Design of AC Machines (2170909) MCQs

MCQs of Design of Single-Phase Induction Motors

Showing 51 to 56 out of 56 Questions
51.

In the design of single-phase induction motor. The flux density in stator theeth is ____________ for For high torque machine and ____________ for general purpose machine.

 

(a)

1.8 Wb/m2     &       1.4 to 1.7 Wb/m2   

(b)

1.7 Wb/m2     &       1.1 to 1.7 Wb/m2   

(c)

2.1 Wb/m2     &       1.4 to 1.7 Wb/m2   

(d)

1.8 Wb/m2     &       1.1 Wb/m2   

Answer:

Option (a)

52.

In the design of single-phase induction motor. The flux density in stator teeth is given by 

(a)

𝑩𝒕𝒔 = ∅π’Ž / ( 𝑾𝒕𝒔  × (𝑺𝒔/𝒑)  × π‘³π’Š )

(b)

𝑩𝒕𝒔 = ∅π’Ž  ×  𝑾𝒕𝒔  × (𝑺𝒔/𝒑)  × π‘³π’Š 

(c)

𝑩𝒕𝒔 = ∅π’Ž  ×  𝑾𝒕𝒔  × (𝑺𝒔/𝒑)  × 2π‘³π’Š 

(d)

𝑩𝒕𝒔 = ∅π’Ž / ( 𝑾𝒕𝒔  × (𝑺𝒔/𝒑)  × 2π‘³π’Š )

Answer:

Option (a)

53.

In the design of single-phase induction motor. The maximum flux density in stator core is ____________ for and the normal range is _________

(a)

1.8 Wb/m2     &       1.4 to 1.7 Wb/m2  

(b)

1.5 Wb/m2     &       0.9 to 1.4 Wb/m2  

(c)

2.1 Wb/m2     &       1.4 to 1.7 Wb/m2  

(d)

1.8 Wb/m2     &       1.1 Wb/m2

Answer:

Option (b)

54.

In the design of single-phase induction motor. The flux density in the stator core is given by 

(a)

𝑩𝒄𝒔 = ∅π’Ž / ( 𝟐 × π’…π’„π’”  × π‘³π’Š )

(b)

𝑩𝒄𝒔 = ∅π’Ž / ( 𝟐 × π’…π’„π’”  × π‘³ )

(c)

𝑩𝒄𝒔 = ∅π’Ž × πŸ × π’…π’„π’”  × π‘³π’Š 

(d)

𝑩𝒄𝒔 = ∅π’Ž × πŸ × π’…π’„π’”  × π‘³

Answer:

Option (a)

55.

In the design of single-phase induction motor. The length of a mean turn of each coil per pole, 

(a)

π‘³π’Žπ’• = [πŸ–.πŸ’ (𝑫+𝒅𝒔𝒔)/𝑺𝒔 ] × π’”π’π’π’•π’” 𝒔𝒑𝒂𝒏 + πŸπ‘³ 

(b)

π‘³π’Žπ’• = [πŸ–.πŸ’ (𝑫+𝒅𝒔𝒔)/𝑺𝒔 ] + 𝒔𝒍𝒐𝒕𝒔 𝒔𝒑𝒂𝒏 + πŸπ‘³ 

(c)

π‘³π’Žπ’• = [πŸ–.πŸ’ (𝑫+𝒅𝒔𝒔)/𝑺𝒔 ] × π’”π’π’π’•π’” 𝒔𝒑𝒂𝒏 × πŸπ‘³ 

(d)

π‘³π’Žπ’• = [πŸ–.πŸ’ (𝑫+𝒅𝒔𝒔)/𝑺𝒔 ] × π’”π’π’π’•π’” 𝒔𝒑𝒂𝒏 + 𝑳 

Answer:

Option (a)

56.

In the design of single-phase induction motor. The Length of air gap is given by 

(a)

π‘³π’ˆ = (𝟎.πŸŽπŸŽπŸ• × π’“π’π’•π’π’“ π’…π’Šπ’‚π’Žπ’†π’•π’†π’“) /√𝒑

(b)

π‘³π’ˆ = (𝟎.πŸŽπŸŽπŸ• + 𝒓𝒐𝒕𝒐𝒓 π’…π’Šπ’‚π’Žπ’†π’•π’†π’“) /√𝒑

(c)

π‘³π’ˆ = (𝟎.πŸŽπŸŽπŸ• - 𝒓𝒐𝒕𝒐𝒓 π’…π’Šπ’‚π’Žπ’†π’•π’†π’“) /√𝒑

(d)

π‘³π’ˆ = 𝟎.πŸŽπŸŽπŸ• × π’“π’π’•π’π’“ π’…π’Šπ’‚π’Žπ’†π’•π’†π’“ x √𝒑

Answer:

Option (a)

Showing 51 to 56 out of 56 Questions