1. SOURCE TRANSFORMATIONS TECHNIQUE
Lecture 11
SOURCE TRANSFORMATIONS TECHNIQUE

2. Learning Outcomes
After completing this module, you will be able to:
convert a voltage source into an equivalent current source,
convert a current source into an equivalent voltage source,
Use source transformation to simplify and analyze circuits.

3. Equivalent Practical Sources
Two sources are equivalent if they produce identical values of vL and iL when they are connected to identical values of RL, no matter what the value of RL may be. Since RL =∞ and RL = 0 are two such values, equivalent sources provide the same open-circuit voltage and short-circuit current.
Cct 1 RL
v(t)
i(t)
i(t)
v(t)
Vo/c
Is/c
Cct 2 RL
v(t)
i(t)

4. Thevenin's Theorem for Linear Resistive Circuits
Any combination of batteries and resistances with two terminals can be replaced by a single voltage source VTH and a single series resistance RTH.
Linear Circuit A B
VTH
RTH

5. The value of VTH is the open circuit voltage at the terminals.
In practice, the Thevenin voltage, VTH, of the network is obtained by measuring the open-circuit voltage VAB.
By definition,
VTH = VAB
Note
In circuit analysis, VTH is found by computing the voltage drop across the terminals A and B. A
Linear Circuit
Voltmeter B

6. Measure the open-circuit voltage VAB.
The value of RTH is obtained by dividing VAB by the short circuit current ISC.
In practice, the Thevenin resistance, RTH, of the network is obtained as follows:
Measure the open-circuit voltage VAB.
Short the output terminals with an ammeter and read off the short circuit current displayed by the ammeter. Let the short-circuit current reading displayed by the ammeter be ISC (say).
Divide the open-circuit voltage by the short circuit current. The ratio obtained is the Thevenin resistance, RTH; that is, A
Linear Circuit
Isc B

7. Worked Example
Find the Thevenin equivalent of the circuit in the figure below. A B
3 A
3 Ω
Solution
By inspection, we see that the Thevenin voltage VTH that we want to find is given by the voltage drop across the 3 Ω resistor; that is
VTH = VAB = V3Ω
Therefore,

8. Finding the Thevenin resistance RTH.
One way to find the Thevenin resistance is to first find the short circuit current through A and B when the two terminals are short-circuited by a wire, as shown in the following figure. A B
3 A
3 Ω
ISC
Thus,
Therefore,

9. Alternative method for finding the Thevenin resistance RTH.
An alternative method for finding the Thevenin resistance of the given circuit is as follows:
Step 1. Replace the current source with an open circuit.
Step 2. Find the equivalent resistance of the circuit, looking into the circuit from the terminals A and B. The resistance obtained is the Thevenin resistance we are looking for.
Thus, replacing the current source in the previous figure with open circuit leads us to the following circuit. A B
3 Ω
RTH
Thus, from inspecting the figure on the left, we see that

10. A B
3 A
3 Ω
3 Ω A
9 V B

11. Find the Thevenin equivalent of the circuit in the following figure.
Worked Example
Find the Thevenin equivalent of the circuit in the following figure. A B
2 A
5 Ω
Solution
By inspection, we see that the Thevenin voltage VTH that we want to find is given by the voltage drop across the 5 Ω resistor; that is
VTH = VAB = V5Ω
Therefore,

12. Finding the Thevenin resistance RTH.
One way to find the Thevenin resistance is to first find the short circuit current through A and B when the two terminals are short-circuited by a wire, as shown in the following figure. A B 2A 5Ω
ISC
Thus,
Therefore,

13. Alternative method for finding the Thevenin resistance RTH.
An alternative method for finding the Thevenin resistance of the given circuit is as follows:
Step 1. Replace the current source with an open circuit.
Step 2. Find the equivalent resistance of the circuit, looking into the circuit from the terminals A and B. The resistance obtained is the Thevenin resistance we are looking for.
Thus, replacing the current source in Figure x with open circuit leads us to the following circuit.
Thus, from inspecting the figure shown on the left, we see that A B
5 Ω
RTH

14. A B
2 A
5 Ω
5 Ω A
-10 V B

15. Find the Thevenin equivalent of the circuit in the following figure.
Worked Example
Find the Thevenin equivalent of the circuit in the following figure. A B
2 Ω
3 V
Solution
By inspection, we see that the Thevenin voltage VTH that we want to find is given by the voltage drop across the 2 Ω resistor; that is
VTH = VAB = V2Ω
Therefore,

16. Finding the Thevenin resistance RTH.
One method for finding the Thevenin resistance of the given circuit is as follows:
Step 1. Replace the voltage source with a short circuit.
Step 2. Find the equivalent resistance of the circuit, looking into the circuit from the terminals A and B. The resistance obtained is the Thevenin resistance we are looking for.
Thus, replacing the voltage source in the previous figure with a short circuit leads us to the following circuit.
Thus, from inspecting the figure shown on the left, we see that A B
2 Ω
RTH

17. A B
2 Ω
3 V
0 Ω A
3 V B

18. Norton's Theorem for Linear Resistive Circuits
Any collection of batteries and resistances with two terminals is electrically equivalent to an ideal current source IN in parallel with a single resistor RN. B IN RN A
Linear Circuit A B

19. The value of IN is the short circuit current at the terminals.
In practice, the Norton current, IN, of the network is obtained by measuring the short circuit IAB.
By definition,
IN= IAB
Note
In circuit analysis, IN is found by computing the short circuit current flowing through the shorting wire connecting the terminals A and B.
Linear Circuit A B
Ammeter

20. Measure the open-circuit voltage VAB.
The value of RN is obtained by dividing VAB by the short-circuit current, ISC.
Like the Thevenin resistance, RTH, the Norton resistance of the network is obtained as follows:
Measure the open-circuit voltage VAB.
Short the output terminals with an ammeter and read off the short circuit current displayed by the ammeter. Let the short-circuit current reading displayed by the ammeter be ISC (say).
Divide the open-circuit voltage by the short circuit current. The ratio obtained is the Norton resistance, RN; that is,
Linear Circuit A B
Isc

21. Find the Norton equivalent of the circuit in the figure shown below.
Worked Example
Find the Norton equivalent of the circuit in the figure shown below.
5 Ω A B
7 V
Solution
By inspection, we see that the Norton current IN that we want to find is given by the short current flowing through terminals A and B when they are shorted by a wire, as shown in the following figure.

22. 5 Ω A B
7 V
ISC
The Norton resistance RN is obtained by finding the resistance looking into the A-B terminals when the voltage source is replaced by a short circuit.
5 Ω A B
RTH

23. A
7/5 A
5 Ω B
Both Thevenin’s and Norton’s Theorems can also be used in ac circuit analysis.

24. Source Transformations Technique for Linear AC Circuits
Source transformations is a technique of analysing circuits which involve transforming part of a circuit into its Thevenin equivalent or into its Norton equivalent, while retaining the terminal characteristics of the original circuit. By transforming a subcircuit into its Thevenin or Norton equivalent, impedances or sources which then appear as being connected in parallel or in series can thus be combined and simplified. By doing repeated source transformations, we can eventually simplify a circuit to its simplest form for solution.
The following examples show how we can use source transformations technique to solve a circuit.

25. Worked Example
Find the steady-state expression for io(t) if ig(t) = 125cos(500t) mA
R1 = 50 Ω
R2 = 250 Ω
L = 1 H
C = 20 F R1 R2 C L

26. Solution Obtain the phasor equivalent circuit. ZR1 = R1 = 50 Ω
ZL = jL) = j500 Ω
ZC = 1/(jC) = - j100Ω
Ig = 0.1250 A
ZR1
ZR2 ZC ZL Ig Io

27. Replace the subcircuit on the left with its Thevenin equivalent.
ZAB A B A B
VAB

28. Applying KVL to the circuit leads us to the equation
Calculate Io.
Applying KVL to the circuit leads us to the equation
ZAB
giving
VAB
Hence
Therefore,

29. Worked Example
Find the steady-state expression for vo(t) if vg(t) = 64cos(8000t)

30. Solution Obtain the phasor equivalent circuit. ZR = R = 2000 Ω
ZC = 1/(jC) = - j4000 Ω
ZL = jL = j4000 Ω
Vg = 640 V ZR ZC ZL
Vg = 640 V Vo

31. Replace the subcircuit on the left with its Thevenin equivalent.
ZAB A
VAB B

32. Applying voltage divider rule to the circuit leads us to the equation
Calculate Vo.
Applying voltage divider rule to the circuit leads us to the equation ZL
Hence,

33. Exercise
If the current source iS outputs a current (1cos2t) A, find the voltage drop vo across the 2-Ω resistor. iS

34. Worked Example
Use source transformations to solve for the steady-state part of vo. The sinusoidal voltages are:
v1(t) = 240cos(4000t ) V
v2(t) = 96sin(4000t) V

35. Solution Obtain the phasor equivalent circuit. ZR2 = R2 = 20 Ω
ZC = 1/(jC) = 1/(j4000 x 25 x 10-6) = - j10 Ω
ZL = jL = j4000 x 15 x 10-3 = j60 Ω
V1 = 24053.13 V
ZR2 = R2 = 20 Ω
V2 = 96-90 V

36. Phasor equivalent circuit:
ZR1
ZR2 ZL ZC V1 V2 V0
ZR2 = R2 = 20 Ω
ZR1 30 Ω ;
ZL = j60 Ω ;
ZC = - j10 Ω
V2 = 96-90 V
V1 = 24053.13 V

37. Replace the subcircuit on the left of ZC with its Norton equivalent.
IAB
ZAB A B
IAB B

38. Replace the subcircuit on the right of ZC with its Norton equivalent.
ZAB
IAB
IAB ;

39. Draw the simplified circuit by replacing the left and right subcircuits with the Norton equivalents.Z1 Z2 I2
where

40. Combine the two current sources into one equivalent source
Combine the two current sources into one equivalent source. Combine the three impedances to one equivalent impedance.
IEQ = I1 + (– I2) = 4-36.87 - (j4.8) = 436.87 A
ZEQ = 6.845-56.88 Ω ZC =
ZEQ Z1 Z2
IEQ
ZEQ
Therefore,
and

41. Worked Example
Find currents Ib, Ic and Id.

42. Solution
Method I
Replace the subcircuit to the right of the j5Ω impedance with its Thevenin equivalent.

43. Therefore, we get
KVL gives
Solving the above equation for Ib, we obtain

44. Replace the subcircuit to the right of the -j5Ω impedance with its Thevenin equivalent.
VAB
ZAB

45. Therefore, we get
KVL gives
Solving the above equation for Ic, we obtain

46. Calculate Id using KCL.
Therefore,

47. Method II
Temporarily remove the j5 Ω inductive impedance and replace it with an open circuit. From the resulting circuit obtained, find the open circuit voltage VAB and the terminal impedance ZAB.

48. VAB
V5Ω
Voltage drop across the 5 Ω resistance is

49. Therefore, voltage drop across terminals A-B is
Thevenin impedance ZAB is equal to the impedance seen when looking into terminals A-B with the voltage sources replaced with short-circuits.
ZAB

50. Connect the j5 Ω inductive impedance (which was removed earlier from the original circuit) across the A-B terminals of the Thevenin equivalent circuit and compute the current Ib.

51. Temporarily remove the -j5 Ω capacitive impedance from the original circuit and replace it with an open circuit. From the resulting circuit obtained, find the open circuit voltage VAB and the terminal impedance ZAB.

52. VAB
V5Ω
Voltage drop across the 5 Ω resistance is

53. Therefore, voltage drop across terminals A-B is
Thevenin impedance ZAB is equal to the impedance seen when looking into terminals A-B with the voltage sources replaced with short-circuits.
ZAB

54. Connect the -j5 Ω capacitive impedance (which was removed earlier from the original circuit) across the A-B terminals of the Thevenin equivalent circuit and compute the current Ic. Ic

55. Calculate Id using KCL.
Therefore,

56. Worked Example
Find the phasor voltage Vo.

57. Solution
Temporarily remove the –j3 Ω capacitive impedance and replace it with an open circuit. From the resulting circuit obtained, find the open circuit voltage VAB and the terminal impedance ZAB. A B

58. A B
Vj3Ω VA VB
Voltage drop across the j3 Ω inductive impedance is

59. Potential of terminal A w.r.t. ground,
Potential of terminal B w.r.t. ground,

60. Therefore,
Thevenin impedance ZAB is equal to the impedance seen when looking into terminals A-B with the voltage source replaced with a short-circuit and the current source replaced with an open-circuit.

61. From the circuit on the right, we getA B

62. The Thevenin equivalent circuit between terminals A and B is as shown below.
Applying voltage divider rule to the circuit above, we obtain Vo.

63. END