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Circuits - Variables dependent on R? |
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| Feb5-13, 02:35 PM | #1 |
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Circuits - Variables dependent on R?
The circuit of interest is attached.
The question is to solve for I1, I2, I3, I4, and V_diode. My question is will the variables come out to be numbers or dependent on R? This is how I did it and it depends on R: I3=0 b/c of the open. The branch with the open is pretty much not there. Thus, KCL: I1+I2=I4 KVL gives 2 equations: 1-2R*I1+R*I2=0 -R*I2-R*I4+1=0 So I guess I1, I2, I4, and V_diode will not solve to be numbers and will depend on R? Or did I do this all wrong? |
| Feb5-13, 03:06 PM | #2 |
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How is the diode hooked up? Cathode to + or the other way?
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| Feb5-13, 03:16 PM | #3 |
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| Feb5-13, 04:46 PM | #4 |
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Circuits - Variables dependent on R? |
| Feb5-13, 06:08 PM | #5 |
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| Feb5-13, 06:24 PM | #6 |
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| Feb5-13, 06:40 PM | #7 |
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I don't see why it can't ask for I3.
I_3 is simply 0. That's the answer for I3, only 1 of the 5 variables to be solved. I still need to solve for I1, I2, I4, v_diode. |
| Feb5-13, 07:29 PM | #8 |
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OK, so the first thing is to simplify the circuit by removing irrelevant components.
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| Feb5-13, 07:31 PM | #9 |
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Do you know the superposition theorem?
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| Feb5-13, 08:09 PM | #10 |
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Wouldn't superposition give me the same answer as my original post? I attached a new drawing with the resistor removed. I kept the V_diode in there though. |
| Feb5-13, 08:35 PM | #11 |
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Good first step. Now, looking at the original schematic, can you identify one of your new nodes with Vdiode(-)? Let us call the node Vdiode(+) = zero volts, shall we.
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| Feb5-13, 09:19 PM | #12 |
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The node would be the wire between the 3 resistors. But I still think you answer will not solve to be a number and will still depend on R. |
| Feb5-13, 10:41 PM | #13 |
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| Feb5-13, 10:48 PM | #14 |
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i.e Vdiode maybe something like Vdiode=2R instead of something like 10V |
| Feb6-13, 04:21 AM | #15 |
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Recognitions:
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The equations for the currents will all involve variable R.
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| Feb6-13, 10:15 AM | #16 |
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What level are you at in your studies? |
| Feb6-13, 09:23 PM | #17 |
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3rd year undergrad. Why?
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