SOURCE TRANSFORMATION AC CIRCUIT

SOURCE TRANSFORMATION

Source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind:

Example:

Calculate Vx in the circuit using source transformation

First we transform the voltage source to a current source  where

Is=20angle-90/5

Is=4angle-90 or -j4A

The parallel combinations of 5ohms resistance and (3+j4) impedance gives

z1=5(3+j4)/8+j4

z1=2.5+j1.25ohms

We must convert the current source to a voltage source yields the circuit in the figure where

Vs=IsZ1=-j4(2.5+j1.25)=5-j10v

By voltage division,

Vx=(10/10+2.5+j1.25+4-j13)(5-j10)

Vx=5.519angle-28V

SUPERPOSITION IN AC CIRCUIT

SUPERPOSITION

Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. It becomes important if the circuit has sources operating at different frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency. The total response must be obtained by adding the individual responses in the time domain. It is incorrect to try to add responses in the phasor or frequency domain. Because the exponential factor e^jwt is implicit in sinusoidal analysis, and that factor would change for every angular frequency w. It would therefor not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.

Example we gather from the book:

Use The superposition theorem to find Io in the circuit in the figure below:

So First we let 

Io=I'o + I''o

Where I'o and I''o are due to the voltage and current sources, respectively. To find I'o consider the circuit in figure a. If we let Z be the parallel combination of -j2 and 8+j10, then

Z=-j2(8+j10)/-2j+8+j10=0.25-j2.25

And current I'o is 

I'o=j20/4-j2+Z

=J20/4.25-j4.25

or I'o= -2.353 + j2.353

To get I''o, consider the circuit in figure b. For mesh 1,

(8+j8)I1-j10I3+j2I2=0

For mesh 2, 

(4-j4)I2+J2I1+J2I3=0

For mesh 3.

I3=5

From equations of mesh 2 and mesh 3

(4-j4)I2+j2I1+j10=0    =>eq 1

Expressing I1 in terms of I2 gives

I1=(2+j2)I2-5     =>eq 2

Substituting equation of mesh 3 and equation 2 into equation of mesh 1 we get

(8+j8)[(2+j2)I2-5]-j50+j2I2=0

or I2=90-40/34 =2.647-j1.176

Current I''o is obtained as

I''o=-I2=-2.647+j1.176

From the quation I'o=-2.353+j2.353 and I''o=-I2 we write

Io=I'o+I''o=-5+j3.529

=6.12angle144.78 A

MESH ANALYSIS IN AC CIRCUIT

MESH ANALYSIS

Kirchhoff's voltage law (KVL) forms the basis of mesh analysis. We must keep in our mind that the very nature of using mesh analysis is that it is to be applied to planar circuits.

EXAMPLE:

Determine current Io in the circuit using mesh analysis

Applying KVL to mesh 1,

(8+j10-j2)I1-(-j2)I2-j10I3=0

Mesh 2,

(4-j2-j2)I2-(-j2)I1-(-j2)I3+20angle90=0

Mesh 3, I3=5 We must substitue this in equation 1 and 2,

(8+j8)I1+j2I2=j50

j2I1+(4-j4)I2=-j20-j10

Put it in matrix form as

   | 8+j8            j2  |     I1=|j50|

         |         j2         4-j4|  I2=|-j30 |

from wiche we obtain the derminants

delta=68

delta 2=416.17angle -35.22

where I2=Delta2/delta=416.17angle-35.22/68

I2=6.12angle-35.22 A      

Supermesh

-supermesh analysis is a better technique instead of using mesh analyis such a complex electric circuit where two meshes have a current source as a common element. This is the same with supernode analysis which we also use when voltage source has a common node in the circuit.