Saturday, December 25, 2010

Module 6: AC Measurements

Dear Studnets
  1. Refer to the multimeter reading shown in the figure below and answer the questions:
    a. Is the multimeter displaying the rms voltage or the peak voltage?
    _____________________________________________________
    b. What is the peak voltage?
    _____________________________________________________
  2. If the frequency set on the function generator = 500 Hz and if the oscilloscope reading gives 550 Hz, calculate the percent error in frequency.

    __________________________________________________________

    __________________________________________________________

20 comments:

  1. AC Measurments
    1) a. Displaying the rms voltage
    b. Peak Voltage = rms Voltage / 0.707 = 5.85 V
    2) Percentage error = (|500-550|/ 500) x 100 = 10%

    Easa Tariq Al Hammadi
    1011090089
    11-02

    ReplyDelete
  2. AbdulRahman Al-Blooshi_11-02

    1.
    a-rms.
    b-Vp=Vrms/0.707=585V

    2.
    500-550/500 x100
    =|-10|

    ReplyDelete
  3. 1) 4.14 V
    2) 4.14 * 0.707= 0.926 V

    3) Percentage error = (|500-550|/ 500) x 100 = 10%

    Abdulrhman Abdullah AL Marzouqi
    - 90106 -
    11 - 02

    ReplyDelete
  4. This comment has been removed by the author.

    ReplyDelete
  5. M6

    Q1- Displaying the rms voltage

    Q2- Peak Voltage is the maximum value reached by an ac wave. Vp = Vrms / 0.707 =
    4.14 / 0.707 = 5.85 V

    Q3- Percentage error = (|500-550|/ 500) x 100 = 10%

    Ahmed Amer Al-Kathiri
    1011090095
    11-02

    ReplyDelete
  6. Hamad Al-Ansari 11-02 1011090056

    1)a. Displaying the rms voltage.

    b. Vp= rms V / 0.707 = 5.85 V

    2) Percentage error= 550-500/500 x100

    =10%

    ReplyDelete
  7. 1-
    a- the rms voltage
    b-4.14/0.707=5.8557 v
    2-
    550-500=50
    50/500=0.1
    0.1*100=10 %
    Mohammed Hassan Hadi
    11-02

    ReplyDelete
  8. AC Measurments
    1) a.rms voltage

    b. Peak.V = rms Voltage / 0.707 = 5.85 V

    2) % error = (|500-550|/ 500) x 100 = 10%

    abdullah ali albalooshi

    ReplyDelete
  9. 1) a. Displaying the rms voltage

    b. Peak Voltage = rms Voltage / 0.707 = 5.85 V

    2) Percentage error = (|500-550|/ 500) x 100 = 10%

    abdulaziz ahmed
    90063
    11-02

    ReplyDelete
  10. 1)
    a.its displaying rmsV

    b.Peak Voltage = rmsV / 0.707 = 5.85 V
    2)
    Percentage error = (500-550/ 500) x 100 = 10%

    Hamza Mohammed
    11-02
    1011090082

    ReplyDelete
  11. rms voltage

    5.85

    10

    salem abdula

    abdlatif hussain

    ReplyDelete
  12. 1-
    a) a. Displaying the rms voltage

    b. Peak Voltage = rms Voltage / 0.707 = 5.85 V

    2- Percentage error = (|500-550|/ 500) x 100 = 10%

    Abdulla Khames AL-Ameri

    11-03
    90072

    ReplyDelete
  13. AC Measurments
    1) a. Displaying the rms voltage
    b. Peak Voltage = rms Voltage / 0.707 = 5.85 V
    2) Percentage error = (|500-550|/ 500) x 100 = 10%

    made by : Talal Al Hamily
    Class:11-03

    ReplyDelete
  14. the rms voltage

    4.14/0.707=5.86


    SAEED TALAL ALJUNAIBI
    11-03

    ReplyDelete
  15. 1
    a) RMS voltage
    b) 4.14/0.707=5.86V

    2
    |500-550|/500 x 100% = 10%

    ReplyDelete
  16. Q1- Displaying the rms voltage

    Q2- Peak Voltage is the maximum value reached by an ac wave. Vp = Vrms / 0.707 =
    4.14 / 0.707 = 5.85 V

    Q3- Percentage error = (|500-550|/ 500) x 100 = 10%
    Awad Juma 11-3

    ReplyDelete
  17. 1-
    a) 4.14
    b)4.14/0.707=5.8

    2-
    (50/500)x100%=10%

    Saeed Zaid Abdulla Saeed Mohammed AL Menhali
    11-03

    ReplyDelete
  18. Q1. Displaying the RMS voltage

    Q2. The peak voltage is the maximum value reached by an AC wave:
    Vp = Vrms / 0.707
    Vp = 4.14 / 0.707 = 5.85v

    Q3. The Percentage error = (|True value-Measured value| / True value) x 100
    The percentage error = (|500-550| / 500) x 100
    = 10%

    Mohammed Khalid Almeraikhi
    11-03. 90217

    ReplyDelete
  19. quastion 1
    a) 4.14
    b)4.14/0.707=5.8
    quastion 2
    2-
    (50/500)x100%=10%

    ReplyDelete