10-20-2012, 09:30 AM | #1 |
FFR Veteran
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Circuit Problem "FINALE"
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 Last edited by ~Zero~; 10-21-2012 at 08:00 AM.. Reason: New Problem |
10-20-2012, 04:06 PM | #2 |
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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. |
10-20-2012, 04:38 PM | #3 |
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Re: Circuit Problem "2"
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? Last edited by ~Zero~; 10-20-2012 at 04:49 PM.. |
10-20-2012, 04:43 PM | #4 |
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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) |
10-20-2012, 04:48 PM | #5 | |
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Re: Circuit Problem "2"
Quote:
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10-20-2012, 05:49 PM | #6 |
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Re: Circuit Problem "2"
Note: everything is rounded weird and I'm not even sure if I did this right lmao
Someone please check if this is right hehu |
10-20-2012, 06:12 PM | #7 |
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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
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10-21-2012, 07:39 AM | #8 |
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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 Last edited by ~Zero~; 10-21-2012 at 07:42 AM.. |
10-22-2012, 05:59 AM | #9 |
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Age: 33
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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 Last edited by LongGone; 10-23-2012 at 03:27 PM.. |
10-23-2012, 03:09 PM | #10 |
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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 |
10-23-2012, 03:26 PM | #11 |
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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 |
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