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Circuits  Variables dependent on R? 
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#1
Feb513, 02:35 PM

P: 398

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: 12R*I1+R*I2=0 R*I2R*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? 


#2
Feb513, 03:06 PM

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PF Gold
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How is the diode hooked up? Cathode to + or the other way?



#3
Feb513, 03:16 PM

P: 398




#4
Feb513, 04:46 PM

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PF Gold
P: 4,828

Circuits  Variables dependent on R?



#5
Feb513, 06:08 PM

P: 398




#6
Feb513, 06:24 PM

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PF Gold
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#7
Feb513, 06:40 PM

P: 398

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. 


#8
Feb513, 07:29 PM

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PF Gold
P: 4,828

OK, so the first thing is to simplify the circuit by removing irrelevant components.



#10
Feb513, 08:09 PM

P: 398

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. 


#11
Feb513, 08:35 PM

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PF Gold
P: 4,828

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.



#12
Feb513, 09:19 PM

P: 398

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. 


#13
Feb513, 10:41 PM

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PF Gold
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#14
Feb513, 10:48 PM

P: 398

i.e Vdiode maybe something like Vdiode=2R instead of something like 10V 


#15
Feb613, 04:21 AM

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P: 5,251

The equations for the currents will all involve variable R.



#16
Feb613, 10:15 AM

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PF Gold
P: 4,828

What level are you at in your studies? 


#17
Feb613, 09:23 PM

P: 398

3rd year undergrad. Why?



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