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

thanks in advance

 

i just need some tips come on guys

 

For first question, you can take one voltage source at a time and short circuit other one. Analyse and add the voltage.
For second question you can use this thread: http://forum.allaboutcircuits.com/t...rk-with-dependant-sources.101166/#post-761586

 

Hi
i am a student of computer engineering but my college so insisted to teach me physics

Well guess what?

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

 

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
http://forum.allaboutcircuits.com/threads/thevenins-theorem.102247/#post-771380

 

Thank you so much ! you are a life saver

 

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
http://forum.allaboutcircuits.com/threads/thevenins-theorem.102247/#post-771380

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?

 

View attachment 76439
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?

Nothing is wrong, Isc is equal to 1/20 = 50mA
So tell me why you think that your analysis is wrong ?

 

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

 

Nothing is wrong, Isc is equal to 1/20 = 50mA
So tell me why you think that your analysis is wrong ?

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.
Thanks in advance.

 

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

Please check your math again because the correct answer for V is 100V
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

You got I =2.5mA that's why

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.

Moreover I did not understand how did you use superposition here.

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)
Also please read this
http://forum.allaboutcircuits.com/t...rrent-and-voltage-sources.101488/#post-764893

 

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.

 

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.

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.

 

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.

Thank you, WBahn for helping me out.