261. The applied voltage of a certain transformer is increase by 50% while the frequency is reduced to 50%. The maximum core flux density will

1. 1. become three times

   
   
   

   2. become1.5 times

   
   
   

   3. become half

   
   
   

   4. remain the same

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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262. If P_c and P_sc represent core and full-load ohmic losses respectively, the maximum kVA delivered to load corresponding to maximum efficiency is equal to rated kVA multiplied by

1. 1. P_c/P_sc

   
   
   

   2. √P_c/P_sc

   
   
   

   3. (P_c/P_sc)²

   
   
   

   4. (P_sc/P_c)²

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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263. Building steel core out of stampings reduces eddy current loss because it

1. 1. increases core resistivity.

   
   
   

   2. increases the effective length of eddy current paths thereby increasing effective resistance to the flow of eddy currents.

   
   
   

   3. increases core permeability.

   
   
   

   4. reduces the effective length of eddy current path, thereby reducing effective resistance to the flow of eddy currents.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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264. A coil would on a magnetic core is excited from an ac voltage source. The source voltage and its frequency are both doubled. The eddy current loss in the core will become

1. 1. half

   
   
   

   2. remains same

   
   
   

   3. double

   
   
   

   4. four times

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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265. The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.
     
     Assuming the transformer to be ideal, the value of the reactance X to improve the input power factor to unity is ___________.

1. 1. 24

   
   
   

   2. 12

   
   
   

   3. 28

   
   
   

   4. 18

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given, V₂ = 110
   Load power, P₂ = 4 kW
   pf = cos θ₂ = 0.89
   

   I2 =	load power
   	V2 × cos θ2

   
   

   =	4000	= 40.858 A
   	110 × 0.89

   
   primary current, I₁ = K I₂
   

   =			N2		I2
   			N1

   
   

   =	1	× 40.858 = 20.429 A A
   	2

   
   To improve input power factor units,
   

   Reactive power φ =	V12
   	X

   
   

   ∴ Reactance , X =	V12
   	φ

   
   

   =	2202	= 23.6 Ω
   	220 × 20.429 × sin (27)

   Correct Option: A

   Given, V₂ = 110
   Load power, P₂ = 4 kW
   pf = cos θ₂ = 0.89
   

   I2 =	load power
   	V2 × cos θ2

   
   

   =	4000	= 40.858 A
   	110 × 0.89

   
   primary current, I₁ = K I₂
   

   =			N2		I2
   			N1

   
   

   =	1	× 40.858 = 20.429 A A
   	2

   
   To improve input power factor units,
   

   Reactive power φ =	V12
   	X

   
   

   ∴ Reactance , X =	V12
   	φ

   
   

   =	2202	= 23.6 Ω
   	220 × 20.429 × sin (27)

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