Electrical and electronics measurements miscellaneous

Electrical and electronics measurements miscellaneous

Easy Questions
Moderate Questions
Difficult Questions

Electrical and Electronics Measurements

Electrical and electronics measurements miscellaneous
  1. An unfinished moving iron voltmeter is used to measure the voltage in an a.c. circuit. If a stray d.c. magnetic field having a component along the axis of the meter coil appears, the meter reading would be








  1. View Hint View Answer Discuss in Forum

    The meter reading will either decrease or increase depending upon the direction of flux.

    As shown in figure let FM be the flux due to meter coil and F, is due to stray magnetic field, then resultant will be F. Similarly when F is in opposite direction. Thus the reading will either increase or decrease.

    Correct Option: D

    The meter reading will either decrease or increase depending upon the direction of flux.

    As shown in figure let FM be the flux due to meter coil and F, is due to stray magnetic field, then resultant will be F. Similarly when F is in opposite direction. Thus the reading will either increase or decrease.

  1. Which one of the following types of indicating instruments is an electrometer?








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: C

    NA

  1. Which one of the following types of indicating instruments is an electrometer?








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: C

    NA

  1. For the voltmeter circuit shown in the given figure, the basic D’Arsonval meter used has fullscale current of 1 mA and meter resistance (Rm) of 100 ohms. The values of the series resistance R1 and R2 required for 10 V range and 50 V range will be respectively








  1. View Hint View Answer Discuss in Forum

    Full scale meter current = 1 mA

    For 10 V range, 10–3 =
    10
    (100 + R1)

    or R1 = 9.9 × 10³ Ω
    For 10 V range, 10–3 =
    50
    (100 + R1 + R2)

    or R1 + R2 = 49.9 × 10³
    ∴ R2 = 40 kΩ

    Correct Option: A

    Full scale meter current = 1 mA

    For 10 V range, 10–3 =
    10
    (100 + R1)

    or R1 = 9.9 × 10³ Ω
    For 10 V range, 10–3 =
    50
    (100 + R1 + R2)

    or R1 + R2 = 49.9 × 10³
    ∴ R2 = 40 kΩ

  1. In the case of power measur ement by t wo wattmeter method in a balanced 3-phase system with a pure inductive load








  1. View Hint View Answer Discuss in Forum

    Two wattmeter readings are,
    W1 = VL IL cos (30 – φ)
    = VL IL cos (30 – 90°) = VL IL cos 60°
    W2 = VL IL cos (30 + φ)
    = VL IL cos 120° = – VL IL cos 60°

    Correct Option: A

    Two wattmeter readings are,
    W1 = VL IL cos (30 – φ)
    = VL IL cos (30 – 90°) = VL IL cos 60°
    W2 = VL IL cos (30 + φ)
    = VL IL cos 120° = – VL IL cos 60°