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Voltage query -- A power supply, resistors and a voltmeter...

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data
    a 12v dc supply is connected across 2 100k resistors in series.a voltmeter having sensitivity of 10,000 ohms per volt is switched to its 10v range and is connected to its 10v range and is connected to measure the voltage of 1 resistor.

    2. Relevant equations
    1)sketch circuit
    2)calculate voltage reading
    3)calculate percentage error of the meter reading in 2

    3. The attempt at a solution

    I can draw the circuit that part is fine.im using v=ir to try and workout voltage so 100,000/10000 = 10
    and for the percentage error 100/10 =10%

     
    Last edited by a moderator: Nov 6, 2015
  2. jcsd
  3. Nov 6, 2015 #2

    gneill

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    Um, those aren't relevant equations. That's just more of the problem statement.

    For relevant equations you want to state things like Ohm's law, or the formulas for combining resistors, or the voltage divider equation, and so forth. Whatever you think might be applicable to this type of problem.
    Can you show your work in detail? How did you determine the resistance of the meter? How did you come up with "100,000/10000 = 10", and what does it mean? What are the units?
     
  4. Nov 6, 2015 #3
    sure thing,i would be using ohms law.100,000r ohms /10,000 I =10volts reading

    for the percentage error I would say 100r / 10v =10 % error
     
  5. Nov 6, 2015 #4

    gneill

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    Sorry, I'm not understanding where the values are coming from, and I'm not certain what the r, v, and I are meant to mean.

    Can you draw a circuit diagram and label the parts with their values?
     
  6. Nov 6, 2015 #5
    attach
     

    Attached Files:

  7. Nov 6, 2015 #6

    gneill

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    Okay, what resistance value are you going to assign to the meter (Let's call it Rm)? Remember, it's rated at 10,000 Ohms per volt and it's on the 10 Volt scale...

    Fig1.png
     
  8. Nov 6, 2015 #7
    ok rm I would say 10,000 x 12dc v = 120,000/10v =12,000r
     
  9. Nov 6, 2015 #8

    gneill

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    Not quite. It's on the 10 V scale. The source voltage V1 (12 V) doesn't play a part in the meter's internals. The meter's resistance will depend upon its Ohms-per-Volt rating and the scale selected. So

    ##R_m = \frac{10 kΩ}{V} 10 V = 100 kΩ##

    That is the resistance that the meter will place across R2 when it takes a reading there.

    So now, what calculations will you perform to find the voltage that the meter reads?
     
  10. Nov 6, 2015 #9
    I would work out the volt across the 1 resistor battery voltage/total resistance x r1 = voltage
     
  11. Nov 6, 2015 #10

    gneill

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    Okay, go ahead. Show us the details.

    What will you do with that voltage across R1?
     
  12. Nov 6, 2015 #11
    12v /200k x 100k =6volts across r1

    I would be reading that with the voltmeter as my voltage reading
     
  13. Nov 6, 2015 #12

    gneill

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    How did you arrive at 200k for the total resistance?
     
  14. Nov 6, 2015 #13
    100 +100 =200k resistors in series rt=r1+r2
     
  15. Nov 6, 2015 #14

    gneill

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    Nope. You've left out the meter resistance. It's in parallel with R2.
     
  16. Nov 6, 2015 #15
    yes your absolutely right. 100k + 1/100 +1/100=150k ohms
     
  17. Nov 6, 2015 #16

    gneill

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    I understand what you're trying to convey, but your math exposition is a tad sloppy. If I ran that expression through my calculator as it is written then I'd arrive at 100.02 k Ohms, not the 150 k Ohms that you intend. You can use parentheses to gather terms, and the edit panel provides icons for doing things like superscripts and subscripts:

    100k + (1/100 +1/100)-1 = 150k ohms

    Now, 150 k Ohms for the total resistance is good. What resistance do you want to know the potential across? You could find the potential across R1, or take the more direct route and find the potential difference across the R2||Rm parallel pair. After all, you've calculated the resistance of that pair, right?
     
  18. Nov 6, 2015 #17
    well the question is asking for the voltage reading across r1

    so 12v / 150k ohms x 100k ohm =0.008v so this would be the reading
     
  19. Nov 6, 2015 #18

    gneill

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    Well, no, it's asking for the voltage reading on the meter. The meter is connected across R2 (according to your figure):
    fig1-png.91429.png
    Use parentheses to gather your terms to force the correct order of operations:

    (12v / 150k ohms) x 100k ohm = ?

    That would be the voltage across R1. Of course, you want the voltage across the R2||Rm pair. You could recalculate that or use KVL to figure it out from V1 and this VR1.
     
  20. Nov 6, 2015 #19
    (12v / 150 ohms) x 100k ohm =8volts

    kvl??

    (12v / 100k) ohms x 100k ohms = 12volts
     
  21. Nov 6, 2015 #20

    gneill

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    Right. That's the voltage drop across R1.
    Kirchhoff's Voltage Law. You should look that up. It's very important in circuit analysis! You'll use it very often indeed, along with KCL, Kirchhoff's Current Law.
    I don't know what that calculation is.... The 100k in the denominator is not the total resistance, and the 100k looks like R1 again...

    It would be preferable if you would first present the calculation as symbols rather than numbers. That way we can tell what it is the calculation represents.
     
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