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Finding voltage in a circuit

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data
    PD_Capacitor.png


    Link: http://s359.photobucket.com/albums/oo40/jsmith613/?action=view&current=PD_Capacitor.png
    How is cell E3 calculated?

    2. Relevant equations





    3. The attempt at a solution

    now I know I can use Q=CV BUT another possible method is to say:
    Voltage-Capacitor + Voltage-resistor = 5V (voltage supply)

    V-capacitor: 5-(1.11*3) = 1.67 (the m and k powers of +3 and -3 cancel)
    But this gives the value in cell E2....why does this method not work for E3?
     
  2. jcsd
  3. Jun 5, 2012 #2

    NascentOxygen

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    Staff: Mentor

    Column E seems suspiciously in step with column D. :smile:
    The current changes during each 0.1 sec increment. The current listed is that at the start of each time period; the current at the end of each time period is listed as for the start of the next time period.
     
    Last edited: Jun 5, 2012
  4. Jun 5, 2012 #3
    thats because of Q=CV
    but what I want to know is why my method does not work :S
    thanks
     
  5. Jun 5, 2012 #4

    gneill

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    Staff: Mentor

    You should pay attention to the units associated with the values and not just the numerical values. You may be comparing apples and oranges.

    Think about what the "Δ" in ΔQ stands for.
     
  6. Jun 5, 2012 #5
    well it means "change in charge"...
     
  7. Jun 5, 2012 #6
    why would I use the end current (i.e the one in the next row down) to find the voltage across the resistor...surely the p.d given is that at the start too??
     
  8. Jun 5, 2012 #7
    well it means "change in charge"...
    the bigger question is though, why must I use the current in the row below to find the voltage across the resistor...?
     
  9. Jun 5, 2012 #8

    gneill

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    Staff: Mentor

    The currents given pertain to the initial current at the beginning of each time step, which is to say, they also correspond to the current at then END of the previous time step. So you would expect I*R to give the voltage drop at the END of the last time step (and the beginning of the current time step). On the other hand, the total charges and the potentials on a given line correspond to the end of the current time step. So the currents are one time step out of sync with the potential differences displayed on a given line.
     
  10. Jun 5, 2012 #9
    ok...how was this obvious from the data given...perhaps I am missing something big but this bit is clearly what is confusing me
     
  11. Jun 5, 2012 #10

    gneill

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    It's not so much obvious as having to figure out how the table was built using a time step model. What values pertain to which instants in time?
     
  12. Jun 5, 2012 #11
    I have never studies time-step models...I presume I will just stick to the Q=CV method (which is what was on the MS...my other method appears not to work very well with this table, based on what I know)
     
  13. Jun 5, 2012 #12
    I agree with all that gneill has written and will add that for charging a capacitor at t=0
    I is max = E/r, Vc = 0 and Qc = 0.
    At t= 0 you have measured I = 1.67mA which means the voltage across r = 1.67mA x 3000 ohms = 5V.... confirming that this is a capacitor at the start of charging.
    Row 2 for t = 0 should have Q = 0 and V = 0
     
  14. Jun 5, 2012 #13
    ok so the examiners were clearly looking for use of Q=CV ONLY as this is the ONLY method that apparently works....never mind...I guess you have to give 'em what they want :)
     
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