Source Transformation

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Presentation transcript:

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.

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.

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

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

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

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

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

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

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

Example We want to calculate the voltage Vo.

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

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

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.

2k is in series with 2k so

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.

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

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

4k and 4k are in series so

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

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

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.

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

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

3k is in series with 4k so

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

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

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

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.

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

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

The current I can be found using KVL -7.5 + 3.5I – 51Vx + 28I +9 =0 Where Vx = 2I

Solving I = 21.28mA