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Homework Help: Potential difference and ohms law

  1. Feb 8, 2004 #1
    A 3Ohm and a 1.5 ohm resistor are wired in paralles and the combination is wired in series to a 4 ohm resistor and a 10 V emf device. The potential difference on across the 3 ohm resistor is?

    I reduced the 2 in parrallel to 1.875 ohms and then addes the 4 ohm, which gave me a total resistance of 5.875 ohms. Then measured the current using ohms law, which gave me 1.7 amps. Then multiplied that times the 3 ohms to get 5.1 V.

    But that is not one of the answer choices, where did I go wrong.

    these are the choices: 2V,6V,8V,10V,12V

  2. jcsd
  3. Feb 8, 2004 #2


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    Wouldn't the equivalent resistance be 1? making the total resistance 5 ohms? Then you'd have a (10/5)*1 volt drop across the 2 resistors in parallel and a (10/5)*4 volt drop across the 4 ohm resistor.

    Look at the resistors in parallel again:

    [tex] R_{eq} = \left(\frac 1 3 + \frac 1 {1.5} \right)^{-1} = \left(\frac{1+2}{3}\right)^{-1} = 1 [/tex]
  4. Feb 8, 2004 #3
    nice, maybe I need to go back to 3rd grade math.
  5. Feb 8, 2004 #4
    I am still not coming up with the answer.

    If the total resistance is the 1 ohm, which i reduced, plus the 4 ohm in series. That gives me a total resistance of 5 ohms. Then I plug that into ohms law, which is V=IR and it gives me a total current of 2 amps.

    To find the potential difference on the 3 ohm, I would have to cut the current in half, which is 1 and then use ohms law, which is V=IR so 1 x 3, which gives me 3, but that is wrong.

  6. Feb 8, 2004 #5


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    I'm getting the answer as being 2V

    Resistance of parallel things

    [tex]R = (3^-^1 + 1.5^-^1)^-^1[/tex]

    [tex]R = 1 \Omega[/tex]

    resistance total

    R = 1 + 4
    R = 5[tex]\Omega[/tex]

    total current:

    [tex]I = \frac{V}{R}[/tex]

    [tex]I = \frac{10}{5}[/tex]

    I = 2A

    voltage drop over the 4[tex]\Omega[/tex] resistor:
    V = IR
    V = (2)(4)
    V = 8V

    voltage remaining:
    V = 10 - 8
    V = 2V
  7. Feb 8, 2004 #6
    So how is that split between the remaining 3 and 1.5 ohm on the parrallel
  8. Feb 8, 2004 #7


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    It isn't. When dealing with a parallel, the voltage is the same in each path. That's why hooking lightbulbs in parallel makes power consumption much higher than hooking them in series.
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