Results 1 to 9 of 9

Math Help - DC Electrical Circuit

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    13

    DC Electrical Circuit


    I need to find the equivalent resistance and the current at I

    I'm not sure where to start, I need to use Thevenins theorem? And it's a Wheatstone bridge?

    If I take out the diagonal 3ohm resister which represents the load (I think this is the first step?) I'm not sure what to do next?

    Please help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    277
    Thanks
    2
    Use Kirchhoff's laws to write down 3 loop equations for voltage drops, and 3 node equations for currents.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    13
    OK hears what I've attempted to do;

    Is any of that correct so far?

    If it looks like I don't know what I'm doing please point me in the right direction. I'm starting to dislike this Kirchhoff chap!

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Here's a way to cut down on the work a bit. Use the rules for combining resistors in series or in parallel to simplify the circuit. For a start, the two 3Ω resistors that I have labelled in blue in the attachment are equivalent to a single 6Ω resistor. That can be combined with the green 3Ω resistor, using the rule for combining resistors in parallel, to form a single resistor. That single resistor is in series with the yellow 2Ω resistor, so again you can reduce them to a single resistor.

    You then have a simplified circuit which has two links going from c to d in the diagram: there is a current I_2 through the 1Ω resistor R_3, and a current I_3 through the equivalent resistor that you have formed from the blue-green-yellow resistors. Now you start to use Kirchoff's laws, but with only three variables I_1,\,I_2,\,I_3 instead of five.
    Attached Thumbnails Attached Thumbnails DC Electrical Circuit-resistors.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2009
    Posts
    277
    Thanks
    2
    Alex - I'm glad you're making a game attempt at this, but your equations are somewhat off.

    First,
    I1 != I2 + I3 + I4 + I5 ( where != means does not equal )

    Is it clear that segments DA, AB and BC all have I1 flowing in them? If not, think of a bus leaving node D and going counterclockwise. There is no place for the current to change until it hits node C. That's why the current is the same in all those segments. Once you hit node C, you see that all the current going into node C must come out of there. That gives:

    I1 = I2 + I3.

    Likewise the node E equation is:

    I3 = I4 + I5

    If you ask what the node F equation is, it is:

    I5 = I5 (I5 going in, I5 going out).

    Can you write down what the node D equation is?

    Your loop equations ( for voltage ) are better. However you need to correct what current is running in R2 and R4. Also, you must make sure that the sign of the voltage drop is consistent (is a positive drop with or against current?).

    You can also use OpAlg's suggestion to simplify the circuit somewhat - combine resistors in series (R4 and R7), then in parallel (R5 and the effective (R4/R7)), etc. Still, you do need to get familiar with writing down general Kirchhoff equations for a general circuit.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2008
    Posts
    13
    Brilliant! I'm making sense of it all now :-) I may have cracked it, thanks very much!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2008
    Posts
    13
    If you can combine resisters 4, 5, 6 and 7, can you not then combine that equivalent resister with R3 useing the parallel rule then with R1 and R2 in series? Then just use ohms law?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by AlexTweed View Post
    If you can combine resisters 4, 5, 6 and 7, can you not then combine that equivalent resister with R3 useing the parallel rule then with R1 and R2 in series? Then just use ohms law?
    That depends what you're looking for. I thought that the original problem was to find the current flowing through the bottom right-hand segment of the diagram (the current I that you subsequently called I_3, or actually -I_3 since it goes in the opposite direction). If you combine all the resistors into a single one then you can only find the current I_1, because that would be the only one left.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Nov 2009
    Posts
    277
    Thanks
    2
    It looks like you've gotten the hang of it. Yes, you can continue to simplify the circuit in the way you described. However, as OpAlg points out, you've just simplified your problem out of existence! So back up one step and I think you're there. Good work!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circuit Analysis - Redrawing Circuit
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: September 13th 2011, 06:20 PM
  2. Electrical Circuit problem?
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: April 25th 2010, 11:17 PM
  3. Electrical power
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: October 22nd 2009, 06:49 AM
  4. electrical resistance
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 13th 2008, 10:40 PM
  5. electrical question help
    Posted in the Business Math Forum
    Replies: 0
    Last Post: April 3rd 2007, 01:46 AM

Search Tags


/mathhelpforum @mathhelpforum