Thursday, January 8, 2015

Series Circuits—Resistance and Impedance : Resistance and inductance in series.

RESISTANCE AND INDUCTANCE IN SERIES

Unit 3 discussed series circuits containing only resistance and pure inductance. It is not possible for a circuit to have pure inductance because the coils of inductive equipment all have resistance. To obtain high values of inductance, iron cores are usually required. However, eddy currents and hysteresis losses in iron cores increase the ac resistance considerably. This effective resistance must be considered in all problems involving inductance. The discussion in this section will be confined to the study of circuits with resistance and inductance in series. The impedance section of each circuit considered consists of a reactor having a relatively large inductive reactance compared to the effective resistance.

Vector Resolution

The solution of most of the following problems uses a method called vector resolution. This method is widely used and its application should be clear to the student.

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PROBLEM 1

Statement of the Problem

Figure 4–7 shows a reactor connected directly across a 120-V, 60-Hz source. Meter readings for the line voltage, current, and power are shown on the circuit diagram. The impedance of the coil is the result of the combination of the effective resistance and inductive reactance. The impedance can be determined by considering the coil as a series circuit with a resistance component and an inductive reactance component.

The ohmic values of the reactor coil are determined as follows:

• Impedance of coil:

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3. The current, in amperes

4. The true power, in watts, taken by

a. the resistor

b. the coil

c. the entire series circuit

5. The apparent power in volt-amperes taken by the entire series circuit

6. The power factor of

a. the coil

b. the entire series circuit

Then construct a vector diagram for the series circuit.

Solution

1. The total resistance of the series circuit is obtained by adding the resistance values. The resistor value is 36 n. The effective resistance, determined earlier, is 4 n. Therefore, the total resistance is

R = 36 + 4 = 40 n

2. The inductive reactance of the entire series circuit is contained in the reactor coil.

When the reactor was connected across a 60-Hz source, its inductive reactance was  29.8 n. The total resistance of the series circuit is 40 n. Thus, the total series circuit impedance is

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4. a. The true power taken by the resistor rated at 36 n is

P = I2R = 52 X 36 = 25 X 36 = 900 W

b. The true power expended in the coil is

P = I2R = 52 X 4 = 25 X 4 = 100 W

c. The total true power for the entire series circuit can be obtained by adding the power expended in the resistor and the reactor. The formula I2R can also be used to find the total true power. R is the total effective resistance of the series circuit: .

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The value 0.1333 is the cosine of an angle that is equal to 82.3°. Thus, the reactor current lags the impressed voltage across the reactor by 82.3°.

b. The power factor of the entire series circuit is the ratio of the total true power in watts to the total apparent power in volt-amperes. The power factor of the circuit is also the ratio of the total series circuit resistance to the total circuit impedance:

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The angle for a cosine value of 0.8000 is 36.9°. This means that the line current lags the line voltage by 36.9°.

7. Figure 4–9 shows one method of constructing the vector diagram for this series circuit. The line current, 5 A, is drawn as a horizontal line. A convenient scale is used. V , the component of the reactor voltage caused by the effective resistance of the coil, is placed directly on the current line. V  is drawn from point 0 along the current line  and has a magnitude of 20 V. The voltage drop across the 36-n resistor is 180 V. This value is added to the 20-V drop on the current vector. The total voltage drop caused by resistance in the series circuit is 200 V in phase with the current. The voltage component caused by inductive reactance in the reactor is drawn in a vertical direction from point 0. This component, VXL  is at an angle of 90° with the current. The current lags  the voltage loss, caused by the inductive reactance, by an angle of 90°. The two voltage components of the reactor can be added vectorially to obtain the reactor voltage.

In Figure 4–9, the voltage across the reactor is shown as the vector sum of the two voltage components of the coil. These components are caused by the effective resistance and the inductive reactance. The phase angle between the voltage across the reactor and the current is 82.3°. This angle is indicated by the small Greek letter alpha (a). Note the right triangle formed by VXL and Vcoil.

Vector addition is used to find the sum of the total voltage drop due to the resistance in the circuit and the voltage drop due to inductive reactance. This vector sum is the line voltage. The angle e between the line voltage and the current is 36.9°, lagging.

Figure 4–10 shows a second method of constructing a vector diagram for the same series circuit. In this figure, the voltage drops are arranged differently because of the two resistance components of the series circuit. This change in placement on the current vector of the two resistance voltage losses means that the voltage components for the reactor must also be shown in a location different from that in Figure 4–9.

Although the two vector diagrams are drawn differently, the magnitude and phase relationships of all voltages, in reference to the current, are exactly the same. Therefore, either vector diagram may be used to analyze a series circuit containing R and Z components.

PROBLEM 2

Statement of the Problem

Figure 4–11 shows a noninductive heater unit connected in series with a reactor. The reactor controls the current in the series circuit. This means that the reactor also controls the temperature of the heater unit. The wattmeter indicates the total true power in watts taken by the circuit.

1. Determine the series circuit power factor and the phase angle.

2. Determine the true power, in watts, taken by the noninductive heating load.

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3. Determine the loss in watts in the coil.

4. Determine the effective resistance of the coil.

5. What is the power factor and angle of lag for the coil?

6. Determine the inductance, in henrys, of the coil at the circuit frequency of 25 Hz.

7. Draw a vector diagram, an impedance triangle, and a triangle of power values. Each diagram is to be drawn to scale and properly labeled.

Solution

1. The total true power taken by the circuit is 1650 W. The power factor is the ratio of the total true power in watts to the total apparent power in volt-amperes:

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The angle of lag of the line current behind the line voltage is 41.4°.

2. The true power in watts taken by the heater unit is the product of the voltage across the heater unit and the current. These two values are in phase:

P = VR X I = 150 X 10 = 1500 W

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5. The impedance of the reactor must be determined before the power factor for the reactor can be found. The effective resistance of the reactor was determined in step 4. Thus, the power factor for the reactor can be determined by the ratio R/Z:

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The angle of lag of the current behind the reactor voltage is 84.1°.

6. The inductance of the reactor can be obtained once the inductive reactance is known. The effective resistance and the impedance of the coil were calculated in steps 4 and 5, respectively. The inductive reactance is given by the expression

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Figure 4–12 shows the impedance triangle, the triangle of power values, and the vector diagram. A study of the three diagrams shows that they are similar right triangles.