31. Hay’s bridge:
    

1. 1. Is particularly suited for the measurement of capacitance over a wide range of values
      

   
   
   

   2. Is particularly suited to the measurement of inductances having high Q-values
      

   
   
   

   3. Is suited for the measurement of capacitances having high Q-values
      

   
   
   

   4. Is suited for the measurement of inductances having low Q-values

   
   
   

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1. View Hint View Answer Discuss in Forum

   Hay’s bridge is particularly suited to the measurement of inductances having high Q-values.

   Correct Option: B

   Hay’s bridge is particularly suited to the measurement of inductances having high Q-values.

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32. The power in a resistor R is estimated by measuring the voltage and current using the voltmeter-ammeter method. Two different arrangements can be used as shown in circuits I and II. Less erroneous results are obtainted by adopting:
    
    
    
    

1. 1. circuit I for low values of R
      

   
   
   

   2. circuit II for low values of R
      

   
   
   

   3. circuit I for high values of R
      

   
   
   

   4. circuit II for low values of R

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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33. In the ac bridge shown in the figure:
    where,
    Z₁ = 450 ohm,
    Z₂ = (300– j600) ohm,
    Z₃ = (200 + j100) ohm.
    The value of unknown impedance Z₄ at balance is:
    
    
    
    

1. 1. (100 + j150) ohm
      

   
   
   

   2. (100 – j150) ohm
      

   
   
   

   3. (300 + j150) ohm
      

   
   
   

   4. (0 + j150) ohm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given, Z₁ = 450Ω
   Z₂ = (300 – j600)Ω
   Z₃ = (200 + j100)Ω
   Z₄ =?
   under balanced condition
   

   Z1	=	Z4
   Z2		Z3

   
   or
   

   Z4 =	Z1Z3
   	Z2

   
   or
   

   Z4 = 450	(200 + j100)
   	(300 – j600)

   
   or
   Z₄ = (0 + j150)Ω (After simplifying)

   
   

   Correct Option: D

   Given, Z₁ = 450Ω
   Z₂ = (300 – j600)Ω
   Z₃ = (200 + j100)Ω
   Z₄ =?
   under balanced condition
   

   Z1	=	Z4
   Z2		Z3

   
   or
   

   Z4 =	Z1Z3
   	Z2

   
   or
   

   Z4 = 450	(200 + j100)
   	(300 – j600)

   
   or
   Z₄ = (0 + j150)Ω (After simplifying)

   
   

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