Where did I go wrong?

Discussion in 'Homework Help' started by Loki4000, Nov 14, 2011.

  1. Loki4000

    Thread Starter Member

    Sep 29, 2011
    22
    0
    Greetings,

    Recently I try to calculate simple circuit:

    [​IMG]

    First I calculated resistances on the right corner

    1/((1/1)+(1/3)) = rx1

    then

    1/((1/rx1)+(1/2)) = rx2

    then

    1/((1/rx2)+(1/1.5)) = rx3

    then

    1+rx3+1 = rxfinal

    I enden up with 2.4

    but according to this programm total resistance must be: 2970,29703ohm
    (V/I)

    [​IMG]

    I want to know where did I go wrong in my calculations?
     
  2. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
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    Instead of using the 1/((1/R1)+(1/R2)....), try starting at the right hand side (right-most resistors are in parallel, see?) and try (R1*R2)/(R1+R2).

    The first formula is only used to find the total resistance of a bunch of resistors only in parallel.

    Der Strom
     
  3. RiJoRI

    Well-Known Member

    Aug 15, 2007
    536
    26
    I'm lazy. Why memorize two formulae when 1 will do?

    While in tech school I was told that the safest way of doing this is to redraw the circuit every time you collapse a branch into its equivalent. When you collapse R4&R5, and redraw the circuit, you will see where your problem lies. Remember, do the serial calculations before the parallel.

    --Rich
     
  4. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
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    Unfortunately, the one formula you memorized wouldn't work. You'll need the other one for this problem. As you said, you need to collapse the resistors down into their equivalent, and for that you will need (RxRy)/(Rx+Ry).
     
  5. Loki4000

    Thread Starter Member

    Sep 29, 2011
    22
    0
    Thanx for the reply, but now I am even more confused.

    Can you show me what you mean using real numbers?

    P.S. I already try starting at the right hand side. First I calculate total resistance of 1 and 3 kOHM resistor using 1/((1/1)+(1/3)) = rx1 , then treating answer as one new resistor plug it back and calculate this new resistor with 2kOHM resistor... and so on untill you move from right back to left, to voltage source. At least that's how I was told its supposed to be done. But double check on circuit calculating problem show that my calculation was wrong...
     
  6. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
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    Ok, let's look at the following circuit for this example. It is a simplified version of yours, but you solve it the same way:

    parallelresistors.PNG

    What is the first thing you notice about R2 and R3?

    If you said that they're in series, you're right. So just add them up to get their equivalent resistance:

    R2+R3 = 2+3 = 5 ohms. Let's call this value R' (R prime).

    Now that you have the equivalent resistance of R2 and R3, you substitute in a 5 ohm resistor for them.

    Now what do you notice about R1 and R'?

    If you said that they're in parallel, then you're right again. So use the formula (RxRy)/(Rx+Ry) to find the equivalent resistance of both of them together.

    (RxRy)/(Rx+Ry) = (R1*R')/(Rx+R') = (1*5)/(1+5) = 5/6 ohms. So that would be your Rt for this circuit. Now, if you simply put on more "stages", then after doing this, you would end up with a circuit looking very much like the one you started with, only there would be one less "stage". You would simply repeat the process over again.

    Does this help?

    Der Strom
     
  7. RiJoRI

    Well-Known Member

    Aug 15, 2007
    536
    26
    r1 = 1k
    r2 = 2k

    r1 || r2

    1/((1/1000)+(1/2000))= 666.6666666667

    (1000*2000)/(1000+2000)= 666.6666666667
     
  8. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
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    Whoops, sorry. Thanks Rich, you're right. That formula will work, but you just need to be careful of the values. If it is in kilohms, you have to scale the result, or convert to ohms. I have a feeling that may have been your mistake--you didn't scale it properly.
     
  9. Loki4000

    Thread Starter Member

    Sep 29, 2011
    22
    0
    Thanx for replies, but I figured the problem, turn out that after calculating 1 and 3 kOHM resistors and treating this sum as separate resistors, I add this separate resistor to 2 kOHM resistor in paralel rather than in series.
     
  10. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    This was somewhat cryptic to me. Can you elaborate?
     
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