Finding Thevenin Equivalent Circuit for Complex Network

[Lunat1c]

Homework Statement

I wish to obtain the Thevenin equivalent for the network show in the attachment between the terminals a and b.

2. The attempt at a solution

First I considered the voltage source alone (replaced current source by an open circuit).

$$I = \frac{V}{Z_T} = \nfrac{20}{5+10+j5+j10+ (2*j4)} = 0.728A \angle-56.89\degrees$$

$$\therefore V_{AB}^1 = (0.728A\angle-56.89\degrees) * (5+10+j5+j4) = 12.74V\angle-25.92\degrees$$

Then I considered the current source alone (replaced the voltage source by a short):

Current through upper branch = $$\frac { 2A\angle90\degrees (10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.282A\angle29.83\degrees$$

$$\therefore V_{AB}^2 = (0.282\angle29.83\degrees)(j10 -j4 + j4) = 2.82V\angle119.83\degrees$$

$$V_{AB} = V_{AB}^1 + V_{AB}^2 = 10.529V\angle-17.25\degrees$$

Can someone please check this for me? I'm not sure whether I did some mistake and I need to know whether I understood these.

Thank you for your time,

Lunat1c

 

[The Electrician]

Homework Statement

I wish to obtain the Thevenin equivalent for the network show in the attachment between the terminals a and b.

2. The attempt at a solution

First I considered the voltage source alone (replaced current source by an open circuit).

$$I = \frac{V}{Z_T} = \nfrac{20}{5+10+j5+j10+ (2*j4)} = 0.728A \angle-56.89\degrees$$

$$\therefore V_{AB}^1 = (0.728A\angle-56.89\degrees) * (5+10+j5+j4) = 12.74V\angle-25.92\degrees$$

Then I considered the current source alone (replaced the voltage source by a short):

Current through upper branch = $$\frac { 2A\angle90\degrees (10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.282A\angle29.83\degrees$$

$$\therefore V_{AB}^2 = (0.282\angle29.83\degrees)(j10 -j4 + j4) = 2.82V\angle119.83\degrees$$

$$V_{AB} = V_{AB}^1 + V_{AB}^2 = 10.529V\angle-17.25\degrees$$

Can someone please check this for me? I'm not sure whether I did some mistake and I need to know whether I understood these.

Thank you for your time,

Lunat1c

Your first calculation should be:

$$I = \frac{V}{Z_T} = \frac{20}{5+10+j5+j10-j4+ (2*j4)}$$

Also, in this calculation, there is an error in your arithmetic:

Current through upper branch = $$\frac { 2A\angle90\degrees (10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.282A\angle29.83\degrees$$

I get:

Current through upper branch = $$2A\angle90\degrees \frac {(10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.4131A\angle38.29\degrees$$

And, of course, these errors propagate into subsequent calculations.

 

[Lunat1c]

Current through upper branch = $$2A\angle90\degrees \frac {(10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.4131A\angle38.29\degrees$$

And, of course, these errors propagate into subsequent calculations.

Ok, I agree with that part, found the mistake. However, do you agree with the rest of the working (especially the way I considered the mutual inductances)? I really wish to confirm whether I have the correct final answer or not. Now I got $$V_{AB}=9.225V\angle-13.79$$

 

[The Electrician]

For the final answer, I get $$11.1156V\angle {-9.722} ^ \circ$$

Your handling of the mutual inductance seems essentially ok, but without showing all the calculations you showed in your first post, with the errors I pointed out corrected, I can't tell where you went wrong.

 

[Lunat1c]

$$I = \frac{V}{Z_T} = \frac {20} {5+10+j5+j10+ (2*j4)} = 0.728A \angle-56.89\degrees$$

$$\therefore V_{AB}^1 = (0.728A\angle-56.89\degrees) * (5+10+j5+j4) = 12.74V\angle-25.92\degrees$$ $$2A\angle90\degrees \frac {(10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.4131A\angle38.29\degrees$$

$$V_{AB}^2 = (0.4131A\angle38.29\degrees)(10 +j5 + j10 -j4 + 8j) = 8.87\angle100.53$$

$$V_{AB} = V_{AB}^1 + V_{AB}^2 = 10.33V\angle17.76$$

I can't see where I'm wrong...

 

[The Electrician]

Go have a look in post #2 where I said:

"Your first calculation should be:"

You're missing a -j4 term in the denominator of that V/Zt calculation.

 

[The Electrician]

$$I = \frac{V}{Z_T} = \frac {20} {5+10+j5+j10+ (2*j4)} = 0.728A \angle-56.89\degrees$$

$$\therefore V_{AB}^1 = (0.728A\angle-56.89\degrees) * (5+10+j5+j4) = 12.74V\angle-25.92\degrees$$


$$2A\angle90\degrees \frac {(10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.4131A\angle38.29\degrees$$

$$V_{AB}^2 = (0.4131A\angle38.29\degrees)(10 +j5 + j10 -j4 + 8j) = 8.87\angle100.53$$

$$V_{AB} = V_{AB}^1 + V_{AB}^2 = 10.33V\angle17.76$$

I can't see where I'm wrong...

I see another error. You should have:

$$V_{AB}^2 = (0.4131A\angle38.29\degrees)(j10 -j4 + 4j) = 4.131\angle128.29$$

 

[Lunat1c]

You're right. Thanks a lot, I really appreciate you taking the time to go through my working.

 

[The Electrician]

You can also solve the system by writing a couple of loop equations. Replace the j2 current source in parallel with the 5 ohm resistor with a j10 voltage source in series with a 5 ohm resistor.

Consider two loop currents, I1 and I2. I1 is a clockwise loop passing through the j10 voltage source, the 5 ohm and 10 ohm resistors, the j5 inductor, and the short across the a-b terminals. I2 is a clockwise loop going around the entire circuit.

Then to solve for the two currents, solve this system. I1 will be the short circuit current through the shorted a-b terminals.

$$\left[ \begin{array}{2}15+j5 & 15+j9\\15+j9 & 15+j19\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}j10\\-20+j10\\\end{array}\right]$$

To find Zth, the Thevenin impedance, short the 20 volt source and the j10 source; connect a 1 volt source to the a-b terminals. The system then becomes:

$$\left[ \begin{array}{2}15+j5 & 15+j9\\15+j9 & 15+j19\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}1\\0\\\end{array}\right]$$

The solution to this system will give a current I1 whose reciprocal is numerically equal to the Thevenin impedance.

Multiply the current I1 from the first system solution, by the Thevenin impedance, which is 1/I1 from the second system, and you will have Vth, the Thevenin voltage.

This is how I solved your problem initially, and then examined your method to see where your errors were.

Edit:
It looks like the tex engine didn't properly format the 2x2 matrix on the left of the two systems above. Where you see a 4x1 matrix, convert it to a 2x2 by taking the terms row wise, two at a time.