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Use node voltage method to solve i2 vbe and vce

  • Thread starter Ninja_IPGO
  • Start date
  • #1
I have attached the problem and the worked out solution can a few people go over it and post if it's right and if not what is wrong with it.

Using node voltage method
 

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Answers and Replies

  • #2
264
0
Hey man i am not sure where you went wrong in your work but it isn't correct. I would first combine all those resistors that are in parallel to get an Req of 5000/17 ohms. Then the problem is really easy to set up. Then to split them back you will need to use current or voltage division to find the proper current through or voltage drop of the resistors. (Parallel resistors have the same voltage across them), hope that helps.
 
  • #3
the problem is he wants me to solve it using node voltage analysis I am going through some class notes now and i'll post a new version later on tonight I ll see what i can do with the information you gave me thank you
 
  • #4
vk6kro
Science Advisor
4,081
40
I get 263 ohms for the parallel combination.

You can work out the voltages and currents with just a calculator and then at least you will know if what you are getting makes sense when you do the node analysis.

I would add a LOT more explanation and diagrams. I couldn't see where most of that was coming from and I shouldn't have to be guessing.

Draw the part of the diagram you are dealing with and justify each step before you go on.
 
  • #5
264
0
Interesting vk6kro because:

[tex] R_{eq}^{-1} = 2500^{-1} +500^{-1}+1000^{-1} =.0034 [/tex]

and

[tex] .0034^{-1}=294.1176471=\frac{5000}{17} [/tex]

how'd you get 263?

@Ninja IPGO, I am currently taking electrical fundamentals for the first time (i dont know what class you are taking but it looks much the same) and we can use any 'tools' we have to solve any of the problems. Why would there be a problem combining resistors in this problem? It is what you would do in a similar real life situation.

edit: what i really meant is if the problem says 'use mesh analysis to solve for i0 an v0' we could most certainly combine resistors to solve the problem, just not use node voltage. I don't see using voltage, current division or series/parallel combination as 'not using the method specified'.
 
  • #6
vk6kro
Science Advisor
4,081
40
You have in parallel:

(1500 + 1000), 2500, 500, 1000 ohms

And this gives 263.158 ohms.

You put this in series with the 750 ohms across 100 volts to get the voltage at node b.

Then you have 2500 ohms in series with 1000 ohms across this voltage to get the voltage at node c.
 
  • #7
264
0
oh i see, i was just combining the 2.5k 500 and 1k resistors, to combine the rest does not seem to be a productive way to getting to the end quickly
 
  • #8
vk6kro
Science Advisor
4,081
40
But, you have to include the two series resistors to get the first voltage divider resistors.

You have 750 ohms then in series with 263 ohms across 100 volts. It is easy to work out the voltage across the 263 ohms. (100 times 263/1013)

Now that you know this voltage, it is across all the resistors in parallel including the two that are in series with each other.
So, they form a new divider and you can work out the voltage across the 1000 ohm resistor.
(100 times 263/1013) times 1000/2500.

It is so easy, you just write down the answers off the calculator.
 
  • #9
264
0
While your method is not wrong i maintain mine is much quicker and more efficient. I did a schematic in LTspice of this circuit and using my method works just fine. No need to combine them all man.
 
  • #10
i redid the problem there was a few algebraic errors but the method i used is correct
 

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