Simple (?) EE Question

[Tony Clark]

> Hi Guys -
> 
> OK, so this is kind of a lame question - but I'm having a bit of trouble
> figuring it out. I wouldn't be surprised if many of you who took EE classes
> in college didn't see this question on an exam at one time or another.....

   Freshman year.  :)

> Say you got a voltage divider, connected to V1, and Ground, R1 on top, R3
> on bottom.
> 
> The voltage at the divider point is V1 * (R3/(R1+R3)). Fine.
> 
> Now suppose you connect another resistor, R2, to the middle of the divider
> and connect it to voltage V2?
> What's the voltage at the divider point?

   What you need to do is figure out the Thevenin equivilant of the 
circuit.  This is not so easy to do with some circuits.  Of course 
someone may be able to help you a little better than me since I haven't 
used this in a long while, but here goes:
   Say you have a circuit as you described above:

   7V 0--,/\/\/`-----0-----,/\/\/`--0  5V
           20K       |       50K
          ---->      |      <----
           Ia        /  |     Ib
                     \  |
                100K /  |  Ic
                     \ \ /
                     |
                    ~~~
                     ~
                    GND

   To start off with, we can determine that Ic = Ia + Ib.

   All of the currents can be determined by ohm's law, I = V/R;

   So then Ic = Va/Ra + Vb/Rb.

   We also know that Va + Vc = 7V, and Vb + Vc = 5V

   This means that Ia(Ra) + Vc = 7V and Ib(Rb) + Vc = 5V

   To further extrapolate these two equations, we can write the following:

   Ia(Ra) + (Ia + Ib)Rc = 7V and Ib(Rb) + (Ia + Ib)Rc = 5V

   Ia(20K)+(Ia+Ib)100K = 7V and Ic(50K)+(Ia+Ib)100K = 5V

   Ia(120K)+Ib(100K) = 7V and Ib(150K)+Ia(100K) = 5V

   Ia(120K) = 7V - Ib(100K) and Ib(150K) = 5V - Ia(100K)

   Ia = 58.33uA - 0.833(Ib) and Ib = 33.33uA - 0.666(Ia)

   Now we can simply do substitution:

   Ia = 58.33uA - 0.833(33.33ua - 0.666(Ia))

   Ia = 58.33uA - 27.76uA + 0.554(Ia)

   This solves to Ia = 68.66uA

   Plug this into the equation for Ib...   Ib = -12.399uA

   The negative current for Ib shows that the arrow I have in the circuit 
above really should be pointing the opposite direction.  However, the 
equations still hold true.  Now solve the equation for Ic.

   Ic = 68.66uA + (-12.399uA) = 56.261uA

   Therefore, Vc (which is the voltage we wanted to know all along, is 
simply:

   Vb = IcRc = 56.261uA(100K) = 5.626V

   Just to make sure my math was right, I plugged up the circuit and got 
a reading of 5.82V.  As my electronics teacher would say, 
"Ehhhhh...that's good enough!"
  
> More generally, if you have a star of resistors R1....Rn connected to
> voltages V1....Vn, is there a general formula for the voltage at the middle
> of the star?

   I'll let you work out the math for that!  Yikes!

> - Thanks,
> CList

   Hope I've been of help.

   Tony

-------------------------------
I can't drive (my Moog) 55!
-------------------------------
Tony Clark -- clark at andrews.edu 
http://www.andrews.edu/~clark
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