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Electrical and Electronic Measurements Test 6
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Electrical and Electronic Measurements Test 6
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  • Question 1/10
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    A 3 ½ digit digital multimeter can display
    Solutions

    In a 3 ½ digit digital multimeter, there are three full digits and one half digit.

    A full digit counts from 0 to 9 whereas an half digit counts from 0 to 1.

    Therefore, from the given options only 1.999 is a valid display for a 3 ½ digital multimeter.
  • Question 2/10
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    A digital voltmeter has a read out range from 0 to 999 counts. If the full scale reading is 9.99 V, the resolution is
    Solutions

    Concept:

    The resolution in a N bit DVM is given by =110N×rangeofvoltage

    Where N is the number of full digits.

    In a DVM, a full digit counts 0 to 9 and half digit counts from 0 to 1.

    Calculation:

    The full-scale reading = 9.99 V

    It is a 3-digit voltmeter.

    Range of voltmeter = 10 V

    Resolution for the given DVM is =1103×10=0.01

  • Question 3/10
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    An integrator type DVM (digital voltmeter) contains a 100 kΩ and 1 μF capacitor. If the voltage applied to the integrator input is 1 volt, what voltage will be present at the output of the integrator after 1 second?
    Solutions

    Resistance (R) = 100 kΩ

    Capacitance (C) = 1 μF

    Integrator input voltage (V) = 1 V

    For an integrator type DVM, output voltage is given by

    V0=1RC0tVidt

    =1100×103×1×106011dt=10V

  • Question 4/10
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    The rms reading meter shows value of 60 V when fed from a triangular wave. If the same input is supplied to an average reading voltmeter. Its reading will be _______ (in V)
    Solutions

    We know that,

    Average value of triangular wave = Vm2

    rms value of triangular wave =Vm3

    Vm30=60Vm=603

    Averagevalue=Vm2=6032=303
  • Question 5/10
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    A 20 V dc voltage is measured by analog and digital voltmeters. The analog instrument is on its 25 V range, and its specified accuracy is ±2%. The digital meter has 3 ½ digit display and an accuracy of ±(0.6% of reading + 1 count). The measurement accuracy in both the cases respectively are:
    Solutions

    Analog instrument:

    Voltage error = ±2% of 25 V = ±0.5 V

    Error =±0.520×100=±2.5%

    Digital instrument:

    For 20 V displayed on 3 ½ digit display, 1 digit = 0.01 V

    Voltage error = ±(0.6% of reading + 1 count)

    = ±(0.12 + 0.01) = ±0.13 V

    Error =±0.1320×100=±0.65%
  • Question 6/10
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    A dual slope ADC has R = 100 kΩ and C = 0.01 μF. The reference voltage is 10 volts and the fixed integration time is 10 ms. The conversion time for a 6.8-volt input is ________ (in ms)
    Solutions

    Concept:

    For a dual-slope analog to digital converter,

    De-integration time, T2=ViVRef×T1

    Where Vi = input voltage

    VRef = reference voltage

    T1 is the first integration time

    Total conversion time = first integration time (T2) + de-integration time.

    Calculation:

    Given that, input voltage (Vi) = 6.8 mV

    Reference voltage (VR) = 10 mV

     Integration time (T1) = 10 ms

    De-integration time, T2=6.810×10=6.8ms

    Total conversion time = 10 + 6.8 = 16.8 ms
  • Question 7/10
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    In a dual slope DVM, an unknown signal voltage is integrated over 150 cycles of the clock. If the signal has a 80 Hz pick up the max clock frequency is _______ (in KHz)
    Solutions

    T1 = nTs

    150clks=n×1fs

    150×Tclk=n×1fs

    150×1fclk=n×1fs

    fclk=fs×150n

    For maximum clock frequency, the value of n should be i.

    ⇒ flck = fs × 150 = 80 × 150 = 12000 = 12 kHz
  • Question 8/10
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    A rectifier type PMMC voltmeter has a sensitivity of 20 kΩ/V. A reading of 4.5 V is obtained when measuring a voltage source with an internal resistance on it’s 5 V scale. When the scale is changed to 10 V full scale, a reading of 6 V is obtained. The value of the voltage and its internal resistance R are
    Solutions

    Let the voltage is VS and the internal resistance is R

    For 5 V scale:

    Internal resistance = full scale voltage × sensitivity

    = 5 V × 20 kΩ/V = 100 kΩ

    Reading = 4.5 V

    VS×(100R+100)=4.5

    For 10 V scale:

    Internal resistance = full scale voltage × sensitivity

    = 10 V × 20 kΩ/V = 200 kΩ

    Reading = 6 V

    VS×(200R+200)=6

    By solving the above two equations, we get

    VS = 9 V and R = 100 kΩ
  • Question 9/10
    1 / -0

    A rectifier type of instrument uses a basic PMMC movement of 100 μA and a resistance 2 kΩ. It employs a half wave rectifier circuit with forward resistance of the diode being 500 Ω. The reverse resistance of the diodes is infinite. The range of the instrument is 0 - 20 V. a.c. sinusoidal. The value of series multiplier is – (in kΩ)
    Solutions

    The dc sensitivity =1100×106=10kΩ/v

    For a halfwave rectifier,

    Ac sensitivity = 0.45 × dc sensitivity

    = 0.45 × 10 = 4.5 kΩ/v

    Series multiplier = (Sac) (V) - Rm - Rd

    = (4.5)(20) – 2 – 0.5

    = 87.5 kΩ
  • Question 10/10
    1 / -0

    A rectifier type instrument uses a bridge rectifier and has its scale calibrated in terms of rms values of a sine wave. It indicates a voltage of 2.77 V. When measuring a voltage of triangular wave shape. The magnitude of percentage error will be____ (in %)
    Solutions

    The meter uses a full wave rectifier circuit it indicates value of 2.77 V the form factor for full wave rectified sinusoidal waveform is 1.11

    Average value of voltage, Vavg = 2.49 V

    For triangular wave shape, peak value of voltage Vm = 2 Vavg = 4.98 V

    Vrms=Vm3=4.983=2.87V%error=2.772.872.87×100=3.48%

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