# thevenin's Voc

Discussion in 'Homework Help' started by stupid, Oct 26, 2009.

1. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
hi,
this is a slight modification to one of the threads posted previously with a 2Ω added as attached.
is Voc=4/3 Vx?

thanks

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2. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
stupid,

You honestly can't expect the more technical guys to answer all your homework problems. You have to think this all through and spend some time and effort before resorting to these forums.

Austin

3. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
hi electronerd,
i have thought thru the circuit with respect to the original one.
i notice the dependent current source is the same value at all its sides irrespective of the components in that mesh loop.

i want to make sure if my assumption is correct.
by that logic Voc = 2 x (2/3)Vx
is that correct?

i m told by hitmen my mesh loop is wrong as quoted:
Hi!
Your mesh 2 is wrong! : for mesh i2, anticlockwise dir
5(i2+i1)=0
5(i2) + 5(i1) =0------------eq2
wrong wrong wrong!

thanks
stupid

Last edited: Oct 26, 2009
4. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Have a look at the system I gave in post #13 of the other thread:

Code ( (Unknown Language)):
1. [  8 -5  0 ]   [ I1 ]   [ 1 ]
2. [  2  1 -1 ] * [ I2 ] = [ 0 ]
3. [ -5  5  15]   [ I3 ]   [ 0 ]
To compute Voc, we know that I3 is zero, so we can delete I3 from this system and get a new system for the case where the output is open circuited. You can see that the second equation is 2*I1 + I2 = 0. This is the same as hitmen gave you in post #17 of that other thread, with a sign difference because I've taken all the currents to be clockwise:

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

I1 = 1/18
I2 = -1/9

Voc will be the voltage across the 5Ω resistor (because the 15Ω resistor has no effect if the output is open circuited). To calculate the voltage across the 5Ω resistor, we need to know the current through it. Since I have taken all the currents to be clockwise, the current in the 5Ω resistor is just I1 - I2 = 1/18 - (-1/9) = 1/6. Multiplying by 5, we get Voc = (1/6)*5 = 5/6 volts.

With the extra 2Ω resistor you're proposing, you just have to add the drop across that resistor to the 5/6 volts across the 5Ω resistor, since the presence of the 2Ω resistor doesn't change I2. You should get Voc = 5/6 + 2/9 = 19/18 volts.

To calculate more at once, modify the larger system I gave to account for a load resistor across the a-b terminals of value R. Then you will get a new system:

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

I1 = (R+20)/(18*R + 285)
I2 = -(2*R + 25)/(18*R + 285)
I3 = 5/(6*R + 95)

The total resistance to the right of the dependent source is R+15. Let's multiply I3 (because I3 passes through the series combination R+15, our "load" to the right of the dependent source) by this value; we get:

5/(6*R + 95)*(R + 15) = (5*R + 75)/(6*R + 95)

This is the voltage across the dependent source, and across the 5Ω resistor, and across the series combination R+15.

Examining this expression, we can see that Voc is given by the value of the expression as R -> ∞, which is 5/6 volts (remember that the 15Ω resistor has no effect when the output is open circuited). But, we can also see that if the output is shorted across the terminals a-b, the voltage across the 15Ω resistor is given by the expression when R = 0; that value is 75/95 volts. From this we can calculate Isc as (75/95)/15 = 1/19 amps.

With your extra 2Ω resistor, the system becomes:

Code ( (Unknown Language)):
1. [  8 -5   0 ]   [ I1 ]   [ 1 ]
2. [  2  1  -1 ] * [ I2 ] = [ 0 ]
3. [ -5  7 R+15]   [ I3 ]   [ 0 ]
Solving, we get:

I1 = (R + 22)/(18*R + 301)
I2 = -(2*R + 25)/(18*R + 301)
I3 = 19/(18*R + 301)

Again, multiplying I3 by (R + 15), we get an expression for the voltage across the dependent source (and across the series combination R+15, our load). We get:

(19*R + 285)/(18*R + 301)

We can see that in this case Voc (that's when R -> ∞) is 19/18 volts.

When R = 0, then we can get Isc as (285/301)/15 = 19/301 amps

Last edited: Oct 27, 2009
5. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
hi electronerd,
i m wondering is if my concept is correct.

i m unfamiliar about yur solution by shorting the terminal to find Voc. I would leave that aside.

coming back, could u verify if following methods r correct.

by mesh analysis, assuming mesh loops begin from left,
at i2 loop, 5(i2-i1)+2i2=0

by node analysis, again assuming V1 at left node & V2 at right node,
at V1 node, (1-V1)/3 + V1 + (V1-V2)/2 = 0

at V2 node, (V2-V1)/2 = (2/3)Vx

assuming also, 1-Vx=V1

thank u.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
In this thread, electronerd has not offered any advice about "...shorting the terminal to find Voc." Are you confusing me with electronerd?

Furthermore, I didn't say to short the terminal to find Voc. Shorting the output terminals was done to find Isc.

You are not including the effect of the dependent current source. Whenever there is a current source in a mesh, it dominates the mesh current; it is the mesh current. The mesh current is just equal to the value of the current source. Expressions like 5(i2-i1)+2i2 don't apply.

You can obtain an expression for I2 like this:

I2 = -(2/3)*Vx
I2 = -(2/3)*(3*I1)
I2 = -2*I1

I'm assuming all mesh currents to be clockwise.

The first equation should be:

at V1 node, (V1-1)/3 + V1/5 + (V1-V2)/2 = 0

7. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Indeed, The Electrician is a far better teacher when It comes to this stuff; I haven't done much of this.

8. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
my apology to the electrician & electronerd for my misidentification.
i got it mix up jumping around the threads.

by using node analysis,
i got V2=19/13
v1=-5/13
Vx=18/13
i2=(2/3)Vx = 12/13

is Voc=V2?
if so Voc=19/13

now compare above with mesh analysis,
if the 2Ω has no impact on the loop 2 current, ie i2=-(2/3)Vx

at mesh V1,
-1+3i1+5(i1-i2)=0

i1=1/18
i2=-1/9
vx=5/6

we know i2=(v2-v1)/2
i2=12/13

how come node's Vx is dirrent from the mesh's Vx?

now from mesh current loop analysis,
how to compute Voc?

thanks & regards,
stupid

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
That's not what I get. I get:

V1 = 5/6
V2 = 19/18

Please show the two equations in full that you're using for the node analysis.

That's not correct for Vx. Vx is the voltage across the 3Ω resistor. Since I1 = 1/18 the voltage across the 3Ω resistor is 3*(1/18) = 1/6 volts.

Recheck the rest of your calculations.

10. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
hi the electrician,
sorry as i have been away on a holiday.

my node analysis as below:
at node v1,
(v1-1)/3 + v1/5 + (v1-v2)/2=0
31v1 - 15v2 = 10 -----------eq1

at node v2,
(v2-v1)/2 = (2vx)/3
(v2-v1)/2 = 2/3 *(1-v1)
3v2 +v1 =4---------------eq2

solving both eqs,
v2 = 19/13

our denominator is different.
where did i go wrong?

btw,
is Voc = v2?

thank u.
stupid

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
I don't know where you went wrong. When I solve the two equations, I get:

V1 = 5/6
V2 = 19/18

Perhaps you should show your steps in solving the two equations. You must have made a mistake in your algebra or your arithmetic.

Yes, Voc = V2, as long as the 15Ω resistor's right end is floating.

12. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
oh..i indeed erred on arithmetic...
problem solved...thanks the electrician.

regards,
stupid