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