Circuit Problem "2"

[~Zero~]

So this next problem I got the answer of 2.5 ohms by using online sources and such, but I am still not understanding the process that easily. If someone could explain it to me that would be great.

[SOLVED] Problem



[SOLVED] New Problem



New Problem

http://imgur.com/NtLuq

[Evnoir]

Re: Circuit Problem "2"

If there are no dependent sources, you can just kill off all of the sources (i.e. short circuit the voltage sources and open circuit the current sources) and solve the equivalent resistance using whatever method you choose. I use MNA as this method works every single time in a circuit like this.



Now that you have this circuit, you can see there are 2 ways to get from A to B; I labelled their paths I1 and I2 (you don't have to). Since the (3+2) ohms path is in parallel with the 5 ohms path, the Thevenin Resistance is (3+2)//(5) = 2.5 ohms.

I hope this helped.

[~Zero~]

Re: Circuit Problem "2"

Circuit Stuff

Thank you this helped a lot, I will be posting another problem that I could use some help with (obviously, i know nothing about circuits lol)

Here is the problem


and I believe the first step was to remove one of the sources
and then I am suppose to calculate the equivalent resistance? Not sure how to go about the next step.
Would 3 and 2 ohm resistors be in series and parallel to the 4 and the other 2 ohm resistor?

[Evnoir]

Re: Circuit Problem "2"

You can't really solve that circuit without knowing the polarities of the voltage sources.

Superposition is the method of keeping just one source in the circuit and removing all of the others, doing that for every single source (i.e. a circuit with a voltage source and a current source will need two circuits: one with just the voltage source and one with just the current source)

[~Zero~]

Re: Circuit Problem "2"

You can't really solve that circuit without knowing the polarities of the voltage sources.

Superposition is the method of keeping just one source in the circuit and removing all of the others, doing that for every single source (i.e. a circuit with a voltage source and a current source will need two circuits: one with just the voltage source and one with just the current source)

Oh okay one sec - gonna be fixed right now

[Evnoir]

Re: Circuit Problem "2"

Note: everything is rounded weird and I'm not even sure if I did this right lmao

Resistance (going from right to left):
2Ω // wire = 0Ω (Note: the 2 ohms on the far right is ignored since the wire that it is in parallel with is shorting it. Current will tend to flow through the path of least resistance)
3Ω + 0Ω = 3Ω ...[*]
3Ω // 2Ω = 1.2Ω
1.2Ω + 4Ω = 5.2Ω

Current of whole circuit:
V(total) / R (total) = I (total)
5V / 5.2Ω = 0.96A

Current through the 2Ω we need for Vo:
I (total) x (Current divider) = Current through 2Ω
**Current divider = The other resistor you are in parallel with / The sum of the resistors in parallel =[*] / ([*] + 2Ω)**
0.96A x (3 / 5) = 0.58A

Vo:
0.58A x 2Ω = 1.15V ... [1]

Resistance (going from left to right):
4Ω // 2Ω = 1.333...Ω
1.333...Ω + 3Ω = 4.333...Ω ...[*]
4.333...Ω // 2Ω = 1.37Ω

Current of whole circuit:
10V / 1.37Ω = 7.31A

Current through the 2 ohms we need for Vo:
7.31A x (2Ω / ([*] + 2Ω)) = 2.31A (Current going to the left side of the voltage source)
2.31A x (4Ω / (2Ω + 4Ω)) = 1.54A

Vo:
1.54A x 2Ω = 3.08V ...[2]

Vo (total) = [1] + [2]
Vo = 1.15V - 3.08V = -1.93V
**The second circuit has the current going from (-) to (+) through the 2Ω resistor**
In other words, the voltage is higher on the bottom of that 2Ω resistor than the top of it.

Someone please check if this is right hehu

[~Zero~]

Re: Circuit Problem "2"

I appreciate all your help evnoir, this will help be able to solve similar problems that require these methods. I will be sure to look over them again and make sure I understand everything. Thanks again

[~Zero~]

Re: Circuit Problem "2"

bumping: You guys taught me about a lot of things and I am very grateful for you guys taking the time to show me how its done. And these last problems involve another new concept and i need help once again

Ugh I took a picture from the book and its too huge, so here is an link to it

http://imgur.com/NtLuq

[LongGone]

Re: Circuit Problem "2"

Do you have the answers to doublecheck? I might be wrong at this but..

For questions like these, important things to remember about an ideal op amp
-Infinite gain
-Infinite input resistance, thus no current flows into the op amp inputs
-Zero output resistance

Since the Op Amp is ideal, Vout=12V, the limit of the linear range of the power rails supplied (+-12V). An Ideal Op Amp has infinite gain; thus any difference in values between the two inputs will cause an output. This value is usually taken to be positive (the output is denoted as a + in this example)



Now we have to find Vi. Since an ideal op amp has infinite input resistance, we can write a simple equation to find Vi using Kirchoff's Current Law(KCL)

(12-Vi)/Rf = (Vi-2)/Rin
Vin = 4V

With that found, we can now find the output current, i(out). Again, by KCL, lets look at the output

i(out) = i1+i2 = (4-12)/100000 + (0-12)/20000 = -0.68mA, i.e. current of 0.68mA flows out of the Op Amp

[~Zero~]

Re: Circuit Problem "2"

My professor gave us the answers today to help us out, but not steps on how to do them

Envioir Superposition was correct and so was his Thenevin equation, except we forgot to calculate for Vo

Long Gone Voltage node method was correct, but not the OP amp one

Answers for OP amp
Vo = 7V
Io = -0.39 mA

and he gave us this hint

For an ideal op amp, there is no current flow into the amp at the inputs and the input potentials have to be equal. Therefore there is no voltage drop across the 5 ohm resistor connected to the non-inverting input, so vp = vn = 3V.

Anyone care to take another shot at that xD

[LongGone]

Re: Circuit Problem "2"

Ok ignore my previous post then; my bad lol. Output will not be 12V because of the feedback

Since vp = vn = 3V,



We can find Vo by simply looking at the vn terminal, using KCL

(3-2)/Rin = (Vo-3)/Rf

Vo=7V

Again KCL at the output,

i(out) = i1+i2 = (3-7)/100000 + (0-7)/20000 = -0.39mA, i.e. current of 0.39mA flows out of the Op Amp