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How to calculate detector voltage from resistances?

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    nJME4b7.png

    The resistances:
    Ra = 10 ohm
    Rb = 10 ohm
    Rc = 10 ohm
    Rd = 8 ohm

    Battery = 10 volts

    Voltage across detector = ?

    2. Relevant equations
    V = IR
    (series) R = R1 + R2 + ...
    (parallel) 1 / R = (1/R1) + (1/R2) + ...

    3. The attempt at a solution

    So to get the voltage, I need to find the total resistance and current. I tried to simplify the problem by calculating Ra and Rb in parallel, Rc and Rd in parallel, then adding up these resistances in series. So I get the total resistance to be:

    Ra + Rb = ##\frac{1}{\frac{1}{10} + \frac{1}{10}}= 5## ohm
    Rc + Rd = ##\frac{1}{\frac{1}{8} + \frac{1}{10}} = \frac{40}{9}## ohm

    Ra + Rb + Rc + Rd = ##5 + \frac{40}{9} = \frac{85}{9}## ohm

    So to get the current, I use V/R = I
    Since battery is 10V:

    ##\frac{10}{\frac{85}{9}} = \frac{18}{17}A##

    So now I need to find the voltage through this resistance, but V = RI doesn't work because I just get back the total voltage from the battery. Not sure what else to do here
     
  2. jcsd
  3. Oct 20, 2013 #2

    mfb

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    Ra and Rb (and Rc and Rd) are not in parallel. The (ideal) voltage detector does not conduct current.
     
  4. Oct 20, 2013 #3
    Is Ra and Rc (Rb and Rd) in series?
     
  5. Oct 20, 2013 #4

    mfb

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    Right.
     
  6. Oct 20, 2013 #5
    OK, so if I add up the series first, can I then calculate the remaining 2 resistances in parallel?
     
  7. Oct 20, 2013 #6

    mfb

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    Sure.
     
  8. Oct 20, 2013 #7
    You can. but it won't help. You have to calculate the voltage at the point between Ra and Rc, and at the point between Rb and Rd.

    Ra and Rc form a voltage divider. You should know a formula for that . If yoiu don't, caculate the current throug Ra and Rc, and then the voltage drop across Rc.

    Same thing for Rb and Rd
     
  9. Oct 20, 2013 #8
    I'm not sure I understand the question. If there's no current through the detector, how can there be a voltage?
     
  10. Oct 20, 2013 #9

    CWatters

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    No current flows through an ideal voltmeter but it can still measure voltage. An ideal meter has very large resistance (aka infinite) so negligible current causes a voltage drop - if you insist on thinking about it that way.

    I prefer to look at your diagram like this... You now have two Potential Divider chains. What is vd ?
     

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