# Can't find Rth or I short circuit

Discussion in 'Homework Help' started by canuck13, Oct 18, 2009.

1. ### canuck13 Thread Starter New Member

Oct 18, 2009
4
0
I'm having trouble finding the I short circuit for this circuit. I've tried doing mesh current and got 1/16A for Is.c. but the answer says its 1/19A. I solved this for V(open circuit) the answer is 5/6v.

Note: i1 is the first mesh in clockwise direction, i2 is second mesh in clockwise direction. We know that i2 = Is.c. (or so I think). ( also I see no current going through the 5ohm resistor so therefore it acts like it isn't there possibly my error).

For the mesh current I am doing this:

-1 + Vx + 15(i2)
Vx = 3(i1)
(2/3)Vx = i2 - i1

Can you please point me in the right direction?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
218
Keep in mind that 2Vx/3 is a voltage-controlled current-source.

When you place a short across terminals A and B, you then end up with the 15 ohm resistor in parallel with the 5 ohm resistor. The 5 ohm resistor does have current in it.

hgmjr

3. ### canuck13 Thread Starter New Member

Oct 18, 2009
4
0
Alright so if I put those two in parallel and get a resistor of 15/4 ohms. Should I then do a mesh equation for the two loops?

4. ### hgmjr Moderator

Jan 28, 2005
9,030
218
If you instructor did not specifically dictate which circuit analysis method you were to use on this problem then I would say that you can use whichever method is the most comfortable to you.

hgmjr

5. ### canuck13 Thread Starter New Member

Oct 18, 2009
4
0
alright so I did two meshes and got the equations

-1 + 3(i1) + 15/4(i1 - i2) = 0

and 15/4(i2 - i1) = -2(Vx)/3

Vx = 3(i1)

Then solved it just to get i2 = 63/100 then tried to do current division getting me 63/400 which is obviously not the right answer. Gah....

6. ### Ratch New Member

Mar 20, 2007
1,068
4
canuck13,

All right. I would do it by the node method, but if you insist on doing it by the mesh method, then OK. You can combine mesh 2 and 3 into a supermesh. Here's how. The dependent current source I = I3-I2 = 2*Vx/3 = (2/3)(-I1*3) =- 2*I1 =====> I3-I2 = -2*I1

then the voltage around mesh 1 is (I1-I2)5+3*I1-1=0
and finally the voltage around the combined mesh 2 and mesh3 is (I2-I1)5+15*I3=0

solving gives I1 = 0.07017543857 ,I2 = -0.08771929825, I3 = 0.05263157895 = 1/19

I3 is the short circuit current you are looking to find.

Ratch

Last edited: Oct 20, 2009
7. ### hgmjr Moderator

Jan 28, 2005
9,030
218
HINT: I2 = 2Vx/3 and Vx = 3*I1
This relationship permits you to find I2 as a function of I1. That will allow you to replace I2 in your first equation with its I1 equivalent.

Then you can solve for I1.

hgmjr

8. ### canuck13 Thread Starter New Member

Oct 18, 2009
4
0
okay so now I have

i2 = 2Vx/3
Vx = 3(i1)
Therfore i2 = 2(i1)

the third equation is -1 + Vx + 15/4(i2)

substituting and solving I get i1 = 2/21
therfore i2 = 4/21

then using current division (5/20 = 1/4
1/4*4/21 = 1/21A

Which is close but not correct. Anything I miss?

9. ### hgmjr Moderator

Jan 28, 2005
9,030
218
Just so that I am clear on your assumptions:

Can you tell me if your assumption of current direction in the left loop (i1) is clockwise or counter-clockwise? Also what is your assumption of the direction of current in the right loop?

hgmjr

10. ### hgmjr Moderator

Jan 28, 2005
9,030
218
I just figured out why things are not working out.

I think you are assuming that the direction of i1 is counter-clockwise. That means that Vx = -3*i1 instead of 3*i1.

Try rerunning your earlier calculation with Vx = - 3*i1. I think that will get you the right answer.

hgmjr.

11. ### stupid Active Member

Oct 18, 2009
81
0
hi ratch,
how does the combined mesh 2 & 3 formulate?

using the logic above, could mesh 1,2 & 3 combine as follow?
-1+3i*1+5(i1-i2) +5(i2-i1) +15*i3=0

thanks...

12. ### Ratch New Member

Mar 20, 2007
1,068
4
stupid,

The are 3 meshes in this circuit. Mesh 2 and mesh 3 contain a common current source, so those two are candidates for a supermesh calculation. A loop calculation for mesh 1 is made, and a second calculation for meshes 2 and 3 are made. Mesh 1 cannot be folded into meshes 2 and 3 because it does not share a common current source. It is important that you understand the concept and the equations involved. If you do not, ask for further clarification.

Ratch

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,548
430
It's possible to systematize the procedure:

First create an impedance matrix, assuming that there are 3 mesh currents, I1, I2 and I3, left to right, each clockwise. Symbolically:

Code ( (Unknown Language)):
1. [ Z11 Z12 Z13 ]   [ I1 ]   [ V1 ]
2. [ Z21 Z22 Z23 ] * [ I2 ] = [ V2 ]
3. [ Z31 Z32 Z33 ]   [ I3 ]   [ V3 ]
The diagonal terms, Znn are the total impedances around each loop and V1, V2 and V3 are the voltage sources in each loop. The off-diagonal terms, Zjk, are the impedances common to loop j and loop k (taking into account the relative directions of the currents in loop j and loop k). The column vector on the right side of the equals sign contains the voltage sources in each loop.

Current sources are dealt with as constraint equations.

When there are supermeshes, I first create the impedance matrix with the supermesh current source replaced with a wire.

Using the numerical values in the given problem (terminals a and b shorted together), and with the dependent source replaced with a wire, we have the system:

Code ( (Unknown Language)):
1. [  8 -5  0 ]   [ I1 ]   [ 1 ]
2. [ -5  5  0 ] * [ I2 ] = [ 0 ]
3. [  0  0  15]   [ I3 ]   [ 0 ]
Next, we have to modify it to account for the supermesh current source. Leave the first row alone, since the left hand mesh is not part of the supermesh. Replace the second row with an equation defining the constraint on I2 and I3 imposed by the dependent source. That equation is I2-I3=-(2/3)*(I1*3).

Replace the third row by the sum of the 2nd and 3rd rows. These are the rows corresponding to the meshes that make up the supermesh

When we make these modifications, we get the system:

Code ( (Unknown Language)):
1. [  8 -5  0 ]   [ I1 ]   [ 1 ]
2. [  2  1 -1 ] * [ I2 } = [ 0 ]
3. [ -5  5  15]   [ I3 ]   [ 0 ]
Solving this system, we get:

I1 = 4/57
I2 = -5/57
I3 = 1/19

Now, as a bonus, having gone to the trouble to create the full impedance matrix, we can obtain Rth by simply inverting the matrix (the result is an admittance matrix, mostly; the second row has non-standard meaning); we get:

Code ( (Unknown Language)):
1. [  4/57  5/19 1/57  ]
2. [ -5/57  8/19 8/285 ]
3. [  1/19 -1/19 6/95  ]
Rth is the reciprocal of the 3,3 element of the inverse. Rth = 95/6 Ω.

And, of course, Vth = Isc * Rth = I3 * Rth = 1/19 * 95/6 = 5/6

Last edited: Oct 27, 2009
14. ### hgmjr Moderator

Jan 28, 2005
9,030
218
Millman's Theorem Solution.
hgmjr

15. ### stupid Active Member

Oct 18, 2009
81
0
hi,
i tried to compute Voc by mesh theory but cant obtain 5/6V, a known answer.

for mesh i1, clockwise dir
-1+3(i1)+5(i1-i2)=0
8(i1)-5(i2)=1 ----------eq1

for mesh i2, anticlockwise dir
5(i2+i1)=0
5(i2) + 5(i1) =0------------eq2

solve with eq1 with eq2,
i1=1/13
i2 also 1/13

current thru 5Ω=1/13 + 1/13 = 2/13
Voc at 5Ω = 2/13 * 5 =10/13 which is incorrect...

i have also tried by assuming i2=(2/3)Vx but in vain too.

pls help..thanks

stupid

16. ### hitmen Active Member

Sep 21, 2008
159
0
Here is the equation using nodal analysis:

(V1 - 1)/3 + V1 / 5 - 2/3 Vx + V1 / 15 = 0 -- eqn 1
Vx = 1 - V1 --------- eqn 2

solve to get V1 = 15/19
I sc = 1/19 A
Clue: multipy everything by 15

...
substep
9V1 - 5 -10 +10V1 = 0

17. ### hitmen Active Member

Sep 21, 2008
159
0
Hi!
Your mesh 2 is wrong! : for mesh i2, anticlockwise dir
5(i2+i1)=0
5(i2) + 5(i1) =0------------eq2
wrong wrong wrong!

Equation 2 is I2 = -two-thirdVx
since Vx = 3I1
I2 = -2I1 ..... this is equation 2 !

I am taking this this sem

18. ### stupid Active Member

Oct 18, 2009
81
0
thank u hitmen for helping to enlighten me.
i m amazed how simple the work out & how it escapes me.

btw, whats the problem with my mesh i2?

regards,
stupid

19. ### sabinoli New Member

Oct 28, 2009
1
0
i couldnot clearly understand numerical value in the circuit.
if u made it clear , i guess i could help u with exaplanation.
please paste the same question as given in exercise