1. Today
Course overview and information
09/16/2010
© 2010 NTUST

2. Sinusoidal Response of RL
When both resistance and inductance are in a series circuit, the phase angle between the applied voltage and total current is between 0 and 90, depending on the values of resistance and reactance.

3. Impedance of Series RL

4. Impedance of Series RL
In a series RL circuit, the total impedance is the phasor sum of R and XL.
R is plotted along the positive x-axis.
XL is plotted along the positive y-axis. Z Z
Z is the diagonal XL XL R R
It is convenient to reposition the phasors into the impedance triangle.

5. Examples

6. Impedance of Series RL Example
Sketch the impedance triangle and show the values for R = 1.2 kW and XL = 960 W.
Z = 1.33 kW
XL = 960 W
39o
R = 1.2 kW

7. Analysis of Series RL
Ohm’s law is applied to series RL circuits using quantities of Z, V, and I.
Because I is the same everywhere in a series circuit, you can obtain the voltage phasors by simply multiplying the impedance phasors by the current.

8. Examples

9. Analysis of Series RL Example
Assume the current in the previous example is 10 mArms. Sketch the voltage phasors. The impedance triangle from the previous example is shown for reference.
The voltage phasors can be found from Ohm’s law. Multiply each impedance phasor by 10 mA.
x 10 mA =
Z = 1.33 kW
VS = 13.3 V
VL = 9.6 V
XL = 960 W
39o
39o
R = 1.2 kW
VR = 12 V

10. Phase Relationships

11. Variation of Phase Angle
Phasor diagrams that have reactance phasors can only be drawn for a single frequency because X is a function of frequency.
As frequency changes, the impedance triangle for an RL circuit changes as illustrated here because XL increases with increasing f. This determines the frequency response of RL circuits.

12. Variation of Impedance and Phase Angle

13. Phase Shift
For a given frequency, a series RL circuit can be used to produce a phase lead by a specific amount between an input voltage and an output by taking the output across the inductor. This circuit is also a basic high-pass filter, a circuit that passes high frequencies and rejects all others. R
Vout
Vin
Vout
Vin
Vin
Vout L VR

14. RL Lead Circuit

15. Phase Shift
Reversing the components in the previous circuit produces a circuit that is a basic lag network. This circuit is also a basic low-pass filter, a circuit that passes low frequencies and rejects all others. L VL
Vin
Vin
Vout
Vin R
Vout
Vout

16. RL Lag Circuit

17. Examples

18. Sinusoidal Response of Parallel RL
For parallel circuits, it is useful to review conductance, susceptance and admittance, introduced in Chapter 10.
Conductance is the reciprocal of resistance.
Inductive susceptance is the reciprocal of inductive reactance.
Admittance is the reciprocal of impedance.

19. Admittance

20. Sinusoidal Response of Parallel RL
In a parallel RL circuit, the admittance phasor is the sum of the conductance and inductive susceptance phasors.
The magnitude of the susceptance is
The magnitude of the phase angle is G VS G BL BL Y

21. Sinusoidal Response of Parallel RL
Some important points to notice are:
G is plotted along the positive x-axis.
BL is plotted along the negative y-axis.
Y is the diagonal G VS G BL Y BL

22. Sinusoidal Response of Parallel RL
Example
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
G = 1.0 mS VS R
1.0 kW L
25.3 mH
f = 10 kHz
BL =
0.629 mS
Y =
1.18 mS

23. Analysis of Parallel RL Circuit
Ohm’s law is applied to parallel RL circuits using quantities of Y, V, and I.
Because V is the same across all components in a parallel circuit, you can obtain the current in a given component by simply multiplying the admittance of the component by the voltage as illustrated in the following example.

24. Analysis of Parallel RL Circuit
Example
Assume the voltage in the previous example is 10 V. Sketch the current phasors. The admittance diagram from the previous example is shown for reference.
The current phasors can be found from Ohm’s law. Multiply each admittance phasor by 10 V.
G = 1.0 mS
x 10 V =
IR = 10 mA
BL =
0.629 mS
Y =
1.18 mS
IL =
6.29 mA
IS =
11.8 mA

25. Phase Angle of Parallel RL
Notice that the formula for inductive susceptance is the reciprocal of inductive reactance. Thus BL and IL are inversely proportional to f:
As frequency increases, BL and IL decrease, so the angle between IR and IS must decrease as well. IR q IS IL

26. Examples

27. Phase Relationships

28. Examples

29. Series-Parallel RL
Series-parallel RL circuits are combinations of both series and parallel elements. The solution of these circuits is similar to resistive combinational circuits but you need to combine reactive elements using phasors.
The components in the yellow box are in series and those in the green box are also in series. R2 R1 Z1 Z2 L1 L2
and
The two boxes are in parallel. You can find the branch currents by applying Ohm’s law to the source voltage and the branch impedance.

30. The Power
Recall that in a series RC or RL circuit, you could multiply the impedance phasors by the current to obtain the voltage phasors. The earlier example from this chapter is shown for review:
x 10 mA =
Z = 1.33 kW
VS = 13.3 V
VL = 9.6 V
XL = 960 W
39o
39o
R = 1.2 kW
VR = 12 V

31. The Power Triangle Example
Multiplying the voltage phasors by Irms gives the power triangle (equivalent to multiplying the impedance phasors by I2). Apparent power is the product of the magnitude of the current and magnitude of the voltage and is plotted along the hypotenuse of the power triangle.
Example
The rms current in the earlier example was 10 mA. Show the power triangle.
x 10 mA =
VR = 12 V
VL = 9.6 V
VS = 13.3 V
39o
Pa = 133 mVA
Pr =
96 mVAR
39o
Ptrue = 120 mW

32. The Power Triangle

33. Examples

34. Power
The power factor was discussed in Chapter 15 and applies to RL circuits as well as RC circuits. Recall that it is the relationship between the apparent power in volt-amperes and true power in watts. Volt-amperes multiplied by the power factor equals true power.
Power factor is defined as
PF = cos 

35. Apparent Power
Apparent power consists of two components; a true power component, that does the work, and a reactive power component, that is simply power shuttled back and forth between source and load.
Power factor corrections for an inductive load (motors, generators, etc.) are done by adding a parallel capacitor, which has a canceling effect.
Ptrue (W)
Pr (VAR)
Pa (VA)

36. Selected Key Terms Inductive susceptance (BL)
The ability of an inductor to permit current; the reciprocal of inductive reactance. The unit is the siemens (S).

37. Quiz
1. If the frequency is increased in a series RL circuit, the phase angle will
a. increase
b. decrease
c. be unchanged

38. Quiz
2. If you multiply each of the impedance phasors in a series RL circuit by the current, the result is the
a. voltage phasors
b. power phasors
c. admittance phasors
d. none of the above

39. Quiz 3. For the circuit shown, the output voltage
a. is in phase with the input voltage
b. leads the input voltage
c. lags the input voltage
d. none of the above
Vin
Vout

40. Quiz
4. In a series RL circuit, the phase angle can be found from the equation a. b.
c. both of the above are correct
d. none of the above is correct

41. Quiz
5. In a series RL circuit, if the inductive reactance is equal to the resistance, the source current will lag the source voltage by
a. 0o
b. 30o
c. 45o
d. 90o

42. Quiz 6. Susceptance is the reciprocal of a. resistance b. reactance
c. admittance
d. impedance

43. Quiz
7. In a parallel RL circuit, the magnitude of the admittance can be expressed as a. b.
c. Y = G + BL d.

44. Quiz 8. If you increase the frequency in a parallel RL circuit,
a. the total admittance will increase
b. the total current will increase
c. both a and b
d. none of the above

45. Quiz 9. The unit used for measuring true power is the a. volt-ampere
b. watt
c. volt-ampere-reactive (VAR)
d. kilowatt-hour

46. Quiz 10. A power factor of zero implies that the
a. circuit is entirely reactive
b. reactive and true power are equal
c. circuit is entirely resistive
d. maximum power is delivered to the load

47. Quiz
Answers:
1. a
2. a
3. c
4. a
5. c
6. b
7. d
8. d
9. b
10. a