Transcript Source Transformation

2.7 Source Transformations
A method called Source Transformations will allow the transformations of a voltage
source in series with a resistor to a current source in parallel with resistor.
R
a
a
vs
is
+
-
R
b
b
The double arrow indicate that the transformation is bilateral , that we can start with either
configuration and drive the other
R
a
a
vs
+
-
is
RL
iL
R
RL
iL
b
b
Voltage source with a series resistor
Current source with a parallel resistor
R
iL 
is
R + RL
vs
iL 
R + RL
Equating we have
,
vs
R

is
R + RL R + RL
 is 
vs
R
OR v s  Ri s
Polarity of Vs in one form must be such that it tends to
push in the direction of Is in the other form
+
-
Ex. 2.29: Determine i and ix using source transformation
Single loop
KVL
15V - 10V
i
 1A
2 + 3
ix should be found from the original circuit
i x  - (i - 5)
= 5 - i  4A
Ex 2.30: Determine I and V
Transform the 3V in series with the 2 Ohm resistor
We get a single-node-pair circuit
14
3

3
 4
V   A + 2A  * 2 || 4   A + 2A  *   V
3
2

2
 3
V 7
 A
Using Ohm’s law in the original circuit: I 
4 6
Example (a) find the power associated with the 6 V source
(b) State whether the 6 V source is absorbing or
delivering power
We are going to use source transformation to reduce the circuit, however note that we
will not alter or transfer the 6 V source because it is the objective.
40
 8A
5
(20 || 5)  4Ω
(8A )(4Ω)  32 V
(6 + 4)  10Ω
(10 + 10)  20Ω
32
 1.6A
20
(30 || 20)  12Ω
(4 + 12)  16Ω
(1.6A )(12Ω)  19.2 V
i
i=
-
 0.825 A
+
 P6V  (0.825)(6)  4.95 W
It should be clear if we transfer the 6V during these steps you will not be able to find
the power associated with it