Grade 12 student physics question

Discussion in 'Homework Help' started by stu123, Jan 2, 2011.

  1. stu123

    Thread Starter New Member

    Jan 2, 2011
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    Ok so I'm pretty sure I uploaded a picture of a circuit diagram but say so if you can't access it. What I'm having trouble with is figuring out how to complete the rest of the table. I know the total equivalent values for voltage, current and resistance and then the resistance for R1. I also know that both R2 and R3 will have a sum of 6Ω. Ohm's law I know so R1, R2, and R3 will have a sum of the total equivalent resistance. I know kirchhoff's law as well.
    I just don't know how to continue on with the calculations to figure out the rest of the table. Any help at all would be awesome.

    Also if anyone knows of a quick and easy thing online where I can make drawings and copy the link to post it would be useful.

    I did go reading the stuff on the site here about parallel and series circuits but it didn't give me everything I needed to know.
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Since you already know Ohm's and Kirchhoff's Laws, you could proceed as follows:

    You know the loop current, so you can find the voltage dropped by R1 (by Ohm's law, V=IR.

    You know the battery voltage and also R1 volt-drop, and thus you can find the voltage drop for R2 and R3. Use Kirchhoffs Voltage Law.

    Knowing the voltage across R2 and R3, you can find the currents through each of them (Ohm's Law again, I=V/R).
     
  3. stu123

    Thread Starter New Member

    Jan 2, 2011
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    Alright I think I got it then. Is this correct? I uploaded another picture.
     
  4. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Looks good. 5 ohms + 6 ohms (from parallel of 10 ohm and 15 ohm) is 11 ohms. 9V/11Ω = 818.18mA

    Tip: Set your calculator to "ENG" mode, so it uses "increments of 3" Scientific Notation, so -3 = Milli, -6=micro, -9=nano, -12=pico, etc. Then you can read microfarads or milliamps very easily.
     
  5. stu123

    Thread Starter New Member

    Jan 2, 2011
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    Actually only part of my diagram is correct. Sorry I'm still lost on how to figure out the rest.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    You know that at the point after the 5Ω Resistor, the voltage across the 10Ω and 15Ω in Parallel is about 4.9596 V (9V-(818.18mA x 5Ω)

    Divide each resistance by that voltage, and you should end up with two currents that will add up to the total current (same current that is flowing through the 5Ω Resistor).
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The current through R1 has to be the same as the battery current, because there is no other path.
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Not quite. 9V-(818.18mA x 5Ω) = 9V - 4.0909V = 4.9091V. Double check: the 10Ω and 15Ω in parallel come to 6Ω.

    The voltage across them is given by 9V * 6Ω / ( 5Ω+6Ω) = 9V * 6/11 = 4.9091V. Checks OK.

    Divide that voltage by each of the 10Ω and 15Ω resistances and you should get two currents which will add up to the total.
     
  9. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    My bad. Not sure what I multiplied by to get the result above.
     
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