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#### bertus

Joined Apr 5, 2008
22,121
Hello,

We are not a homework service:

Important

The Homework Help Forum is not a free homework service; we are here to help your understanding, but fully expect the users of this forum to dictate the course of their own learning.

When posting a request for homework/coursework/assignment help, you must provide details of your attempts at the questions. Any thread that just posts up a copy of the questions without any attempts on the part of the opening poster will be directed to this thread and will be given 48 hours to satisfy the requirements detailed within. Help us, help you.
Show us what you have reached and we will give hints.

Bertus

Thread Starter

#### jarednoob

Joined Apr 27, 2010
5
So far i got to whats written on the page. Im having trouble finding the thevenin voltage because im used to dealing with 1 voltage source not 2

#### Pym

Joined Jul 24, 2009
1
For Part 1 There are at least 2 ways to do this.
Probably the best way is to take Va R1 and R2 and make a first Thévenin equivalent circuit. Then you connect this equivalent circuit to R3 and Vb. You can determine the current which flows in this new circuit, and then using ohm's law calculate the voltage of (a) with respect to (b)

Another way
Lets call the current which passes through R1 I1, the current which passes through R3 I2.
So the current which passes through R2 will be I1 + I2
Now you have to write 2 equations:
The first for the current in the loop with R1 and R2
And the second for the current in the loop with R3 and R2
The first equation will be
15=I1*R1 + (I1+I2)*R2
This needs to be rearranged
15=I1*(R1+R2)+I2*R2
I will leave you to determine the second equation and rearrange it
So, 2 equations with 2 unknowns to be solved, which will give you I1 and I2
Using ohm's law will allow you to calculate the Thévenin voltage (a) to (b)
I suggest that you use both methods to check.

BTW, You need to recalculate the Thévenin equivalent resistance
Good luck !

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#### retched

Joined Dec 5, 2009
5,208
I cant believe you didn't at least remove the school name from the assignment.

What if your professor sees this? Or another student? Or a real student that does his own work and stayed awake for 2 days to solve this, happens to e-mail a link to your school?

Jeesh. You should re-think your methods.

After post #4, are you still having trouble? What are you stuck on?

#### JoeJester

Joined Apr 26, 2005
4,390
$5 ... are you kidding me. What does$5 buy anymore?

I'd forward this link to the Chair at your university and see if they would accept payment for the services you wanted.

Thread Starter

#### jarednoob

Joined Apr 27, 2010
5
Ok this is what i have so far. my question is, my question is finding the voltage across a-b. I found the current to be 0.036A so do I do V=(0.036A)(40) to find the voltage across A and V=(0.036A)(100) for B and just add them?

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#### kingdano

Joined Apr 14, 2010
377
$5 ... are you kidding me. What does$5 buy anymore?

I'd forward this link to the Chair at your university and see if they would accept payment for the services you wanted.

lol

got em!

seriously kid, no one is going to do your homework for you

just because we are engineers doesnt mean we want to do thevinin and norton circuits all day long.

do your own homework - or go ask your prof for help - THATS WHAT YOU PAY THEM FOR BUDDY!

/rant

and if you do want to bribe someone, start higher - $50USD minimum. Time is money - and my time is worth more than$5.

#### t06afre

Joined May 11, 2009
5,934
Asking your prof my be a good solution, and more money for beer Just never say "I do not understand it" Just ask like "According to what I understand, the problem can be solved by doing this and this" And then "Can you please confirm if this is correct" This will show that you at least have tried.

#### blueroomelectronics

Joined Jul 22, 2007
1,757
And a University student nonetheless. Sheesh.

Thread Starter

#### jarednoob

Joined Apr 27, 2010
5
well my EE prof is the dean so he doesnt have office hours like a regular professor.

Anyways, i posted a question, along with an img with how far I had gotten, so if you could help with me with that i would appreciate it

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#### t06afre

Joined May 11, 2009
5,934
Thread Starter

#### jarednoob

Joined Apr 27, 2010
5
Ok let me see if i understand this...Is #1 asking for the equivalent circuit thats to the left of a-b?? If thats the case this circuit is cake

#### t06afre

Joined May 11, 2009
5,934
No Vth is the same as you put a voltmeter between a and b. Rth is the resistance you measure then all voltage sources are shorted and current sources are left open

#### JoeJester

Joined Apr 26, 2005
4,390
Jarod,

Visit http://www.allaboutcircuits.com/vol_1/chpt_10/7.html and see how they analyzed a similiar circuit.

Then attempt your circuit and come back here with your work. We will assist you, but not do your work. We will guide you.

The assignment is the prelude to a lab on the same topic. It's important you know what to expect when your doing the lab and then identify the differences between the theory and the measurements, if any.

Vth is the open circuit voltage at the node where the 200 ohm resistor is connected, only the 200 ohm resistor is removed.

You can figure it out using superposition or any of the theorums of your choice, just let us know what one you ended up using.

#### Norfindel

Joined Mar 6, 2008
326
Why would him remove the 200 ohm resistor? I think it's not the load, just part of the circuit. The load would be at points A-B.

#### JoeJester

Joined Apr 26, 2005
4,390
Why would him remove the 200 ohm resistor? I think it's not the load, just part of the circuit. The load would be at points A-B.
That could be true. Since the OP is at a loss, doing the one would gain some confidence and it's not a great leap to include the 200 ohm resistor.

The lab assignment never mentioned a load at points A-B, but it will be discussed if the OP comes back. I'm sure you noticed his attempt at the solution.

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#### salmanshaheen_88

Joined Mar 5, 2009
88
hey jarednoob send me an email at <snip> for your complete solution

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