

1. In each of the diagrams a., b., c., and d., solve for the value of the hypotenuse of the right triangle.

2. Solve for the phase angle in the circuits a., b., c., and d.

3. Draw a phasor diagram for this circuit. Show voltage, current, and power phasor.

4. 

5. Solve for the circuit values specified for this circuit:

6. Solve for the circuit values specified for this circuit.

7. Solve for the circuit values specified for the circuit below.

8. In what units are the values of voltage, current, and power in question 5 expressed?

9. In what units are the values of current and power in question 6 expressed?

10. In what units are the values of voltage, current, and power in question 7 expressed?