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Voltage-divider and d'Arsonval voltmeter problem

  1. May 4, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    The voltage-divider circuit is designed so that the no-load output voltage is 8/10ths of the input voltage. A d'Arsonval voltmeter having a sensitivity of 200 ohms/V and a full-scale rating of 150 V is used to check the operation of the circuit.
    a) What will the voltmeter read if it is placed across the 126 V source?
    b) What will the voltmeter read if it is placed across the 60 kohms resistor?
    c) What will the voltmeter read if it is placed across the 15 kohms resistor?
    d) Will the voltmeter readings obtained in parts (b) and (c) add to the reading recorded in part (a)? Explain why or why not.

    Output voltage is 100.8 V. About d'Arsonval voltmeter: using sensitivity and full-scale rating we can find resistance of the voltmeter? Than it is 30 kohms. I think in part (a) answer is 126 V, is it correct? Have some problems with parts (b)-(d). Thanks for sharing your thoughts!
     

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    Last edited: May 4, 2016
  2. jcsd
  3. May 4, 2016 #2

    cnh1995

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  4. May 4, 2016 #3

    berkeman

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    Your answer for (a) is correct. Please show us your work on the rest of the parts of the problem. What is the equation for a Voltage Divider?
     
  5. May 5, 2016 #4
    Voltage divider equation: Uout = Uin * R2 / (R1 + R2). And for my problem Uout = 126 V * 60 kohms / (60 kohms + 15 kohms) = 100.8 V.
     
  6. May 6, 2016 #5

    CWatters

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    Your comment about calculating the meter resistance is correct.

    Try drawing the equivalent circuit for the set up in b)
     
  7. May 6, 2016 #6
    Here it is.
     

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  8. May 7, 2016 #7

    CWatters

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    OK now replace the real world volt meter with an ideal meter and a resistor
     
  9. May 7, 2016 #8
    Can I use this equations: U2 = I * R;
    I = Uin/(R1 + R2)?
    I = 126 V / (15 kohms + 60 kohms) = 1.68 mA; U2 = 1.68 mA * 60 kohms = 100.8 V. Is that correct?
     
  10. May 8, 2016 #9

    CWatters

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    Thats the voltage that an ideal meter would display not the voltage this meter would display. See my post #7.
     
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