EE 201 ELECTRIC CIRCUITS
Class Notes
CLASS 12
The material covered in this class will be as follows:
⇒
Source Transformation
⇒
Use of Source Transformation in Circuit Analysis
At the end of this class you should be able to:
⇒ Understand the meaning of source transformation
⇒ Apply source transformation
⇒ Recognize when source transformation is not applicable
⇒ Use source transformation to simplify and analyze circuits
Source Transformation:
Given an ideal voltage source Vs in series with a resistor Rs .
⇓ (can we replace them with)
An ideal current source I s in parallel with a resistor R p ?
Figure 1
Connect the same load resistor RL across terminal “a-b” in both circuits.
If circuits “1” and “2” are equivalent ⇒
I1 = I 2
&
V1 = V2
Circuit “1” ⇒ I1 =
Circuit “2” ⇒
I1 = I 2
⇒
Vs
Vs
=
Req Rs + RL
I2 =
Rp
R p + RL
Is
Rp I s
Vs
=
Rs + RL R p + RL
(1)
(2) (CDR is used)
(3)
If we choose R p = Rs ⇒ Vs = I s R p = I s Rs
∴ Rs = R p
&
Vs = I s Rs
⇓
⇔
Vs in series with Rs
I s in parallel with Rs
Figure 2
The above conversion is called Source Transformation (ST).
Circuits (1) & (2) are equivalent. However, they are not the same.
When any load is connected to terminals “ a ” & “ b ” of circuits (1) & (2)
⇓
Load cannot distinguish between the two circuits.
Circuits (1) & (2) are equivalent from the outside when accessed from terminals “a” & “b”.
Circuits (1) & (2) are different from the inside.
Example 1:
Covert the following circuits using ST.
Figure 3
Solution:
Circuit (1):
R p = Rs = 2Ω
Is =
Vs 3
= = 1.5 A
Rs 2
≡
Figure 4
Circuit (2):
R p = Rs = 5Ω
Is =
Vs 4
= = 0.8 A (downwards. Why?)
Rs 5
≡
Figure 5
Example 2:
Covert the following circuits using ST.
Figure 6
Solution:
Circuit (1):
Rs = R p = 3Ω
Vs = Rs I s = 3(3) = 9V
&
≡
Figure 7
Circuit (2):
Rs = R p = 5Ω
&
Vs = Rs I s = 5(2) = 10V (upper voltage polarity is negative. Why?)
≡
Figure 8
You need to be careful when using ST, as we will see in the next example.
Example 3:
Covert the following circuits using ST.
Figure 9
Solution:
Circuit (1):
A resistor in parallel with a voltage source (not in series) ⇒
ST does not apply.
A resistor in parallel with a voltage source ⇒ equivalent to a voltage source.
a
3V
2
≡
a
3V
b
b
Figure 10
Circuit (2):
A resistor in series with a current source (not in parallel) ⇒
ST does not apply.
≡
Figure 11
A resistor in series with a current source ⇒ equivalent to a current source.
Example 4:
Use ST to calculate:
a) i1
b) i2
Figure 12
Solution:
It is a good idea to first label some points on the circuit.
Figure 13
Apply ST to the 2 A & 1Ω combination ⇒ V = IR = 2(1) = 2V
Notice that i2 cannot be drawn. It disappears. Why?
The current through the 1Ω of the transformed circuit is not i2 . Why?
Reason: R p = Rs means the two resistors have the same value. It does not mean we have the same
resistor!!
Q 2Ω & 1Ω in series ⇒ current through the 1Ω of the transformed circuit is i1 .
Figure 14
2Ω & 1Ω in series
8V
⇒
& 2V in series ⇒
a) i1 =
3Ω
8 − 2 = 6V
6
= 2A
3
b) KCL at node “ a ” (of the original circuit)
⇒
i2 = i1 + 2 = 2 + 2 = 4 A
Figure 15
Figure 12
Example 5:
Use ST to calculate V .
Figure 16
Solution:
Apply ST to ( 3V in series with 1Ω ) & ( 18V in series with 2Ω )
I s1 =
3
= 3A
1
&
Is2 =
18
= 9A
2
Figure 17
1Ω 3Ω 2Ω
3A 9 A
⇒
⇒
Req =
1
= 0.545Ω
1 1
1+ +
3 2
I eq = 9 − 3 = 6 A
∴V = I eq Req = 6(.545) = 3.273V
Figure 18
Example 5: Using ST, calculate:
a) i1
b) i2
Figure 19
Solution:
ST ⇒ V = 3 × 2 = 6V
Figure 20
2Ω 8Ω
10Ω
⇒
Figure 21
ST ⇒
I=
6
= 0.6 A
10
Figure 22
1Ω 10Ω
⇒
10 × 1 10
=
= 0.909Ω
10 + 1 11
Figure 23
ST ⇒ V = 0.909 × 0.6 = 0.545V
Figure 24
Req = 5 + 0.909 = 5.909Ω
a) ∴ i1 =
&
Veq = 7 + 0.545 = 7.545V
7.545
= 1.277 A
5.909
b) KVL (in the original circuit) ⇒ −7 + 5i1 + 1i2 = 0 ⇒ −7 + 5(1.277) + 1i2 = 0 ⇒ i2 = 0.615 A
Figure 25
Figure 19