# help for my assingment ?

Discussion in 'Homework Help' started by ubeyd, Nov 30, 2014.

1. ### ubeyd Thread Starter New Member

Nov 30, 2014
4
0
Hi
i am a student of computer engineering but my college so insisted to teach me physics
so i have an assignment..i think questions are pretty easy but i am stuck at few points

so at first question i know how to find the voltage drops if there is only voltage source but there are 2 in this question so any tips ?
and for the second part, i am familliar with KVL but again 2 voltage source so i am not sure how it effects the answer.

about the second question i take any advice cuz i am not familliar with Thevenin equivalent yet

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2. ### ubeyd Thread Starter New Member

Nov 30, 2014
4
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i just need some tips come on guys

Aug 9, 2014
58
6
4. ### MrChips Moderator

Oct 2, 2009
12,447
3,363
Well guess what?

You cannot become a competent engineer if you do not know physics and math.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Q1 - simply try applying KVL around the loop you have in the circuit.
Q2 - use a circuit theory and find Vab voltage, and Vab = Vth, next short A and B together and solve for Isc.
And finally Rth = Vth/Isc.
For example we could find Vth by using the superposition for I current.
I = 10V/(2000 + 200) - 9*I * 200/(2000 +200)
I = 10/2.2 - 9*I*0.2/2.2
2.2*I = 10 - 9*I*0.2
2.2*I = 10 - 1.8*I
4*I = 10
I = 10/4
I = 2.5 and this result is in mA
I = 2.5mA

So Vab = Vth = 10V - 2.5mA*2000 = 10V - 5V = 5V

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Last edited: Nov 30, 2014
6. ### ubeyd Thread Starter New Member

Nov 30, 2014
4
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Thank you so much ! you are a life saver

7. ### dalam Member

Aug 9, 2014
58
6

In the circuit that you uploaded if I take Vb as a reference node then Va becomes zero.
I1=0. And I=10/2000. Isc= 10*I=1/20.
What is wrong in this analysis?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Nothing is wrong, Isc is equal to 1/20 = 50mA
So tell me why you think that your analysis is wrong ?

9. ### dalam Member

Aug 9, 2014
58
6

V/2000+V/200=9I+1
I=-V/2000
So, V/2000+10V/2000=-90V/2000+1
101V/2000=1
V=2000/101
V/I(test source=1A)=Rth=2000/101

10. ### dalam Member

Aug 9, 2014
58
6
You got I =2.5mA that's why. Moreover I did not understand how did you use superposition here.
Can you please testify the answer that I posted in the post above and the approach as well.

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
And Rth = 100V/1A = 100Ohm
http://www.wolframalpha.com/input/?i=V/2000 +V/200=-9V/2000 +1

I got the same result using different approach.
Vth = 5V and Isc = 50mA so Rth= 5V/50mA = 100 Ohm

Where I wrote that Isc is 2.5mA ??
In my post 5 I solve for Vth, but I use superposition to find I current first.

What you do not understand. I simply use a superposition to find "I" current
First I remove current controlled current source from the circuit.
So I' = 10V/(2000 + 200)
Next I remove 10V voltage source and use a current divide rule.
I'' = - 9*I * 200/(2000 +200)
and finally
I = I' + I'' = 10V/(2000 + 200) - 9*I * 200/(2000 +200)

12. ### dalam Member

Aug 9, 2014
58
6
Thanks a lot. My bad, I wrote -90V/2000 instead of -9V/2000. And thanks for explaining about superposition. I was not sure weather dependent sources can also be analysed using superposition.

13. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
Yes they can. As long as the circuit is linear, superposition applies (in fact, that is the definition of a linear system).

Imagine that you have a current I0 that is a function of three different independent sources:

I0 = I1 + I2 + I3

Now imagine you have a dependent source that is

Ix = kI0

Well, that works out to

Ix = k(I1 + I2 + I3) = kI1 + kI2 + kI3

and so we see that it is the sum of the currents that are output by the dependent source in response to the three independent sources individually.

14. ### dalam Member

Aug 9, 2014
58
6
Thank you, WBahn for helping me out.