# solving an AC circuit using superpositon and loop analysis

Discussion in 'Homework Help' started by blalaw, Feb 4, 2014.

1. ### blalaw Thread Starter New Member

Oct 28, 2013
9
0
I have the following circuit and trying to solve it and the problem calls to solve it using superposition and loop analysis.

I have canceled the 5.5cos2t source and replaced it with a short circuit then calculate my impedance's and have gotten four resulting loops.

1: (9.5 + 4j)I1 + (0 - 4j)I2 + (-7 +0j)I3 + (-4I + 0j)I4 = 0
2: (0 -4j)I1 + (3 + 3.5j)I2 +(0 + 0j)I3 + (-3 + 0j)I4 = -7
3: (-7 + 0j)I1 + (0 + 0j)I2 + (12-.125j)I3 + (0 + .125j)I4 =
4: (0 + 0j)I1 + (-3 + 0j)I2 + (0 + .125j)I3 + (4 + 11.875j)I4 = 0

I think I am just lost as how to solve for currents with the imaginary numbers there with the 4 unknowns and am a little lost from there any help or insight would be appreciated thanks.

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,702
555
You have 4 unknowns and you have a system of 4 simultaneous equations. Your choices are:
1) Use substitution.
2) Use matrices.
3) Use calculator to solve the system of 4 simultaneous equations, if your calculator can. (Mine can)

3. ### blalaw Thread Starter New Member

Oct 28, 2013
9
0
I understand I can do that but when using matrcies its been a while since using imaginary numbers would I just create two separate matrices one for the real numbers and another for the imaginary or one complete matrix?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,702
555
One complete matrix.

If you don't mind the skull work, just do substitution.

The first step for me would have been replacing capacitors and inductions with their equivalent impedance. This way you have a circuit that is made of "resistors". That way 1 H inductor is 1s (or 1jw) Ohm resistor. 2 F capacitor becomes 1/(2s) (or 1/(2jw)=-j/(2w)) Ohm resistor. I am using w to represent omega, the angular frequency.

5. ### blalaw Thread Starter New Member

Oct 28, 2013
9
0
ya I did that before I did the loops but I believe it was different for each source when I did superposition since each source has a different omega value.

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,702
555
Ah, ok, I was sort of wondering about the sums in the parenthesis.
Edit.
I was wrong, my calculator does not do complex numbers in solving simultaneous equations and in matrices. I feel a little betrayed.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Be careful. There are two different driving frequencies.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
There are several downloadable applications available which can handle complex matrices and in the case of [non-singular] square forms, their inversion.
I use a free application called Scilab which readily "solves" the set of (4) equations in post #1.
Also use an android app called Addi which handles complex matrices.

BTW I think equation 1. should be

(9.5 + 4j)I1 + (0 - 4j)I2 + (-7 +0j)I3 + (-4 + 0j)I4 = 0

rather than

(9.5 + 4j)I1 + (0 - 4j)I2 + (-7 +0j)I3 + (-4I + 0j)I4 = 0

..... a small notation error.

Last edited: Feb 5, 2014
9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Not sure what you mean. What is the 5th unknown variable?

10. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,702
555
Oops. I looked at the wrong thing.

11. ### WBahn Moderator

Mar 31, 2012
19,141
5,176
The problem seems to be working with complex numbers.

Just replace each coefficient with a symbolic parameter.

1: K11*I1 + K12*I2 + K13*I3 + K14*I4
2: K21*I1 + K22*I2 + K23*I3 + K24*I4
3: K31*I1 + K32*I2 + K33*I3 + K34*I4
4: K41*I1 + K42*I2 + K43*I3 + K44*I4

Now you can forget about the fact that each K is, in general, a complex number. Just solve this set if equations for the four currents. The result for each will be an equation in terms of the K values. Then just plug in the values for K and turn the crank. At the end of the day, you know that all of the currents should be purely real.