text: # Errors in MI Type Instruments MCQ Quiz - Objective Question with Answer for Errors in MI Type Instruments - Download Free PDF
  
  Last updated on Apr 16, 2024
  
  ## Top Errors in MI Type Instruments MCQ Objective Questions
  
  #### Errors in MI Type Instruments Question 1:
  
  The effect of stray magnetic fields on the actuating torque of a portable instrument is maximum when the operating field of the instrument and the stray fields are
  
  1. Perpendicular
  2. Parallel
  3. Inclined at 60°
  4. Inclined at 30°
  
  Option 2 : Parallel
  
  #### Errors in MI Type Instruments Question 1 Detailed Solution
  
  Stray Magnetic Field:
  
  • It is also called demagnetization fields.
  • This is weak at full-scale deflection hence it can easily be distorted.
  • Due to the external magnetic field inside, the operating field inside the meter is disturbed.
  • This will lead to the weakness of the operating magnetic field.
  • The stray magnetic field depends on the magnitude of the external magnetic field and its direction.
  • When the magnitude of the external magnetic field is in phase with the main field, stray magnetic field error will be more.
  
  This error is minimized by the following methods:
  
  • By shielding the instrument with the help of a steel case.
  • By sheet steel lining to the case.
  
  In PMMC instrument permanent magnets are used. Therefore it will have a strong magnetic field.
  
  In an MI instrument, the Air core is used. Therefore it will have a weak magnetic field.
  
  The stray magnetic field error is more in the MI instrument compared to the PMMC instrument.
  
  #### Errors in MI Type Instruments Question 2:
  
  A moving iron instrument is calibrated with full scale deflection on 100 V DC. This instrument is used to measure 100 V at 50 Hz supply. Find the percentage error in reading if the instrument has a resistance of 2 kΩ and an inductance of 0.5 H
  
  #### Errors in MI Type Instruments Question 2 Detailed Solution
  
  Full scale deflection of instrument (VFSD) = 100V
  
  Resistance (Rm) = 2 kΩ
  
  Inductance (Lm) = 0.5 H
  
  XL = 2πfL = 2π × 50 × 0.5 = 157.08 Ω
  
  $${\rm{Full\;scale\;current}}\left( {{{\rm{I}}_{FSD}}} \right) = \frac{{100}}{{2 \times {{10}^3}}} = 50\;mA$$
  
  Current through meter when 100 V at 50 Hz supply is applied
  
  $$I = \frac{V}{Z} = \frac{{100}}{{\sqrt {{{\left( {2 \times {{10}^3}} \right)}^2} + {{\left( {157.08} \right)}^2}} }} = 49.84\;mA$$
  
  $$Voltage\;reading\;of\;meter = \frac{{49.84}}{{50}} \times 100 = 99.68$$
  
  $$\% \;error = \frac{{Measrued - true}}{{true}} \times 100 = \frac{{99.68 - 100}}{{100}} \times 100 = - 0.32\%$$