Electrical and electronics measurements miscellaneous


Electrical and electronics measurements miscellaneous

Electrical and Electronics Measurements

  1. The power factor in a single phase circuit is cos φ = P/VI where P power read by an wattmeter, V = voltage read by a voltmeter and I = current read by an ammeter. All the three instruments have a limiting error of ± 1%. The power factor of the circuits can be known if only one meter i.e. an electrodyna-mometers type power factor having an limiting error ± 1% used. The preferred method is to use a single instrument like a power factor meter rather than using 3 instruments because the error with using wattmeter, voltmeter and ammeter is









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    NA

    Correct Option: B

    NA


  1. The voltage of a circuit is measured by a voltmeter having an input impedance comparable with the output impedance of the circuit thereby causing error in voltage measurement. This error may be called









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    NA

    Correct Option: B

    NA



  1. In the center zero analog ammeter having a range of – 10 A to + 10 A, there is a detectable change of the pointer from its zero position on either side of the scale only if the current reaches a value of 1 A (on either side). The ammeter has a









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    NA

    Correct Option: C

    NA


  1. A d.c. circuit can be represented by an internal voltage source of 50 V with an output resistance of 100 kΩ. In order to achieve 99 percent accuracy for voltage measurement across its terminals, the voltage measuring device should have a resistance of









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    NA

    Correct Option: A

    NA



  1. The power in a 3-phase, 3-wire load is measured using two 100W full-scale wattmeters W1 and W2. W1 is of a accuracy class ± 1% and reads 100 W. W2 is of accuracy class ± 0.5% and reads – 50 W. Then the uncertainty in the computation to total power is









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    W1 = 100W + 1W;
    W2 = – 50 + 0.5W

    ∴ W1 + W2 = 50 ±
    1.5
    × 100 = ± 3%
    50

    Correct Option: D

    W1 = 100W + 1W;
    W2 = – 50 + 0.5W

    ∴ W1 + W2 = 50 ±
    1.5
    × 100 = ± 3%
    50