1. Source Transformation
If we have any source embedded within a network, say this source is a current source having a value I & there exists a resistance having a value R, in parallel to it. We can replace it with a voltage source of value V=IR in series with same resistance R.

2. Source Transformation
The reverse is also true that is a voltage source V, in series with a resistance R can be replaced by a current source having a value I= V/R
In parallel to the resistance R.

3. Source Transformation
Parameters within circuit, for example an output voltage remain unchanged under these transformations.

4. Example
We want to calculate the voltage Vo using
source transformation method. We
proceed as

5. Vo
1k is in series with 2k
so combined effect =3k

6. Now 2mA source is in parallel with 3k
resistor.
So it can be changed to a voltage source
of value

7. 3k resistor will become in series with this source.Vo
= 2m x 3k
=6 Volts.
3k resistor will become in series with this
source.

8. Positive terminal of the 6 volts battery is
connected with the negative terminal of
3 volts battery so they will be summed up.

9. Vo = (6 x 9)/9
= 6 Volts

10. Example
We want to calculate the voltage Vo.

11. 3k is in series with 12 volts battery. So it
can be converted into a current source of
value =12/3k
=4mA

12. Now 3k resistor is in parallel with 6k
resistor so
3k||6k = (3k x6k)/3k +6k
=2k

13. 4mA source is parallel with 2k resistor.
So it can be converted into a voltage
source of 8 volts in series with a resistor
of 2k.

14. 2k is in series with 2k so

15. 8V source is in series with 4k resistor . So
it can be converted into a current source
of value = 8/4k =2mA
in parallel to 4k.

16. Current sources will add up to give value
of 4mA.

17. 4mA source is in parallel with 4k resistor .
So it can be converted into a voltage
source of value 16volts.

18. 4k and 4k are in series so

19. Now applying voltage division rule
V0 = 8k/16k x 16V
=8volts

20. Example
We want to calculate the voltage V0 by source
transformation method.

21. 12 volt source is in series with 3k resistor.
So it can be converted into current source
of value
I = 12/3k =4mA
and resistor 3k in parallel.

22. Now we will combine current sources as
4m – 2m = 2mA

23. 2mA source is in parallel with 3k source so
it can be converted into a voltage source
of value
V=2 x3 = 6volts.

24. 3k is in series with 4k so

25. Now applying voltage division rule
V0 = 2k/9k x 6
=12/9
=4/3 volts.

26. EXAMPLE
we want to calculate the current I by
source transformation.

27. 3 Ohms resistor is in parallel with 5A
source. So it can be converted into
a voltage source 15V.

28. 3 Ohms resistor 4 Ohms are in series
and the resultant is in series with 15V
so it can be converted into a current
Source.

29. 7 ohm is parallel with 7 ohm so,
7||7 = 7 X 7 / = 49 /14 = 3.5 ohms

30. Current source is in parallel with 3.5 ohms
Resistance. So it can be converted into a
Voltage source by using Ohm’s Law
V =IR= 15/7 X 3.5 = 7.5 Volts

31. The current I can be found using KVL
I – 51Vx + 28I +9 =0
Where
Vx = 2I

32. Solving
I = 21.28mA