Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsEngineering · 1 month ago

Electrical engineering: circuits?

So I am solving the total resistance of a given complex circuit. I had an easy time simplifying those with rectangular shape until i came across this one:

So what I did was redraw this into a rectangle, where each two resistors are placed into the top and bottom.

Then, I added all the resistance at the top followed by the bottom since they are just in series. 

R(top)= 1+2 =3 ohms

R(bottom)= 1+1=2 ohms

Now, the top and bottom resistance are parallel. So,

R total = (1/3 +1/2)^-1= 1.2 ohms

Is this correct?

Attachment image

5 Answers

Relevance
  • 1 month ago

    This is the redrawn circuit with the Prod/sum calculation of Req.

    Attachment image
  • 1 month ago

    If you read the first chapter, you could do this in your head.

    (1+2)||(1+1) = 3*2/(1+2+1+1) = 6/5 = 1.2Ω

  • ?
    Lv 7
    1 month ago

    Yes. The first stage with any circuit exercise is to redraw it such that the simplification process becomes obvious.

  • ?
    Lv 7
    1 month ago

    Yes.

    In fact: you've hit upon a great method for dealing with such problems.

    It's common for homework and test problems to have circuits drawn in a non-standard format in an attempt to confuse the student

    and so

    redrawing circuits in a format that is familiar to you is a great method for dealing with those particular problems.

    Note, though, that sometimes the redrawing method doesn't work. Sometimes a circuit just has a strange format that literally cannot be redrawn into a standard format. In such cases you have to use different (more complicated) tricks to figure out the properties of the circuit.

  • 1 month ago

    Yes, that is correct.

    For a mental shortcut to work out parallel resistances like that, imagine each is in turn made up of a value that is a common multiple of both; eg. six ohms for that example, as a multiple of both 2 and 3.

    The upper leg (3R) is like two 6R in parallel, the lower leg is like three 6R in parallel.

    The overall network is equivalent to five, 6R resistors in parallel.

    Another mental shortcut to divide by five; double it and move the decimal point.

    12, 1.2

    1.2 Ohms.

    It works for any number of resistors in parallel & avoids the reciprocal calcs; as an electronics designer it means much less hassle & less calculator use.

Still have questions? Get your answers by asking now.