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Measuring Instruments Test 2
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Measuring Instruments Test 2
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  • Question 1/10
    1 / -0.25

    Mark out the correct option.

    Solutions

    Ammeter:

    • Ammeter is an instrument which is used to measure the current flowing through the circuit
    • It has low resistance, ideally zero but practically it has very low internal resistance.
    • Connecting the ammeter in series allows all of the circuit current to pass through it and hence measure it.


    Voltmeter:

    • Voltmeter is an instrument which is used to measure the voltage across two terminals in a circuit
    • Voltmeter ideally has infinite resistance zero but practically it has very high internal resistance.
    • Connecting a voltmeter with infinite resistance in parallel ensures no current flows through the voltmeter and the voltage drop (IR) remains the same.
  • Question 2/10
    1 / -0.25

    A moving-coil galvanometer can be used as a DC ammeter by connecting
    Solutions
    • A galvanometer can be converted into an ammeter by connecting a shunt resistance parallel to it
    • A shunt is basically a wire of very small resistance
    • A galvanometer can be converted into a voltmeter by connecting a very high resistance in the series
    • A voltmeter is used to measure the potential difference whereas an ammeter is used to measure the electric current
    • A galvanometer is used to measure a small electric current in a circuit

     

    Points to remember:

    A galvanometer can be converted into a voltmeter by connecting a resistance in series with it

  • Question 3/10
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    A rectangular and sinusoidal alternating currents each having a peak value of 150 A are passed through a moving iron ammeter. Then the meter readings are respectively
    Solutions

    Concept:

    Moving iron ammeter:

    Moving iron ammeter is used to measure both AC and DC values in the circuit.

    The reading of the MI instrument is in RMS value.

    Calculation:

    Moving iron ammeter reads the RMS value of alternating current.

    The peak value of the waveform = Im = 150 A

    For rectangular waveform, RMS value (I1)= peak value

    I1 = Im = 150 A

    For a sinusoidal waveform, the RMS value  of the current = Im / √2

     \(I_2 = \frac{{150}}{{√ 2 }} = 106.6\;A\)

    Points to remember:

    • The PMMC instruments measure the quantity in DC only.
    • The reading of the PMMC instruments is an average value.
    • For alternative quantities, PMMC shows zero reading as it is not working on AC circuits

     

  • Question 4/10
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    AC current cannot be measured directly by _______
    Solutions

    PMMC instruments:

    • PMMC or Moving coil instruments are used to measure the DC quantities only.
    • So, a moving coil ammeter is used to measure DC current.
    • The reading of the moving coil instruments shows zero when they are used to measure the AC quantities.
    • The moving iron instrument can be used to measure both AC and DC quantities.

     

    Important Points:

    • Moving coil instrument measures, the average value of quantity and the scale is linear and uniform as the deflection is directly proportional to the current.
    • Moving iron instrument measures, the RMS value of quantity and the scale is non-uniform as the deflection is directly proportional to the square of the current.
    • Hotwire instruments are used to measure the true RMS value of the quantity.
    • The combination of PMMC with the rectifier circuit measures AC as well as DC values.
  • Question 5/10
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    A current of -4 + 3√2 sin (ωt + 30°) A is passed through a centre zero PMMC meter and moving-iron meter. The two meters will read respectively
    Solutions

    Concept:

    1). Moving coil instruments are used to measure the DC quantities only. So, a moving coil ammeter is used to measure DC current.

    2). The moving iron instrument can be used to measure both AC and DC quantities.

    3). Moving coil instrument measures, the average value of quantity and the scale is linear and uniform as the deflection is directly proportional to the current.

    4) . Moving iron instrument measures, the RMS value of quantity and the scale is non-uniform as the deflection is directly proportional to the square of the current.

    Calculation:

    i(t) = -4 + 3√2 sin (ωt + 30°) A

    Reading of PMMC instrument = Average value of i(t) = -4

    Reading of moving iron meter = RMS value of i(t)

    \( = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( {\frac{{3\sqrt 2 }}{{\sqrt 2 }}} \right)}^2}} = 5\;A\)

    Note:

    The rms value of a sinusoidal equation of the form

    f(t) = A0 + A1 sin (ω1t + ϕ1) + A2 sin (ω2t + ϕ­2) + … is given by

    \({f_{rms}} = \sqrt {A_0^2 + {{\left( {\frac{{{A_1}}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{{{A_2}}}{{\sqrt 2 }}} \right)}^2} + \ldots }\)

    The average value = A0

  • Question 6/10
    1 / -0.25

    How are voltmeter and ammeter connected in a circuit?
    Solutions

    Ammeter:

    • An ammeter is an instrument which is used to measure the current flowing through the circuit
    • It has low resistance, ideally zero
    • The connecting ammeter in series allows all of the circuit current to pass through it and hence measure it

    Voltmeter:

    • A voltmeter is an instrument which is used to measure the voltage across two terminals in a circuit
    • Voltmeter ideally has infinite resistance
    • Connecting voltmeter with infinite resistance in parallel ensures no current flows through the voltmeter and the voltage drop (IR) remains the same

     

    Hence, Ammeter should be connected in series and voltmeter in parallel.

  • Question 7/10
    1 / -0.25

    A galvanometer may be converted into ammeter or voltmeter. In which of the following cases the resistance of the device will be the largest?
    Solutions

    Concept:

    A galvanometer can be converted into an ammeter by connecting a low resistance in parallel.

    The effective resistance of device = Rsh in parallel with Rg

    Where Rg is galvanometer resistance

    A galvanometer can be converted into a voltmeter by connecting a high resistance in series.

    The effective resistance of device = Rse in series with Rg

    Where Rg is galvanometer resistance

    Application:

    The voltmeter has high resistance and ammeter has low resistance.

    Voltmeter with high full-scale range has high resistance.

    So, a voltmeter of range 10 V has higher resistance.

  • Question 8/10
    1 / -0.25

    Which device is not used for extending the range of Instrument?
    Solutions
    • Switchboard is not used for extending the range of Instrument
    • There are four common devices used for the range extension of ammeter and voltmeter namely; multipliers, shunts, current transformer and potential transformer
    • An ammeters range can be extended by changing the value of the shunt resistance in the ammeter
    • A lower resistance would increase the maximum current range of the meter
  • Question 9/10
    1 / -0.25

    The range of a dc milliammeter can be extended by using a
    Solutions
    • Shunts are used for the range extension of ammeters.
    • A shunt is a low-value resistance having minimum temperature co-efficient.
    • It is connected in parallel with the ammeter whose range is to be extended. The combination is connected in series with the circuit whose current is to be measured.
    • Shunt provides a path for extra current as it is connected across (in parallel with) the instrument.
    • These shunted instruments can be used to measure currents many times greater than their normal full-scale deflection currents.
    • The ratio of maximum current (with shunt) to the full-scale deflection current (without shunt) is known as the ‘multiplying power’ or ‘multiplying factor’ of the shunt.

    In the figure

    I = Total current flowing in the circuit

    Ish = The current through the shunt resistor

    Rm = The ammeter resistance

    Points to remember:

    The extension of the voltmeter can be done by connecting a high resistance in series.

  • Question 10/10
    1 / -0.25

    Series resistance required to read 0-250 V with a moving coil instrument of internal resistance 2 ohm and full-scale deflection of 50 mA is
    Solutions

    Extension of voltmeter:

    For range extension of voltage measurement in moving coil instrument, a resistance is connected in series with coil resistance.

    Because, for a constant value of current, resistance connected in series connection has a higher voltage drop compared to a parallel or shunt connection.

    The value of the series resistance is given by:

    Rse = R(m – 1)

    \({R_m} = \frac{{{V_m}}}{{{I_m}}}\)

    m = Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

    Calculation:

    Full scale deflection of current (IFSD) = 50 mA

    Internal resistance (Rm) = 2 Ω

    Full scale deflection of voltmeter (Vm) = 0.1 V

    Required full scale deflection of voltmeter (V) = 250 V

    \({R_{se}} = {R_m}\left( {\frac{V}{{{V_m}}} - 1} \right)\)

    \(= 2\left( {\frac{{250}}{{0.1}} - 1} \right) \)

    ∴ Rse = 4998 Ω

     

    Extension of ammeter:

    For range extension of current measurement in moving coil instrument, a resistance is connected in parallel or shunt with coil resistance.

    f current measurement in moving coil instrument, a resistance is connected in parallel or shunt with coil resistance. Because for a constant value of voltage, resistance connected in parallel connection has a higher value of current flow compared to series connection.

    The amount of shunt resistance to be added is given by:

    \({R_{sh}} = \frac{{{R_m}}}{{\frac{I}{{{I_m}}} - 1}}=\frac{R_m}{m-1}\)

    Rm = internal resistance of the coil

    Rsh = Shunt resistance

    I = total current

    Im = Full-scale deflection current

    m = Multiplying factor

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