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Determining resistance of a series resistor in a simple series circuit

  1. Jan 4, 2013 #1
    Background: I learned electronic theory 10 years ago from the Army, and have since worked in the electronics industry as a technician, for about the past 9 years. I'm working towards a BSEET (my interest lies in application, not research) and am required to take a few algebra based physics classes. Below is a problem that I thought I could do in my sleep (classes start monday, found this while perusing online coursework), but it seems like I have forgotten more than I have realized.

    1. The problem statement, all variables and given/known data
    Vt = 8 vdc
    It = 13.7 amps
    R1 = 7Ω
    R2 = unknown

    Find voltage drop across R1 and the value of R2.

    2. Relevant equations
    V=IR, single-loop circuit rule


    3. The attempt at a solution
    1) Since current is constant in a series circuit, I THOUGHT I could find the voltage drop across R1 by multiplying the value of R1 by the current, which amounts to 95.9 volts. Impossible, since the source voltage is only 8 volts. Had this worked, I could have used the resulting voltage drop to calculate the resistance of R2, since the voltage drop across R2 would have been Vt - VR1.

    2) I attempted to determine Rt by using the equation Vt = It(R1 + R2). In doing so, I get -6.4Ω, which is, again, impossible.

    I thought that in order to satisfy ohm's law, Rt has to be 0.58Ω, which makes the value of R1 false, since Rt = R1 + R2. Where am I going wrong?

    These numbers are random numbers based upon student's name and ID number. That being said, is it even possible to achieve 13.7 amps with a voltage source of 8 volts and total resistance of >7Ω? According to my memory, calculations and sim software, it's not. Even with a total resistance of 7 ohms, total current in the circuit is about 1.15 amps at +8 Vdc. We'd need to raise voltage to roughly 95.6 volts to achieve 13.7 amps through a 7Ω circuit, even more if we added a second resistor.
     
    Last edited: Jan 5, 2013
  2. jcsd
  3. Jan 4, 2013 #2

    phinds

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    Well, where you are going wrong at the moment is that you forgot to post the circuit diagram so you likely won't get any help just yet :smile:
     
  4. Jan 4, 2013 #3
    I didn't think I needed a circuit diagram for a simple series circuit... but here ya go:

    attachment.php?attachmentid=54466&stc=1&d=1357361317.jpg
     

    Attached Files:

  5. Jan 5, 2013 #4

    Dick

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    No, I don't think you are likely to get 13.7 amps flowing through that circuit. Must be some mistake.
     
  6. Jan 5, 2013 #5
    That's precisely what I thought. There's no mistake, the numbers are taken from the student's name and ID number. Each student solves the circuit with different numbers. It could work if the supply voltage was greater than 96 volts (allowing max current of 13.7 amps with a resistance of >7 ohms), or the current was less than 1.15 amps (assuming a +8 Vdc source), but it seems that it's not plausible with the original numbers.

    Being a bit rusty on my circuit solving skills, I'm relieved to know that I'm not an idiot; that there really isn't a valid solution for this problem as originally given.

    Thanks for the help!
     
  7. Jan 5, 2013 #6

    Dick

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    If the numbers are completely random, then maybe you can pick a negative number for R2? It's not very physically plausible, but maybe you should just solve it anyway.
     
  8. Jan 5, 2013 #7
    If that's the case, then R2 would be about -6.5 ohms.

    I really hope the professor doesn't expect this, this class is for an applied engineering degree.
     
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